Chapter 2 Orbit Determination from Observations

Transcription

1 Chapter Orbit Determination from Observations.1 Position of the Problem In general terms, the determination of an orbit is an iterative process meant to know and predict the position and velocity (or the orbital elements) of a space object, with respect to a primary celestial body, from observations of that object. For example, the orbit determination for an artificial satellite revolving about the Earth is a series of operations aimed at determining the motion (that is, the six orbital elements or the six Cartesian components of the position and velocity vectors at some given epoch) of the satellite with respect to a reference system having its origin in the centre of mass of the Earth. Likewise, for a natural celestial body revolving about the Sun, an astronomer computes the orbital elements of that body with respect to the Sun on the basis of observations which have been performed at some place on the surface of the Earth. The methods used to this end may be classified into two broad categories. The first category includes the classical (or deterministic) methods, which consider the measurements as free from errors, and use therefore the minimum number of measurements required to determine an orbit. The second category includes the modern (or statistical) methods, which consider the measurements as affected by errors, and use therefore more measurements than those which would be strictly necessary to determine an orbit, with the view of reducing the influence of such errors by means of a suitable mathematical treatment of the data gathered. Let x = x(t) be the state vector, at a given time t, of an artificial satellite orbiting around the Earth, that is, the vector whose six components are the three components of the position vector, r, and the three components of the velocity vector, v, of the satellite at time t with respect to either the true-of-date or the J000.0 geocentric-equatorial reference system XYZ (defined in Sect. 1.9). Let x 0 = x(t 0 )be the known state vector of the same satellite at a given initial time t 0 with respect to the same reference system. When the forces (the central force and its perturbations) acting upon the satellite are known exactly, then it is possible to determine its state Springer International Publishing AG 018 A. de Iaco Veris, Practical Astrodynamics, Springer Aerospace Technology, 19

2 10 Orbit Determination from Observations vector x at any given time t by integrating the equations of motion. However, the state vector x of the satellite at the initial time t 0 is known approximately, not exactly. In addition, the force model used to compute the forces acting upon the satellite at a time t can only provide approximate values of the true forces, because the physical constants (e.g. the gravitational parameter of the Earth) are known approximately and also because the mathematical model used to compute the forces can never be exact. Consequently, the orbit determination of a satellite requires that the observations of the satellite (which are affected by random and systematic errors due to the non-exactness of the force model used) should be mathematically treated in such a way as to obtain the best estimate of the orbital elements representing the satellite motion at any time. To this end, it was customary to operate in two stages. The first stage, called preliminary orbit determination, is the approximate determination of an orbit by means of a minimum number of observations of the orbiting object. The second stage, called differential correction of orbits or orbit improvement, comprises: 1. the collection of more observations than those strictly required, and. the fitting of the data gathered to an orbit by means of some mathematical algorithm, based usually on the method of the least squares. The results of the second stage are differential corrections to the preliminary orbital elements, which have been determined in the first stage. In the first stage, both the main attracting body and the orbiting body are considered as isolated particles subject only to their mutual gravitational forces. Therefore, the motion is governed by the second-order differential equation shown in Sect. 1.1, that is, by r 00 þ l r r ¼ 0 where r is the acceleration vector of the orbiting body, l is the gravitational parameter of the attracting body (resulting from the product of the mass, M, of the attracting body by the gravitational constant, G), and r is the position vector of the orbiting body with respect to an inertial reference system having its origin in the centre of mass of the attracting body, whose mass M is supposed to be much greater than the mass m of the orbiting body. As also shown in Sect. 1.1, the constants of integration of this differential equation are six, that is, the orbital elements of the orbiting body. In the second stage, the perturbations to the primary gravitational force are taken into account, so that the differential equation of motion is expressed as follows r 00 þ l r r ¼ a

3 .1 Position of the Problem 11 where a is the vector sum of the perturbing accelerations acting on the orbiting body. In case of an artificial satellite of the Earth, as will be shown at length in Chap., these accelerations are primarily due to: the non-spherical shape and the non-homogeneous mass density of the Earth; the gravitational attraction exerted on the satellite by other celestial bodies than the Earth (in particular, by the Sun and the Moon); solid Earth and ocean tides; the aerodynamic drag exerted by the Earth atmosphere, particularly important in case of artificial satellites orbiting at low altitudes; solar radiation pressure; and thrusters used for orbital manoeuvres. By the way, solid Earth tides are similar to ocean tides, both of them being due to the gravitational forces exerted upon the Earth by the Sun and the Moon. The difference between the two types of tide resides in the much smaller tidal distortion of the Earth (about 0 cm a day) in comparison with that of the ocean, because of the resistance opposed by the rocks. As a result of the perturbing accelerations, the orbital elements of an artificial satellite revolving about the Earth are not constant, but vary slowly with time. Consequently, the second-order differential equation r +(l/r )r = a is expressed as a system of m first-order differential equations with the initial conditions x 0 ¼ f ðx; tþ xðt 0 Þ¼x 0 where x = x(t) is the augmented m-dimensional, time-dependent state vector, that is, the state vector comprising not only the six position (x, y, z) and velocity (v x, v y, v z ) co-ordinates of the orbiting body but also the physical constants provided by the force model used (that is, the physical quantities which are independent of time); f(x, t) is a vector-valued function (the integrand function) which depends on the state vector and on time; and x 0 is the known state vector at some epoch t 0. On the other hand, the observations can be represented by the following system of n nonlinear algebraic equations z i ¼ gðx i ; t i Þþe i ( i =1,,, n), where z i are the actual observations made at epochs t 1, t,, t n, g(x i, t i ) are the predicted values, and e i are random errors of measurement affecting the observations. Such equations, solved for x i, yield the state vector xðt i Þ¼Hðx 0 ; t 0 ; t i Þ

4 1 Orbit Determination from Observations (i =1,,, n) at times, respectively, t 1, t,, t n. This way of proceeding, comprising the two stages described above, is called batch estimation processing and is based on the least-squares method. Recently, as a result of the mathematical theory due to Kalman and Bucy [44, 45] and also of the advent of advanced computing means, the two stages are no more separated from each other, and each epoch of observations is processed individually by means of a sequential estimation algorithm, which is just the Kalman filter. The preliminary orbit determination comprises several methods meant to determine the orbital elements. These methods will be shown in the following paragraphs. The determination of an orbit is part of a broader process called integrated guidance, navigation, and control. The whole process of spacecraft guidance, navigation, and control is shown in the following scheme, which is redrawn from Thornton and Border [70]. Guidance is the actual steering of a spacecraft travelling through space. This steering may come from sources placed either inside (a human crew or an on-board computer) or outside (a ground station) the spacecraft. An example is provided by the commands given to the rocket motor of a spacecraft to control the thrust. Navigation is the measurement of the motion (position, orientation, and velocity) of a spacecraft in space, obtained by means of observations performed by the crew, or by automatic on-board sensors, or by tracking equipment located on the ground.

5 .1 Position of the Problem 1 Control is the alignment and stabilisation of a spacecraft while the guidance and navigation functions are being performed. These functions make it necessary not only to determine the orbital elements (or the position and velocity vectors) of a spacecraft, but also to adjust its path and attitude by means of control forces and moments obtained by firing the main rocket motor or by routing commands to other on-board devices (such as thrusters, reaction wheels, control moment gyroscopes, or aerodynamic surfaces), which produce the desired forces on the spacecraft. The functions of guidance and control are meant to compute and send a series of commands to the propulsion system to alter the spacecraft velocity or attitude. Following Thornton and Border [70], the orbit determination process requires an a priori estimate (prediction) of the spacecraft trajectory, referred to as the nominal orbit. In this estimate, expected values of the observable quantities are computed on the basis of nominal values for the trajectory and models of such observable quantities. The computed quantities differ from the correspondent observed quantities coming from the tracking system. Such differences form the residuals (computed minus observed quantities), which are due not only to random uncorrelated measurement errors (such as thermal noise in the tracking receiver), but also to errors in the trajectory and the observable models. The errors of the latter type introduce distinctive signatures, that is, characteristic marks, in the residuals. Such signatures make it possible to adjust the model parameters by means of a procedure, called weighted linear least-squares estimation, which yields the set of parameter values corresponding to the minimum weighted sum of the squares of the residuals. This procedure, when the data are weighted by the inverse of their error covariance, yields a minimum-variance estimator. Since this estimator provides a linear solution to a nonlinear problem, then the estimation procedure comprises more iterative steps, each of which uses the parameter estimation of the previous step, until convergence is reached. After the orbit determination process has been completed, the orbital elements are compared with those required by the project. In case of discrepancies, trajectory correction manoeuvres must be planned and executed. For example, when the actual flight path differs from that which is required by the planned mission, it is necessary to compute the magnitude and direction of the Dv vector required to correct to the desired trajectory. In addition, suitable times must be computed, at which the corrective manoeuvres will be executed. At the proper time, the spacecraft attitude will change the direction of the rocket motor axis to the desired one and the thrusters will be fired for a determined interval of time. The magnitude of Dv is usually small (metres or tens of metres per second), because of the limited amount of propellant which can be carried on board. Orbit trim manoeuvres are sometimes necessary in case of a satellite revolving around the Earth, in order to prevent orbital decay due to air drag, and also to perform orbit changes required by the mission objectives.

6 14 Orbit Determination from Observations. Topocentric Co-ordinate Systems A radar station located on the surface of the Earth can measure the position and velocity of an object, for example, of an artificial satellite, with respect to the place where the station is located. However, the position vector r of interest to the observer has its origin in another place, that is, in the centre of mass of the Earth. In addition, the velocity vector v r of interest to the observer relates to the geocentric-equatorial reference system XYZ, which moves with respect to the radar station, because the Earth rotates about its axis. Thus, a series of measurements relating to reference system located in the radar station must be converted so as to provide the correspondent measurements with respect to the geocentric-equatorial system. The measurements performed at a radar station are related to the topocentric-horizon system UVW, which is defined below. The topocentric-horizon reference system UVW, shown in the preceding figure, has its origin in the point on the surface of the Earth where the radar station is located, and its three Cartesian axes U, V, and W are directed so that its fundamental plane UV is the horizon, with the U-axis pointing towards east, the V-axis pointing towards north, and the W-axis perpendicular to the horizon and pointing upwards (that is, towards the zenith). As to the directions forming the horizon, the agreement of the authors is not unanimous. Some of them, for example Bate et al. [5] and Vallado [77], choose the south and east directions. Other authors, for example Montenbruck and Gill [5] and Curtis [0], choose the east and north directions. We follow the latter convention.

7 . Topocentric Co-ordinate Systems 15 Let the unit vectors of the three axes U, V, and W be, respectively, u U, u V, and u W. Unlike the geocentric-equatorial reference system XYZ, which is fixed in space with respect to the stars and does not rotate with the Earth, the topocentric-horizon reference system UVW is evidently a non-inertial system, because it is fixed to the Earth and does rotate with it. Another reference system of the topocentric type is the topocentric-equatorial co-ordinate system, shown on the left-hand side of the following figure. The origin P H of this reference system is the point on the surface of the Earth where the observer (or the radar station) is located. The axes X t, Y t, and Z t of this co-ordinate system are parallel to the axes, respectively, X, Y, and Z of the geocentric-equatorial co-ordinate system having its origin O in the centre of the Earth. The fundamental plane X t Y t of the topocentric-equatorial system X t Y t Z t is parallel to, but not coincident with, the fundamental plane XY of the geocentric-equatorial system XYZ. Both X and X t point towards the vernal equinox. The position vector P H Q q, of magnitude q, goes from the point of observation P H to the space object Q. The position of Q is usually expressed by means of the three co-ordinates q, d, and a, where the two angles d (declination) and a (right ascension) are shown on the right-hand side of the preceding figure, which is an enlarged view of the topocentric-equatorial co-ordinate system. In the topocentric-equatorial system the North and South celestial poles are determined by intersecting the rotation axis of the Earth with the celestial sphere, that is, with the sky as seen from the Earth. The centre of the Earth is also the centre of the celestial sphere, as shown in the following figure, which is due to the courtesy of the University of Tennessee [76].

8 16 Orbit Determination from Observations The celestial poles and equator are the geographic poles and equator projected up onto the celestial sphere. The equivalent of the geographic latitude in the topocentric-equatorial system is called declination and is measured in degrees North (positive numbers) or South (negative numbers) of the celestial equator. The equivalent of the geographic longitude in the topocentric-equatorial system is called right ascension, and may also be measured in degrees, but for historical reasons it is more common to measure it in time (hours, minutes, seconds), 1 hour of right ascension being equivalent to 15 degrees of apparent sky rotation. In addition to the celestial equator, another important plane intersecting the celestial sphere is the ecliptic plane, which is the plane containing the orbit of the Earth about the Sun, inclined of about.4 with respect to the celestial equator, as shown in the preceding figure. The inclination angle e.4 is called the obliquity of the ecliptic. The position of a space object Q in the topocentric-equatorial system is expressed by the following vector q ¼ ðq cos d cos a Þu X þ ðq cos d sin aþu Y þ ðq sin dþu Z where u X, u Y, and u Z are the unit vectors of, respectively, X, Y, and Z. Let us express the position vector q as follows q ¼ q q q ¼ qu q

9 . Topocentric Co-ordinate Systems 17 where q is the magnitude of q and u q = q/q is the unit vector having the direction of q. By so doing, it is possible to write u q ¼ðcos d cos aþu X þðcos d sin aþu Y þðsin dþu Z As shown above, the geocentric-equatorial system and the topocentric-equatorial system have two different origins. Consequently, the direction cosines of the position vectors r and q are not the same. Thus, the topocentric declination and right ascension of a space object are not equal to the geocentric declination and right ascension of the same object. However, when the magnitude of r is much greater than the equatorial radius of the Earth, as is the case with stars and distant planets, then the differences in topocentric and geocentric angles can be neglected.. Orbit Determination from a Single Radar Observation With reference to the following figure, let us consider an artificial satellite inside the field of view of a radar station located on the surface of the Earth. The radar station measures the range, that is, the magnitude of the position vector going from the radar station to the satellite and the direction of this position vector with respect to a reference system having its origin in the place where the radar station is located. Reference systems of this type are called topocentric, because their origin is in the place of observation. One of these is the

10 18 Orbit Determination from Observations topocentric-horizon reference system UVW, defined in Sect.. and also shown in the preceding figure. Let q be the position vector of an artificial satellite with respect to the topocentric-horizon system indicated above. The direction of q is measured by two angles, namely azimuth (A) and altitude (h), which result from the gimbal axes on which the radar antenna is mounted. The azimuth angle A is measured clockwise, for an observer in the direction of the positive W-axis, from the north south direction to the projection of q onto the horizon (U, V) plane, so that 0 A 60 ; the altitude angle h is measured counterclockwise, for an observer in the direction of the positive U-axis, from the projection of q onto the horizon plane to q itself, so that 90 h 90. If the station is equipped with a Doppler radar, that is, of a radar capable of detecting a shift in frequency in the returning echo, then the rate of change of q can also be measured. On the other hand, sensors on the gimbal axes can measure the rate of change of the azimuth and altitude angles. Thus, the radar apparatus can measure six quantities related to a satellite in its field of view, namely range (q), azimuth angle (A), altitude angle (h), rate of change of range (q ), rate of change of azimuth angle (A ), and rate of change of altitude angle (h ). Let U, V, and W be direction cosines of u q = q/q with respect to the topocentric-horizon system UVW, which system has its origin in P and its axes pointing to, respectively, east, north, and zenith. The unit vector u q = q/q is expressible as follows u q ¼ U u U þ V u V þ W u W where u U, u V, and u W are the unit vectors pointing to, respectively, east, north, and zenith. By projecting u q onto, respectively, U, V, andw, there results u q ¼ðcos h sin AÞ u U þðcos h cos AÞ u V þðsin hþ u W Having obtained the components of the position vector q in the topocentric-horizon reference system UVW, it remains to transform the same components, for the purpose of computing the components of the position vector in the geocentric-equatorial reference system XYZ. To this end, it is necessary to determine the projections of the unit vectors u U, u V, and u W of the topocentric-horizon system UVW onto the unit vectors u X, u Y, and u Z of the geocentric-equatorial system XYZ. With reference to the following figure, let u be the geodetic latitude (defined below) of the point P where the observer or the radar station is located, and let h be the angular distance, measured in the equatorial plane, between the X-axis (pointing towards the vernal equinox) and the local meridian passing through P.

11 . Orbit Determination from a Single Radar Observation 19 In the preceding figure, the Earth is represented as an oblate ellipsoid, whose equatorial bulge is exaggerated for the sake of clarity. The geodetic latitude u is the angle between the equatorial plane XY and the normal in P to the surface of the ellipsoid, which approximates the Earth. Since the Earth is not perfectly spherical, then u does not coincide with the geocentric latitude, which is the angle between the equatorial plane XY and the normal in P to the surface of the sphere. Let u X * be the unit vector of OR, where OR lies in the plane of the meridian passing through P and is perpendicular to the Z-axis. Let u W be the unit vector of the zenith direction in P. By projecting u W onto OR and the Z-axis, there results u W ¼ðcos uþ u X þðsin uþu Z where u Z is the unit vector of the Z-axis. On the other hand, by projecting u X * onto X and Y, there results u X ¼ðcos hþ u X þðsin hþ u Y where u X and u Y are the unit vectors of, respectively, X and Y. It follows that u W ¼ ½ðcos hþu X þðsin hþu Y cos u þðsin uþ u Z ¼ðcos u cos hþu X þðcos u sin hþu Y þðsin uþu Z

12 140 Orbit Determination from Observations The unit vector u U, directed towards east, is expressed as a function of u X and u Y, by taking the vector product of u Z and u W, as follows u U ¼ u Z u W ju Z u W j where u Z u W is the magnitude of u Z u W. The following equality holds u X u Y u Z a b ¼ 4 a X a Y a Z 5 b X b Y b Z for any two vectors a = a X u X + a Y u Y + a Z u Z and b = b X u X + b Y u Y + b Z u Z. In the present case, there results u X u Y u Z 6 u Z u W ¼ cos u cos h cos u sin h sin u ¼ðcos u sin hþ u X þðcos u cos hþ u Y h ju Z u W j ¼ ðcos u sin hþ þðcos u cos hþ i1 ¼ cos u u U ¼ u Z u W ju Z u W j ¼ðsin hþ u X þðcos hþu Y Finally, u V results from u X u Y u Z 6 u V ¼ u W u U ¼ 4 cos u cos h cos u sin h 7 sin u 5 sin h cos h 0 ¼ðsin u cos hþ u X ðsin u sin hþ u Y þðcos u cos h þ cos u sin hþ u Z ¼ðsin u cos hþ u X ðsin u sin hþ u Y þðcos uþ uz In summary, we have obtained the following results u U ¼ðsin hþu X þðcos hþu Y u V ¼ðsin u cos hþu X ðsin u sin hþu Y þðcos uþu Z u W ¼ðcos u cos hþu X þðcos u sin hþu Y þðsin uþu Z

13 . Orbit Determination from a Single Radar Observation 141 Let us consider the following scalar products u U u X ¼sin h u U u Y ¼ cos h u U u Z ¼ 0 u V u X ¼sin u cos h u V u Y ¼sin u sin h u V u Z ¼ cos u u W u X ¼ cos u cos h u W u Y ¼ cos u sin h u W u Z ¼ sin u As shown in Sect. 1.9, in case of a transformation from perifocal co-ordinates xyz to geocentric-equatorial co-ordinates XYZ, the rotation matrix R is defined as follows u X u x u X u y u X u z R 4 u Y u x u Y u y u Y u z 5 u Z u x u Z u y u Z u z and this matrix is orthogonal, because it satisfies the condition R T R = I, where R T is the transpose of R, and I is the identity matrix. Likewise, in the present case, the rotation matrix R, which transforms the geocentric-equatorial co-ordinates XYZ into the topocentric-horizon co-ordinates UVW, is defined as follows u U u X u U u Y u U u Z R 4 u V u X u V u Y u V u Z 5 u W u X u W u Y u W u Z Thus, by replacing the scalar products by their respective values which are given above, there results U sin h cos h 0 4 V 5 ¼ 4 sin u cos h sin u sin h cos u W cos u cos h cos u sin h sin u 5 X 4 Y 5 Z where U, V, and W are the direction cosines of the topocentric-horizon co-ordinates, and X, Y, and Z are the direction cosines of the geocentric-equatorial co-ordinates. The preceding expression defines the transformation from geocentric-equatorial co-ordinates XYZ to topocentric-horizon co-ordinates UVW.

14 14 Orbit Determination from Observations As is easy to verify, the square matrix written above, that is, sin h cos h 0 R 4 sin u cos h sin u sin h cos u 5 cos u cos h cos u sin h sin u is an orthogonal matrix, because it satisfies the condition R T R ¼ I where R T is the transpose of R and I is the identity matrix. Therefore, in order to transform topocentric-horizon co-ordinates UVW into geocentric-equatorial co-ordinates XYZ, we use R T, the transpose of R, that is, X sin h sin u cos h cos u cos h 4 Y 5 ¼ 4 cos h sin u sin h cos u sin h 5 U 4 V 5 Z 0 cos u sin u W The matrices R and R T shown above can also be used for transformations from topocentric-equatorial co-ordinates X t Y t Z t to topocentric-horizon co-ordinates UVW, and from topocentric-horizon co-ordinates UVW to topocentric-equatorial co-ordinates X t Y t Z t, because the unit vectors u X, u Y, and u Z of the system XYZ coincide with those of the system X t Y t Z t. The following relation holds between the position vector (q) in the topocentric-horizon system UVW and the position vector (r) in the geocentric-equatorial system XYZ: r ¼ r E þ q where r E OP is the vector going from the centre of mass, O, of the Earth to the position, P, of the radar station on the surface of the Earth. This can be understood by considering the following figure.

15 . Orbit Determination from a Single Radar Observation 14 If the Earth were a perfect sphere having a radius equal to r E, then the local vertical passing through the radar site P would join (extended downward) this site with the centre O of the Earth. In this case, r E would be expressible as follows r E ¼ r E u Z Such is not the case in practice, because the shape of the Earth is not perfectly spherical. The following section of this paragraph takes account of the non-spherical shape of the Earth. In the model described below, the Earth is approximated to an ellipsoid, as shown in the following figure. Consequently, latitude cannot be used any more as a spherical co-ordinate and the radius of the Earth depends on latitude. Thus, a point on the surface of the Earth will be represented by two rectangular co-ordinates (x and z) and the longitude (k), because the latter has the same meaning in a spherical as in an oblate Earth. The ellipsoid is an approximation to the true shape of the Earth. The cross section of the Earth ellipsoid along a meridian is an ellipse, whose major semi-axis (a E ) is equal to the equatorial radius of the Earth and whose minor semi-axis (b E ) is equal to the polar radius of the Earth. Cross sections of the ellipsoid parallel to the equator are circles. The shaded area represents the bulge, which is maximum (1.4 km) at the equator and is zero at the poles. The values considered here for the major semi-axis a E and the flattening (or oblateness) f =(a E b E )/a E of the Earth ellipsoid are those given in Ref. [58], that is, a E ¼ 678:17 km f ¼ 1 0: :5756

16 144 Orbit Determination from Observations Hence, the eccentricity (e E ) and the minor semi-axis (b E ) of the ellipsoid result, respectively, from e E =(f f ) ½ = and b E = a E (1 f) = km. The reference ellipsoid defined above is an approximation to the geoid, that is, to the equipotential surface of the Earth gravity field which best fits, in the least-squares sense, the global mean sea level. In describing the non-spherical shape of the Earth, some authors, Herrick [4, 5] for one, use f instead of e E. The angle u* between the equatorial plane and the radius OP (between the centre O of the Earth and the point P of the Earth surface where the observer is located) is called geocentric latitude. The angle u between the equatorial plane and the normal in P to the surface of the ellipsoid is called geodetic latitude and is used in the maps and charts of the Earth. The normal in P to the surface of the ellipsoid is the direction which a plumb bob would indicate, were it not for local anomalies in the gravitational field of the Earth. The angle u a between the equatorial plane and the actual plumb bob vertical uncorrected for these anomalies is called the astronomical latitude. In practice, since the difference between the reference ellipsoid and the mean sea level is small, then the difference between geodetic latitude u and the astronomical latitude u a is negligible. We must now calculate the station co-ordinates of a point on the surface of the reference ellipsoid, when we are given the geodetic latitude, the longitude, and the height of that point above the mean sea level, which is assumed to be the height above the reference ellipsoid. To this end, with reference to the following figure, we first determine the co-ordinates x and z of a point P on the ellipse, assuming that we know the geodetic latitude u of P; then these co-ordinates will be adjusted for another point of which we know the height above the surface of the ellipsoid in the direction of the normal in P. The angle b, called the reduced altitude, is introduced to express the x and z co-ordinates as functions of the equatorial radius of the Earth and of b itself.

17 . Orbit Determination from a Single Radar Observation 145 By considering the elliptic cross section of the Earth (having a E as its major semi-axis along the equator and b E as its minor semi-axis along the polar axis of the Earth) and the corresponding auxiliary circumference (having a E as its radius), the Cartesian co-ordinates of P are expressible as follows x ¼ a E cos b z ¼ b E a E sin b a E In Sect. 1., we have shown that, for an ellipse, there results b E ¼ a E 1 e 1 E where e E is the eccentricity of the ellipse. Thus, there also results x ¼ a E cos b z ¼ a E 1 e 1 E sin b Now, cos b and sin b must be expressed as functions of the geodetic latitude u and of the constant quantities a E and b E. To this end, since the slope of the tangent to the ellipse is dz/dx, and the slope of the normal is dx/dz = tan u, then the differentials of the expressions written above x ¼ a E cos b z ¼ a E 1 e 1 E sin b are dx ¼a E sin b db dz ¼ a E 1 e 1 E cos b db and consequently tan u ¼ dx dz ¼ tan b ð1 e EÞ 1 and also tan b ¼ 1 e E 1 tan u ¼ 1 e 1 sin u E cos u Now, the last expression can be written in the following form tan b ¼ A B

18 146 Orbit Determination from Observations where A =(1 e E ) ½ sin u and B = cos u. In addition, since tan a sin a ¼ ð1 þ tan aþ 1 1 cos a ¼ ð1 þ tan aþ 1 are trigonometric identities, then there results after simplification sin b ¼ 1 1 e E sin u cos b ¼ 1 e 1 E sin u cos u 1 e E sin u 1 The preceding expressions make it possible to write the x and z co-ordinates of a point P on the surface of the ellipsoid as follows x ¼ a E cos b ¼ z ¼ a E 1 e E a E cos u 1 e E sin u 1 1 sin b ¼ a E 1 e E sin u 1 1 e E sin u Let us consider now another point P H, placed at a height H above the ellipsoid, that is, at a height H above the mean sea level, along the normal in P to the ellipsoid. By inspection of the preceding figure, the x and z components of the height H result Dx ¼ H cos u Dz ¼ H sin u The quantities Dx and Dy are the increments which must be added to the co-ordinates of P derived above, to obtain the co-ordinates of P H as functions of geodetic latitude (u), altitude (H) above the mean sea level, and equatorial radius (a E ) and eccentricity (e E ) of the ellipsoid representing the Earth: a E x H ¼ 4 þ H5 a E 1 e E cos u z 1 e 1 H ¼ 4 E sin u þ H5 sin u 1 e 1 E sin u When the flattening (f) of the Earth ellipsoid is used instead of e E, that is, when e E is replaced by f f, then the expressions derived above become 8 9 < a = E x H ¼ þ H 1 ðf f Þsin 1 : u ; cos u 8 9 < a E ð1 f Þ = z H ¼ þ H 1 ðf f Þsin 1 : u ; sin u

19 . Orbit Determination from a Single Radar Observation 147 The third co-ordinate of P H is the east longitude (k) ofp H. Thus, if we know the Greenwich sidereal time (h G ), it can be used with the east longitude of P H to compute the local sidereal time (h), which is the angle between the X-axis (pointing towards the vernal equinox) of the geocentric-equatorial system XYZ and the local meridian. As will be shown below, the x H and z H co-ordinates plus the angle h, in turn, are used to locate the observer site in the geocentric-equatorial system. In the model representing the Earth as a perfect sphere of radius r E, r E was the vector from the centre of the Earth to the position of the radar station on the surface of the Earth, with respect to the geocentric-equatorial system XYZ. Now, likewise, in the model representing the Earth as an oblate ellipsoid, let r E be the position vector from the centre, O, of the Earth to the point P H, with respect to the same system XYZ. The components of r E are then r E ¼ ðx H cos h Þu X þ ðx H sin hþu Y þ z H u Z where u X, u Y, and u Z are the unit vectors along, respectively, X, Y, and Z. Now, the vectors which have been expressed in the topocentric-horizon (or UVW) co-ordinates must be converted into the geocentric-equatorial (or XYZ) co-ordinates. The angle h G from the unit vector u X (pointing towards the vernal equinox) and the Greenwich meridian is the Greenwich sidereal time. This angle, added to the geographic longitude k of the radar site measured eastward from Greenwich, yields the local sidereal time h, as follows h ¼ h G þ k The angles u and h determine the relation between the topocentric-horizon system and the geocentric-equatorial system. To this end, it is necessary to compute h G at any given time t. When h G is known, h results from the expression (h = h G + k) given above. Let h G0 be the Greenwich sidereal time at some particular time t 0 (e.g. t 0 = 0 h Universal Time on the 1 st of January of the year of interest). When h G0 is known, then the local sidereal time h at any given time t can be determined from h ¼ h G 0 þ x E ðt t 0 Þþk where x E = rad/s is the angular velocity of the Earth about its axis. The following paragraph will show how to compute h G0, that is, h G at 0 h Universal Time for every day of the year of interest. The same paragraph will also provide further information on the concept of sidereal time introduced here. It is to be noted that the vector r E OP H shown above is the position vector from the centre, O, of the Earth to the point P H (where the observer is located), with respect to the geocentric-equatorial system XYZ. The knowledge of r E and q, where q P H Q is the position vector of a celestial body Q in the topocentric-horizon system UVW, suffices for orbits of bodies revolving about the Earth, because r = r E + q, where for such orbits r is the position vector from the centre of mass, O, of the Earth to the body Q revolving in a geocentric orbit.

20 148 Orbit Determination from Observations Following Boulet [1], we consider now bodies revolving about the Sun. With reference to the following figure, let r be the position vector of a body B revolving about the Sun, with respect to a celestial reference system XYZ whose origin is the centre of mass of the Sun, and whose plane XY is the Earth equator. Let E and S be the centres of mass of, respectively, the Earth and the Sun. Let q be the topocentric position vector of the body B. As shown below, the knowledge of r E EP H and s ES is necessary to compute P H S= EP H + s. The vector EP H defines the position of the observer with respect to the centre of the Earth at time t. The local sidereal time h, the latitude u, and the east longitude k are used to compute the rectangular components of EP H. The most common unit of measurement for distances of celestial bodies within the solar system is the astronomical unit (AU), which has been defined in Sect. 1.1.

21 . Orbit Determination from a Single Radar Observation 149 Let a E = AU (from Ref. [1]) be the radius of the Earth expressed in astronomical units (AU). The rectangular components of r E EP H at time t are r EX ¼ a E cos u cos h r EY ¼ a E cos u sin h r EZ ¼ a E sin u As to the components X S, Y S, and Z S of the solar vector s ES, the geocentric rectangular equatorial co-ordinates of the Sun, with respect to the mean equator and equinox of J000.0 (the 1 st of January 000 at noon), are published each year, for every day of the year, in The Astronomical Almanac. The tabular values of the co-ordinates of the Sun can be interpolated so as to obtain the values relating to the time of interest. Alternatively, a simple algorithm can be used for computing the angular co-ordinates of the Sun to an accuracy of about 1 arc-minute within two centuries of 000, which is given by the U.S. Naval Observatory [7]. This and other similar algorithms require the knowledge of concepts on the measurement of time in astronomy which have been not yet introduced to the reader and will be shown at the end of Sect..4. When X S, Y S, and Z S are known, the components of P H S result from ðp H SÞ X ¼a ð E cos u cos hþ þx S ðp H SÞ Y ¼a ð E cos u sin hþ þy S ðp H SÞ Z ¼a ð E sin uþþz S and therefore r is given, in the heliocentric system XYZ, by r ¼P H S þ q.4 The Measurement of Time in Astronomy As has been shown in the preceding paragraph, the determination of the orbit of either an artificial satellite or a natural celestial body by means of observations makes it necessary to record not only the observations themselves but also the times at which they have been performed. The time commonly used in every-day life is the time indicated by ordinary clocks, that is, the mean solar time. The measurement of solar time is based on the apparent motion of the Sun through the sky, due to the rotation of the Earth about its axis, as seen by an observer placed in a fixed point of the surface of the Earth. Thus, a solar day is the interval of time taken by the Sun, in its apparent motion around the Earth, to travel an arc of 60 along the sky. In other words, a solar day

22 150 Orbit Determination from Observations is the time elapsed between two consecutive passages of the Sun across the same meridian of the place of observation. By meridian of the place of observation we mean the great circle passing through the two celestial poles and the zenith of the site. A solar day is measured by the astronomers from noon to noon, and is divided into 4 h, or 4 60 = 1440 min, or = s. However, the rotation of the Earth about its axis does not comprise exactly 60 in a solar day, because the Earth also revolves about the Sun. In order for the Earth to do one revolution about the Sun, that is, in order for the Earth to travel an arc of 60 about the Sun, 65.5 solar days are necessary. Thus, 60 travelled in 65.5 solar days are about per solar day. This means that, in the course of a solar day, the direction of the Sun seen from a point of the Earth changes by about 1. This also means that the Earth must travel an arc of about 61 in order for the Sun to travel (in its apparent motion with respect to the Earth) an arc of 60 along the sky. The astronomers are interested to determine the rotation time of the Earth about its axis not only with respect to the Sun, but also with respect to the distant stars (the so-called fixed stars). In other words, they want to determine how long it takes the Earth to rotate 60 around its axis with respect to the fixed stars. This period of rotation is called sidereal day and can be determined by observing the starry sky by

23 .4 The Measurement of Time in Astronomy 151 night. The difference between solar day and sidereal day is illustrated in the preceding figure (due to the courtesy of Wikimedia, Ref. [1]), where the amplitude of the angle a = has been exaggerated for the sake of clarity. When the Earth is in position A along its orbit, the Sun and a given star of reference are, both of them, overhead. In position B, the Earth has performed a complete rotation about its axis, so that the star of reference (but not the Sun) is overhead again. The time interval taken by the Earth to move from A to B equals one sidereal day. Shortly after time t B, that is, at time t C, the Sun is overhead again. The time interval taken by the Earth to move from A to C equals one solar day. Thus, two consecutive passages of a fixed star (chosen as the star of reference) across the same meridian of the Earth measure one sidereal day, which is also divided into sidereal hours, minutes, and seconds. Since two consecutive passages of the star of reference through the same meridian take about h, 56 min and 4 s of ordinary solar time to occur, then a sidereal day is on the average min and 56 s shorter than a solar day. In other terms, a solar day is longer than a sidereal day by a factor which is about 60 þ 0: ¼ 1: The United States Naval Observatory [75] indicates a value of for this factor. This is the value which will be used below. In general terms, sidereal time is defined by some authors (see, e.g., Refs. [4, 5]) as the hour angle of the vernal equinox, that is, the hour angle of the ascending node of the ecliptic on the celestial equator. This is because the daily motion of the vernal equinox measures the rotation of the Earth with respect to the fixed stars, not to the Sun. By the way, the hour angle of any given point is the angle between the half plane determined by the rotation axis of the Earth and the zenith (half of the meridian plane) and the half plane determined by the rotation axis of the Earth and the given point. This angle is taken with the minus sign if the given point is eastward of the meridian plane, and with the plus sign if the given point is westward of the meridian plane. The hour angle is usually expressed in time units (hours, minutes, and seconds), where 4 h correspond to 60. In particular, a sidereal day is the interval of time taken by the hour angle of the vernal equinox to increase by 60. There are no two solar days having the same duration. This is because the axis of rotation of the Earth is not perpendicular to the ecliptic (that is, to the plane containing the orbit of the Earth around the Sun), and also because the orbit of the Earth is slightly elliptical. Since the areal velocity of the Earth is constant, the Earth moves faster along its orbit at perihelion (early in January) than it does at aphelion (early in July). Thus, a mean solar day is defined by making reference to the Earth revolving about the Sun in a circular orbit placed in the same plane as that of the ecliptic and having the same period as that of the real elliptical orbit [5]. The difference between the true and the mean solar time is called the equation of time. The two causes (that is, the eccentricity of the Earth orbit and the obliquity of

24 15 Orbit Determination from Observations the ecliptic) mentioned above produce effects which overlap with different periods of time, because the eccentricity has a period of one year, whereas the obliquity of the ecliptic has a period of half a year. Consequently, the equation of time has two minima and two maxima per year, as has been shown by Husfeld and Kronberg [8]. As we all know, the Earth comprises 4 time zones, equally spaced in longitude of about 15, so that the mean solar time of each zone differs by ± 1 h from the mean solar times of the two contiguous time zones. The world time zones are shown in the following figure, due to the courtesy of CIA [16]. Of all these mean solar times, that which relates to the Greenwich meridian is called Greenwich Mean Time (GMT) or Universal Time (UT1) or Zulu time (Z). There are several versions of Universal Time, the principal of which are the following: UT0 is the mean solar time of the Greenwich meridian obtained from direct astronomical observations. UT1 is UT0 corrected for the effects of small movements of the Earth relative to the axis of rotation (polar variation). UT is UT1 corrected for the effects of a small seasonal fluctuation in the rate of rotation of the Earth. UT is mainly of historic interest and rarely used. Universal Time is based on the imaginary mean Sun, which takes into account the effects on the solar day of the weakly elliptic orbit of the Earth about the Sun. As shown above, UT1 is a measure of the rotation angle of the Earth around its axis as resulting from astronomical observations, account being taken of slight movements of the Earth poles of rotation. UT1 predicts the solar position with sufficient accuracy for astronomical purposes, but the duration of a second derived from UT1

25 .4 The Measurement of Time in Astronomy 15 varies noticeably because of variations of the Earth rotation (the velocity of the Earth rotation is variable). As has been shown by various authors (see, e.g., McCarthy [5]), these variations may be classified into three types: secular, irregular, and periodic. The secular variation of the Earth rotational velocity depends on the apparently linear increase in the length of the day due mainly to tidal friction. The Moon and (to a lesser extent) the Sun raise tides in the oceans. Friction carries the maximum tide ahead of the line joining the centres of the Earth and Moon. The resulting couple decreases the velocity of rotation of the Earth and increases the orbital momentum of the Moon. In other words, the Earth loses energy and slows down, whereas the Moon gains this energy and increases its orbital period and distance from the Earth. According to McCarthy, the decrease of the Earth rotational velocity results in an increase of the day duration by about to s per century. According to Espenak and Meeus [], the secular acceleration of the Moon implies an increase in the day length of about 0.00 s per century. The irregular variation of the Earth rotational velocity appears to be due to random accelerations, but may be correlated with physical processes occurring on or within the Earth. The resulting variation in the day length is evaluated by McCarthy to s over the past 00 years. Finally, the periodic variation of the Earth rotational velocity is associated with periodically repeatable physical processes affecting the Earth. According to McCarthy, tides raised in the solid Earth by the Moon and the Sun produce changes in the length of the day with amplitudes of the order of s and with periods of 18.6 years, 1 year, ½ year, 7.55 days, 1.66 days, and others. Due to the reasons indicated above, the rotational velocity of the Earth varies in an unpredictable manner. In the most common civil usage, Universal Time is related to a time called Co-ordinated Universal Time (UTC), which provides the basis for Civil Time. UTC is kept by several time laboratories around the world and is measured by high-precision atomic clocks. The length of a UTC second is defined in terms of an atomic transition of Caesium under specific conditions and does not depend on the observation of astronomical phenomena. The international standard UTC is provided by the International Bureau of Weights and Measures on the basis of the data coming from the timing laboratories and is accurate to about s (that is, 1 ns) per day. UTC is made public by various radio stations, which also provide the difference between UTC and UT1, so that the latter can be computed as a function of the former. As mentioned above, UTC is the basis to compute Civil Time in the time zones of the Earth. By international agreements, UTC (which is measured by atomic clocks) cannot differ from UT1 (which is measured by the rotation of the Earth) by more than 0.9 s. When this limit is approached, a one-second change (called leap second) is introduced into UTC. The UTC can be easily computed starting from the local Civil Time (CT), which depends on where the observer is placed, as follows

26 154 Orbit Determination from Observations UTC ¼ CT þ z where z is the number of standard time zones by which the observer is displaced to the west of the Greenwich meridian. Some of the Earth time zones are indicated below: Western European Time (0 h of difference with respect to UTC); Central European Time (+1 h); USA, comprising in turn: Atlantic Standard Time ( 4 h); Eastern Standard Time ( 5 h); Central Standard Time ( 6 h); Mountain Standard Time ( 7 h); and Pacific Standard Time ( 8 h); Moscow Time (+ h); Tokyo Time (+9 h). In particular, there are four standard time zones in the conterminous USA. From east to west they are: Eastern Standard Time (EST), Central Standard Time (CST), Mountain Standard Time (MST) and Pacific Standard Time (PST), as shown in the following figure (due to the courtesy of the National Atlas of the United States [59]).

27 .4 The Measurement of Time in Astronomy 155 Consequently, for any given place within the conterminous USA, UTC can be computed by means of one of the following expressions UTC ¼ EST þ 5 UTC ¼ CST þ 6 UTC ¼ MST þ 7 UTC ¼ PST þ 8 For example, we want to compute the UTC corresponding to 1:1:15 pm EST First, we convert this time to a 4-h clock, as follows 1:1:15 pm EST þ 1 h ¼ 1:1:15 EST Then, adding 5 h (see above) to 1:1:15 EST, we have 1:1:15 EST þ 5 h ¼ 18 h 1 m 15 s UTC The International System of Units (usually abbreviated as SI) defines the second as the duration of 9,19,61,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the Caesium 1 atom. This defines, in the abstract, the Atomic Time [8]. In practice, since atomic clocks do not agree fully with one another, the weighted mean of several hundred atomic clocks, placed in various laboratories on the Earth, is used to define the so-called International Atomic Time (usually abbreviated as TAI). As has been shown above, at about 190, the Earth rotational period was found to be irregular and therefore, for purposes of orbital calculations, time based on Earth rotation was abandoned to choose a more uniform time scale based on the Earth orbit about the Sun. Thus, the Ephemeris Time was defined as the time scale which, together with the laws of motion, predicts correctly the positions of celestial bodies. Because of this property, the Ephemeris Time is used as the argument of the ephemerides, that is, of the tables giving the positions of the Sun, Moon, planets, and respective satellites as a function of time. For this purpose, in 1958, the International Astronomical Union (IAU) decided that Ephemeris Time is reckoned from the instant, near the beginning of the calendar year AD 1900, when the geometric mean longitude of the Sun was 79 degrees 41 min s, at which instant the measure of Ephemeris Time was 1900 January 0, 1 o clock precisely. Ephemeris Time was used for ephemeris calculations for the solar system until 1979, when it was replaced by Terrestrial Dynamical Time (TDT). TDT takes into account relativistic effects and is based on International Atomic Time (TAI), which has been defined above. To ensure continuity with Ephemeris Time, TDT was defined to match ET for the date 1977 January 1. In 1991, the IAU modified the

28 156 Orbit Determination from Observations definition of TDT to make it more precise. It was also renamed Terrestrial Time (TT); however, the old name (Terrestrial Dynamical Time) is still used. According to the United States Naval Observatory [7], Terrestrial Time is effectively equal to International Atomic Time plus.184 s exactly. Thus, the epoch designated J000.0 is specified as Julian date (see definition below) TT, or 000 January 1, 1 h TT. This epoch is also expressed as 000 January 1, 11:59:7.816 TAI, or 000 January 1, 11:58: UTC [74]. As a result of the decrease, at an irregular rate, of the rotational velocity of the Earth, the difference DT ¼ TT UT1 also decreases irregularly. The exact value of DT cannot be predicted and can only result from the historical record and observations. Since this value must be known to predict the correct times of astronomical events (such as eclipses), then a series of polynomial expressions have been created by Morrison and Stephenson [55], Espenak and Meeus [], and Islam et al. [40] to evaluate approximately DT during intervals of time of interest. Some examples (from Ref. []) of such evaluations are given below. Let the decimal year (y) be defined as follows y ¼ year þ month 0:5 1 which expression gives y for the middle of the month of interest. Let t = y 000. Then, the approximate value of DT (in seconds) is given by: DT ¼ 6:86 þ 0:45 t 0:06074 t þ 0: t þ 0: t 4 þ 0: t 5 ðwith year between 1986 and 005Þ DT ¼ 6:9 þ 0:17t þ 0:005589t ðwith year between 005 and 050Þ DT ¼0 þ ½ðy 180Þ=100 0:569 ð150 yþ ðwith year between 050 and 150Þ Section. has shown that a change of reference system from the topocentric-horizon (or UVW) system to the geocentric-equatorial (or XYZ) system requires the latitude (u) and longitude (k) of the observer and the Greenwich sidereal time (h G ). If h G were known on a particular day and at a particular time, then h G could be determined for any future day and time, because we know, as has been shown above, that the Earth turns through rotations on its axis per day. Let h G0 be the value of h G relating to 0 h UT1 on the 1 st of January of a particular year. Let day 0 designate the 1 st of January of the chosen year. Then, day 0 will designate the 1 st of January, day 58 will designate the 8 th of February, and so on, as shown in the following table, where the numbers in parentheses refer to the leap years.

29 .4 The Measurement of Time in Astronomy 157 Date Day No. Date Day No. 1 January 0 1 July 11(1) 8 February 58 1 August 4(4) 1 March 89(90) 0 September 7(7) 0 April 119(10) 1 October 0(04) 1 May 150(151) 0 November (4) 0 June 180(181) 1 December 64(65) In addition, any time can be expressed as a decimal fraction of a day. By so doing, any given set of values specifying a date and a time can be converted into a single floating-point number (D), such that its integral part indicates the number of days and its fractional part indicates the fraction of days elapsed from the chosen origin of times. Consequently, h G is expressible in degrees as follows or in radians as follows h G ¼ h G 0 þ 1: D h G ¼ h G0 þ 1: p D The methods used to determine sidereal time are based on a continuous count of days and fractions of day obtained by means of the Julian Day. The Julian Day or Julian Day Number (JDN) is the number of days that have elapsed since an initial epoch. The initial epoch has been set at noon Universal Time, Monday, the 1 st of January 471 BC in the proleptic Julian Calendar (that is, in the Julian calendar extended earlier in time so as to include dates preceding 4 AD). The initial epoch, which is counted as Julian Day 0, corresponds to the 4 th of November 4714 BC in the proleptic Gregorian calendar (that is, in the calendar obtained by extending the Gregorian calendar backwards to years earlier than 158, using the Gregorian leap year rules). The Julian date (JD) is a continuous count of days and fractions thereof elapsed since the same initial epoch, as follows JD ¼ JDN þ UT1 4 According to the definition given above, a Julian date comprises an integral part and a fractional part. The integral part of a Julian date is the Julian Day Number, whereas its fractional part is the time of day elapsed from noon UT1 expressed as a decimal fraction of a day. Examples of this fractional part are given below. A fractional part equal to 0.5 indicates midnight UT1, because the Julian Day begins at noon. A fractional part of 0.1 indicates UT1/4 = 0.1, that is, UT1 =.4 h (or.4 60 = 144 min or = 8640 s) elapsed from noon. Methods and tables for computing JDN from a date expressed in year, month, and day of our Gregorian calendar have

30 158 Orbit Determination from Observations been indicated by several authors and can also be found in the Internet. Curtis [0] and Boulet [1] use the following method JDN ¼ 67 y INTf1:75½y þ INTðm=1 þ 0:75Þg þ INTð75m=9Þ þ d þ 17101:5 where the function INT(x) means truncation (that is, retaining the integral part and dropping the fractional part of the argument x of the function INT), y is the full four-digit year, m is the month and d is the day of the Gregorian calendar date to be converted. Curtis points out that y, m, and d are integers such that: 1901 y m 1 1 d 1 This means that the formula given above holds only if the year of interest falls between 1901 and 099. These limitations do not affect another method, proposed by Jefferys [41], which method is based on the following rules. As is the case with Boulet s method, Jefferys expresses a Gregorian calendar date as y-m-d, where y is the year, m is the month number (January = 1, February =, etc.), and d is the day in the month. If the month is January or February, then 1 must be subtracted from the year to get a new value of y, and 1 must be added to the month to get a new value of m. Thus, January and February of a given year are considered as being, respectively, the 1 th and the 14 th month of the previous year. Then, the computation follows the following scheme a ¼ INT ðy=100þ b ¼ INTða=4Þ c ¼ a þ b e ¼ INT½65:5 ðy þ 4716Þ f ¼ INT½0:6001 ðm þ 1Þ JDN ¼ c þ d þ e þ f 154:5 This is the Julian Day Number for the beginning of the desired date at 0 h, UT1. To convert a Julian Day Number to a Gregorian calendar date, Jefferys [41] uses the following method, assuming that the JDN is for 0 h, UT1 (so that it ends in 0.5). The necessary calculations are shown below. Jefferys notes that his method does not give dates accurately in the proleptic Gregorian Calendar; in particular, the method fails if y is less than 400.

31 .4 The Measurement of Time in Astronomy 159 z ¼ JDN þ 0:5 w ¼ INT½ðz :5Þ=654:5 x ¼ INT ðw=4þ a ¼ z þ 1 þ w x b ¼ a þ 154 c ¼ INT½ðb 1:1Þ=65:5 d ¼ INT ð65:5cþ e ¼ INT½ðb dþ=0:6001 f ¼ INT ð0:6001eþ day of month ¼ b d f month ¼ e 1ore 1ðthe number obtained must be less than or equal to 1Þ year ¼ c 4715 ðif the month is January or FebruaryÞ or c 4716 ðotherwiseþ The U.S. Naval Observatory has an Internet-based calculator [71], which makes it possible to compute automatically the Julian date. As an example, let us convert the following Gregorian calendar date and time February 011 at 1:15:18 UT1 into the corresponding Julian date, by using Boulet s method. Since the condition 1901 y 099 is satisfied for y = 011, then we can compute the Julian Day Number as follows JDN ¼ INTf1:75 ½011 þ INTð=1 þ 0:75Þgþ INTð75 =9Þ þ þ 17101:5 ¼ 7807 INTf1:75 ½011 þ INTð0:917Þgþ INTð61:111Þþþ :5 ¼ 7807 INT½1:75 ð011 þ 0Þþ61 þ þ 17101:5 ¼ 7807 INTð519:5Þþ171077:5 ¼ :5 days The universal time, expressed in hours, is Finally, the desired Julian date is JD ¼ JDN þ UT1 4 UT1 ¼ 1 þ þ ¼ 1:55 ¼ :5 þ 1: :059 days

32 160 Orbit Determination from Observations Now, let us apply Jefferys method to the same example. Since, in this case, the month is February, then 1 must be subtracted from the year (011) to get a new value of y (011 1 = 010), and 1 must be added to the month () to get a new value of m ( + 1 = 14). Then the Julian Day Number is computed as follows a ¼ INTðy=100Þ¼INTð010=100Þ¼INTð0:10Þ¼0 b ¼ INTða=4Þ¼INTð0=4Þ¼INTð0=4Þ¼5 c ¼ a þ b ¼ 0 þ 5 ¼1 e ¼ INT½65:5 ðy þ 4716Þ ¼ INT½65:5 ð010 þ 4716Þ ¼ INTð456671:5Þ ¼ f ¼ INT½0:6001 ðm þ 1Þ ¼ INT½0:6001 ð14 þ 1Þ ¼ INTð459:0015Þ ¼459 JDN ¼ c þ d þ e þ f 154:5 ¼1 þ þ þ :5 ¼ :5 which is the same value as that computed previously by means of Boulet s method. Now, Jefferys method is applied to the case JDN = to obtain the corresponding Gregorian calendar date. Using the rules indicated above, we have z ¼ :5 þ 0:5 ¼ w ¼ INT½ð :5Þ=654:5 ¼ INTð16:109Þ ¼16 x ¼ INTð16=4Þ¼INTð4Þ¼4 a ¼ þ 1 þ 16 4 ¼ b ¼ þ 154 ¼ 4571 c ¼ INT½ð4571 1:1Þ=65:5 ¼ INTð676:99Þ ¼676 d ¼ INTð65:5 676Þ ¼INTð456671:5Þ ¼ e ¼ INT½ð Þ=0:6001 ¼ INTð15:098Þ ¼15 f ¼ INTð0: Þ ¼INTð459:0015Þ ¼459 day of month ¼ ¼ month ¼ 15 1 ¼ year ¼ ¼ 011 As was expected, the resulting Gregorian calendar date is February 011. The U.S. Naval Observatory also gives algorithms in FORTRAN programming language, which are due to Fliegel and van Flandern [5]. The current Julian epoch has been set to the 1 st of January 000 at noon. This epoch is denoted by J000.0 and corresponds to the Julian Day Number A Julian year has 65.5 days and consequently a Julian century has 655 days. Let T 0 be the time, expressed in Julian centuries, elapsed between a given Julian day J 0 and J Then the time T 0 results from

33 .4 The Measurement of Time in Astronomy 161 T 0 ¼ J :0 655 The time h G0 (that is, the Greenwich sidereal time at 0 h UT1) can be expressed (in seconds) as a function of the dimensionless time T 0 by means of the following formula given by Aoki et al. [4]: h G0 ¼ 4110 s :54841 þ s :81866 T 0 þ 0 s :09104T 0 0 s : T 0 where the superscript s stands for seconds. If, as is often the case, we want to express h G0 in degrees, then the coefficients appearing in the formula given above must be multiplied by 60/(4 600). The same formula expressed in degrees is h G0 ¼ 100: þ 6000: T 0 þ 0: T 0 : T 0 Since the value computed by means of the preceding formula may be outside the interval 0 < h G0 < 60, then the computed value must, if necessary, be brought into that interval by adding or subtracting an integral multiple of 60. This done, the Greenwich sidereal time (h G ) relating to any other universal time than 0 h can be found as follows h G ¼ h G0 þ 60: UT1 4 where = is the number of degrees covered by the Earth in its rotation about its axis in 4 h (solar time). If the (local) sidereal time h is required for a site placed at an east longitude k, then h ¼ h G þ k As shown above, in case of the value of h being greater than 60, it is necessary to bring it into the interval 0 < h < 60 by subtracting an integral number of 60 from the computed value. As an example of application, let us compute the local sidereal time in degrees for Kiruna, Sweden (latitude u = 65.85N; longitude k = 0.167E) on the 1 th of February 01 at :0:00 UT1. First, we use Jefferys method to compute the Julian Day Number, that is, the Julian date relating to 1 th of February 01 at 00:00:00 UT1. Since, in this case, the month is February, then 1 must be subtracted from the year (01) to get a new value of y (01 1 = 011), and 1 must be added to the month () to get a new value of m ( + 1 = 14).

34 16 Orbit Determination from Observations Then the Julian Day Number (J 0 ) is computed as follows a ¼ INTðy=100Þ¼INTð011=100Þ¼INTð0:11Þ¼0 b ¼ INTða=4Þ¼INTð0=4Þ¼INTð5Þ¼5 c ¼ a þ b ¼ 0 þ 5 ¼1 e ¼ INT½65:5 ðy þ 4716Þ ¼ INT½65:5 ð011 þ 4716Þ ¼ INTð45706:75Þ ¼ f ¼ INT½0:6001 ðm þ 1Þ ¼ INT½0:6001 ð14 þ 1Þ ¼ INTð459:0015Þ ¼459 J 0 ¼ c þ d þ e þ f 154:5 ¼1 þ 1 þ þ :5 ¼ :5 Second, the dimensionless time T 0 results from the expression shown above T 0 ¼ J :0 655 ¼ : :0 655 ¼ 0: Thirdly, h G0 is computed by using Aoki s formula expressed in degrees h G 0 ¼ 100: þ 6000:77005 T 0 þ 0: T 0 : T 0 Substituting T 0 = in the preceding formula, there results h G0 ¼ 100: þ 6000: : þ 0: : : : ¼ 446 : Since this value is outside the interval [0, 60], we bring it into that interval by subtracting a multiple of 60 from it. To this end, we observe that INT 446: ¼ 1 60 It follows that h G0 ¼ 446: ¼ 14 : The universal time given in this example is :0:00, that is, UT1 ¼ :5h Thus, the Greenwich sidereal time h G is computed by replacing h G0 with the value obtained above and UT1 with.5 into the expression

35 .4 The Measurement of Time in Astronomy 16 This yields h G ¼ h G0 þ 60: UT1 4 h G ¼ 14 : þ 60: :5=4 ¼ 180 : Finally, the east longitude of Kiruna (k = 0.167) must be added to h G to obtain the local sidereal time, as follows h ¼ h G þ k ¼ 180 : þ 0 :167 ¼ 00 : Orbital Elements from Angle and Range Measurements The orbital elements of a space object Q revolving around the Earth and shown in the following figure are determined when its position (r) and velocity (r ) vectors, with respect to the geocentric-equatorial system XYZ, are known at a given time. Section. has shown how to determine Q as a function of the line-of-sight vector q of Q with respect to the topocentric-horizon system UVW (located at the radar station P) and the position vector r E of P with respect to the geocentric-equatorial system. The geocentric position vector r OQ of the space object Q results from

36 164 Orbit Determination from Observations r ¼ r E þ q ¼ r E þ qu q where u q q/q is the unit vector having the direction of q. The velocity (r ) and acceleration (r ) vectors of Q, with respect to the geocentric-equatorial system XYZ, result from differentiating once and twice the preceding expression with respect to time. This yields r 0 ¼ r 0 E þ q0 u q þ qu 0 q r 00 ¼ r 00 E þ q00 u q þ q 0 u 0 q þ q0 u 0 q þ qu00 q ¼ r00 E þ q00 u q þ q 0 u 0 q þ qu00 q Now, all vectors appearing in the preceding expressions must be expressed in the same reference system, that is, in the geocentric-equatorial reference system XYZ. To this end, it is to be noted that r E OP is a vector which rotates with the Earth at a constant angular velocity x E equal to x E ¼ x E u Z where x E is the magnitude of x E, and u Z is the unit vector of the Z-axis. Since r E rotates with the Earth, then its first (r E ) and second (r E ) time derivatives result from the following expressions r 0 E ¼ x E r E r 00 E ¼ x E ðx E r E Þ Let X, Y, and Z be the direction cosines of PQ with respect to the topocentric-equatorial system X t Y t Z t, having its origin in P and its axes parallel to, respectively, X, Y, and Z. The unit vector u q = q/q is expressible as follows u q ¼ X u X þ Y u Y þ Z u Z Since the unit vectors u X, u Y, and u Z do not change with time, then the first and second time derivatives of the preceding expression are u 0 q ¼ 0 X u X þ 0 Y u Y þ 0 Z u Z u 00 q ¼ 00 X u X þ 00 Y u Y þ 00 Z u Z Now, since u q = (cos d cos a) u X + (cos d sin a) u Y + (sin d) u Z, where d and a are, respectively, the declination and right ascension of Q, then the direction cosines of PQ with respect to the topocentric-equatorial system are expressible in terms of d and a as follows

37 .5 Orbital Elements from Angle and Range Measurements 165 X ¼ cos d cos a Y ¼ cos d sin a Z ¼ sin d Differentiating the preceding expressions with respect to time yields 0 X ¼a0 sin a cos d d 0 cos a sin d 0 Y ¼ a0 cos a cos d d 0 sin a sin d 0 Z ¼ d0 cos d 00 X ¼a00 sin a cos d d 00 cos a sin d ða 0 þ d 0 Þ cos a cos d þ a 0 d 0 sin a sin d 00 Y ¼ a00 cos a cos d d 00 sin a sin d ða 0 þ d 0 Þ sin a cos d a 0 d 0 cos a sin d 00 Z ¼ d00 cos d d 0 sin d Now, let U, V, and W be the direction cosines of PQ with respect to the topocentric-horizon system UVW, having its origin in P and its axes pointing to, respectively, east, north, and zenith. As shown in Sect.., the unit vector of the direction PQ is u q ¼ðcos h sin AÞu U þðcos h cos AÞu V þðsin hþ u W where h and A are, respectively, the altitude angle and the azimuth angle of Q. Consequently, the direction cosines of PQ with respect to the topocentric-horizon system UVW are U ¼ cos h sin A V ¼ cos h cos A W ¼ sin h The direction cosines X, Y, and Z of PQ with respect to the geocentric-equatorial system XYZ can be obtained by means of the co-ordinate transformation shown in Sect.., that is, X sin h sin u cos h cos u cos h 4 Y 5 ¼ 4 cos h sin u sin h cos u sin h 5 U 4 V 5: Z 0 cos u sin u W

38 166 Orbit Determination from Observations replacing X, Y, Z, U, V, and W by their respective values yields cos d cos a sin h sin u cos h cos u cos h cos h sin A 4 cos d sin a 5 ¼ 4 cos h sin u sin h cos u sin h 54 cos h cos A 5 sin d 0 cosu sin u sin h Expanding the matrix product in the preceding equality leads to cos d cos a ¼sin h cos h sin A sin u cos h cos h cos A þ cos u cos h sin h hence hence cos u sin h sin u cos h cos A cos a ¼ ð Þcos h sin h cos h sin A cos d cos d sin a ¼ cos h cos h sin A sin u sin h cos h cos A þ cos u sin h sin h cos u sin h sin u cos h cos A sin a ¼ ð Þsin h þ cos h cos h sin A cos d sin d ¼ cos u cos h cos A þ sin u sin h The preceding expressions of cos a, sin a, and sin d can be simplified by using the hour angle H, defined in Sect..4, which is the angular distance between the object observed and the local meridian. In terms of the variables used above, the hour angle is expressible as follows H ¼ h a The sine and cosine of the hour angle can be expressed as follows sinðh aþ ¼sin h cos a cos h sin a cosðh aþ ¼cos h cos a þ sin h sin a Thus, substituting the following expressions cos u sin h sin u cos h cos A cos a ¼ ð Þcos h sin h cos h sin A cos d cos u sin h sin u cos h cos A sin a ¼ ð Þsin h þ cos h cos h sin A cos d

39 .5 Orbital Elements from Angle and Range Measurements 167 into sinðh aþ ¼sin h cos a cos h sin a yields cos u sin h sin u cos h cos A sinðh aþ ¼sin h ð Þcos h sin h cos h sin A cos d cos u sin h sin u cos h cos A cos h ð Þsin h þ cos h cos h sin A cos d cos h sin A ¼ cos d into Likewise, substituting cos u sin h sin u cos h cos A cos a ¼ ð Þcos h sin h cos h sin A cos d cos u sin h sin u cos h cos A sin a ¼ ð Þsin h þ cos h cos h sin A cos d cosðh aþ ¼cos h cos a þ sin h sin a yields cos u sin h sin u cos h cos A cosðh aþ ¼cos h ð Þcos h sin h cos h sin A cos d cos u sin h sin u cos h cos A þ sin h ð Þsin h þ cos h cos h sin A cos d cos u sin h sin u cos h cos A ¼ cos d This expression makes it possible to compute the hour angle, as follows cos u sin h sin u cos h cos A H ¼ arccos cos d which holds if sin H > 0. Otherwise, if sin H < 0, then cos u sin h sin u cos h cos A H ¼ p arccos cos d

40 168 Orbit Determination from Observations Since the altitude (h) and declination (d) angles range from p/ to p/ radians, then neither cos h nor cos d can be negative. Consequently, the expression cos h sin A sin H ¼ cos d shows that sin H is negative when sin A is positive, which happens when A ranges from 0 to p radians. In summary, let the topocentric angles azimuth (A) and altitude (h) of a space object be known at a given time. Let also the latitude (u) of the tracking station and the sidereal time (h) be known at the same time. Then, the topocentric declination (d) results from the equation derived above which, solved for d, yields sin d ¼ cos u cos h cos A þ sin u sin h d ¼ arcsin ðcos u cos h cos A þ sin u sin hþ This done, the hour angle (H) results from p arccos ½ðcos u sin h sin u cos h cos AÞ= cos d 0\A\p H ¼ ð Þ arccos½ðcos u sin h sin u cos h cos AÞ= cos d ðp A pþ and the right ascension (a) results from a ¼ h H If the azimuth and altitude angles are known as functions of time, then the right ascension and declination angles can be computed as functions of time by means of the expressions given above. Then, these functions are differentiated with respect to time, and the results are introduced into the following expressions u q ¼ X u X þ Y u Y þ Z u Z u 0 q ¼ 0 X u X þ 0 Y u Y þ 0 Z u Z u 00 q ¼ 00 X u X þ 00 Y u Y þ 00 Z u Z where X ¼ cos d cos a Y ¼ cos d sin a Z ¼ sin d

41 .5 Orbital Elements from Angle and Range Measurements X ¼a0 sin a cos d d 0 cos a sin d 0 Y ¼ a0 cos a cos d d 0 sin a sin d 0 Z ¼ d0 cos d 00 X ¼a00 sin a cos d d 00 cos a sin d ða 0 þ d 0 Þ cos a cos d þ a 0 d 0 sin a sin d 00 Y ¼ a00 cos a cos d d 00 sin a sin d ða 0 þ d 0 Þ sin a cos d a 0 d 0 cos a sin d 00 Z ¼ d00 cos d d 0 sin d This makes it possible to compute the unit vector u q and its time derivatives u q and u q. In order to compute a and d from A and h, the expression sin d ¼ cos u cos h cos A þ sin u sin h is differentiated with respect to time, taking into account that u = constant. This yields d 0 cos d ¼h 0 cos u sin h cos A A 0 cos u cos h sin A þ h 0 sin u cos h which in turn, solved for d, yields d 0 ¼ A0 cos u cos h sin A þ h 0 ðsin u cos h cos u sin h cos AÞ cos d Now, the expression sin H = (cos h sin A)/cos d is differentiated with respect to time. This yields H 0 cos H ¼ h0 sin h sin A A0 cos h cos A d0 cos h sin A sin d cos d cos d cos d ð ¼ A0 cos h cos A h 0 sin h sin AÞcos d þ d 0 cos h sin A sin d cos d Since cos u sin h sin u cos h cos A cos H ¼ cos d then substituting this expression of cos H into the expression of H cos H yields H 0 cos H ¼ H 0 cos u sin h sin u cos h cos A cos d ð ¼ A0 cos h cos A h 0 sin h sin AÞcos d þ d 0 cos h sin A sin d cos d

42 170 Orbit Determination from Observations The preceding equation, solved for H, yields H 0 ¼ A0 cos h cos A h 0 sin h sin A þ d 0 cos h sin A tan d cos u sin h sin u cos h cos A Now, since H = h a, then H 0 ¼ h 0 a 0 ¼ x E a 0 where x E is the angular velocity of the Earth about its axis. It follows that that is, a 0 ¼ x E H 0 a 0 ¼ x E þ A0 cos h cos A h 0 sin h sin A þ d 0 cos h sin A tan d cos u sin h sin u cos h cos A As an example of application, we compute the classical orbital elements of a space object, which the Catalina station of the University of Arizona, Tucson (latitude u =.417 N, longitude k = E, and height H = 509 m on the mean sea level) detected on the 5 th of April 004 at 6:00:00 UT1, obtaining the following data: Slant range q ¼ 7148 km Azimuth A ¼ 18 Altitude h ¼ 41 Range rate q 0 ¼ :67 km=s Azimuth rate A 0 ¼1: rad=s Altitude rate h 0 ¼ : rad=s The three components of the position vector (r E ) of the tracking station, with respect to the geocentric-equatorial system XYZ, are r E ¼ ðx H cos h Þu X þ ðx H sin hþu Y þ z H u Z where u X, u Y, and u Z are the unit vectors along, respectively, X, Y, and Z, x H and z H are the two co-ordinates of the point P H representing the position of the tracking station. Let us compute the local sidereal time h by means of the date 5 April 004 at 6:00:00 UT1

43 .5 Orbital Elements from Angle and Range Measurements 171 The Julian Day Number (J 0 ) is computed as follows a ¼ INTðy=100Þ ¼ INTð004=100Þ ¼ INTð0:04Þ ¼ 0 b ¼ INTða=4Þ ¼ INT ð0=4þ ¼ INT ð5þ ¼5 c ¼ a þ b ¼ 0 þ 5 ¼1 e ¼ INT½65:5 ðy þ 4716Þ ¼ INT½65:5 ð004 þ 4716Þ ¼ f ¼ INT½0:6001 ðm þ 1Þ ¼ INT½0:6001 ð4 þ 1Þ ¼ INT ð15:0005þ ¼15 J 0 ¼ c þ d þ e þ f 154:5 ¼1 þ 5 þ þ :5 ¼ 45100:5 The dimensionless time T 0 results from the following expression T 0 ¼ J :0 655 ¼ 45100: :0 655 Now, h G0 is computed by using Aoki s formula ¼ 0:04587 h G0 ¼ 100: þ 6000:77005 T 0 þ 0: T 0 : T 0 ¼ 100: þ 6000: :04587 þ 0: :04587 : :04587 ¼ 16 :6541 Since this value is outside the interval [0, 60], we bring it into that interval by subtracting a multiple of 60 from it. To this end, we observe that INT 16:65 ¼ 4 60 It follows that h G0 ¼ 16: ¼ 19 :65418 The universal time given in this example is 6:00:00, that is, UT1 = 6.0 h. Thus, the Greenwich sidereal time h G is computed by replacing h G0 with and UT1 with 6.0 into the following expression h G ¼ h G0 þ 60: UT1 4 This yields h G ¼ 19:65418 þ 60: ¼ 8 :

44 17 Orbit Determination from Observations The east longitude of Catalina (k = east) must be added to h G to obtain the local sidereal time, as follows h ¼ h G þ k ¼ 8 : :7 17 :14 Now, x H and z H are computed as follows 8 9 < a = E x H ¼ þ H 1 ðf f Þsin 1 : u ; cos u 8 9 < a E ð1 f Þ = z H ¼ þ H 1 ðf f Þsin 1 : u ; sin u where a E = km and f = are, respectively, the equatorial radius and the flattening of the Earth (represented as an ellipsoid), and u is the geodetic latitude (that is, the angle between the equatorial plane and the local vertical) measured at the tracking station. Taking u =.417, x H and z H result from n x H ¼ 678:17= 1 0:0058 0:0058 sin 1= o :417 þ :509 cos :417 ¼ 591:55 km n z H ¼ 678:17 ð10:0058þ = 1 0:0058 0:0058 sin :417 1= þ :509 o sin :417 ¼ 400:91 km Hence, the position vector r E of the tracking station, with respect to the geocentric-equatorial system XYZ, is r E ¼ ðx H cos h Þu X þ ðx H sin hþu Y þ z H u Z ¼ð591:55 cos 17 :14Þu X þð591:55 sin 17 :14Þ u Y þ 400:91u Z ¼55:954 u X þ 64:987 u Y þ 400:91 u Z ðkmþ Let us compute now the declination d of the observed object with respect to the topocentric-equatorial system X t Y t Z t, by means of the following expression d ¼ arcsin ðcos u cos h cos A þ sin u sin hþ ¼ arcsinðcos :417 cos 41 cos 18 þ sin :417 sin 41 Þ ¼ : The given azimuth (A = 18 ) lies between 0 and 180. Consequently, the hour angle H results from the following expression

45 .5 Orbital Elements from Angle and Range Measurements 17 H ¼ 60 cos u sin h sin u cos h cos A arccos cos d ¼ 60 arccos cos :417 sin 41 sin :417 cos 41 cos 18 cosð :Þ ¼ :47 Thus, the right ascension of the observed object results from a ¼ h H ¼ 17 :14 :47 ¼150 : Now, we compute the unit vector u q of the line joining the tracking station with the object, with respect to the topocentric-equatorial system X t Y t Z t, by means of u q ¼ X u X þ Y u Y þ Z u Z where X ¼ cos d cos a Y ¼ cos d sin a Z ¼ sin d This yields u q ¼ ½cosð :Þ cosð150 :Þu X þ ½cosð :Þ sinð150 :Þu Y þ½sinð :Þu Z ¼0:868 u X 0:4946 u Y 0:0405 u Z The position vector r (in km) of the observed object results from r ¼ r E þ qu q ¼55:954 u X þ 64:987 u Y þ 400:91 u Z þ 7148 ð0:868 u X 0:4946 u Y 0:0405 u Z Þ ¼89:848 u X 178:414 u Y þ 01:47 u Z The velocity vector r E of the tracking station, with respect to the celestial geocentric-equatorial system XYZ, results from r 0 E ¼ x E r E where r E = u X u Y u Z is the position vector of the tracking station and x E = x E u Z is the angular velocity of the Earth around its axis with respect to XYZ. Remembering that x E = rad/s, there results

46 174 Orbit Determination from Observations u X u Y u Z r 0 E ¼ x 6 E r E ¼ 0 0 7: :954 64: :91 ¼ 64:987 7: ux þð55:954 7: Þu Y ¼0:047 u X 0:90 u Y ðkm=sþ Now, the declination rate d of the observed object results from d 0 ¼ ½A 0 cos u cos h sin A þ h 0 ðsin u cos h cos u sin h cos AÞ= cos d ¼ 1: cos :417 cos 41 sin 18 þ : ðsin :417 cos 41 cos :417 sin 41 cos 18 Þ= cosð :Þ ¼: ðrad=sþ The right ascension rate a of the observed object results from a 0 ¼ x E þða 0 cos h cos A h 0 sin h sin A þ d 0 cos h sin A tan dþ=ðcos u sin h sin u cos h cos AÞ ¼7: þ½1: cos 41 cos 18 : sin 41 sin 18 þ : cos 41 sin 18 tan ð :Þ =½cos :417 sin 41 sin :417 cos 41 cos 18 ¼6: ðrad=sþ The direction cosine rate vector u q results from where Hence, u 0 q ¼ 0 X u X þ 0 Y u Y þ 0 Z u Z 0 X ¼a0 sin a cos d d 0 cos a sin d 0 Y ¼ a0 cos a cos d d 0 sin a sin d 0 Z ¼ d0 cos d 0 X ¼6: sinð150 :Þ cosð :Þ : cosð150 :Þ sinð :Þ ¼: Y ¼ 6: cosð150 :Þ cosð :Þ: sinð150 :Þ sinð :Þ ¼5: Z ¼ : cosð :Þ ¼ :

47 .5 Orbital Elements from Angle and Range Measurements 175 which, substituted into u q = X u X + Y u Y + Z u Z, yield u 0 q ¼ : u X 5: 10 5 u Y þ : u Z ðrad=sþ The velocity vector r of the observed object, with respect to the geocentric-equatorial reference system XYZ, results from r 0 ¼ r 0 E þ q0 u q þ qu 0 q Hence, the velocity vector r (in km/s) of the observed object is r 0 ¼0:047 u X 0:90 u Y þ :67 ð0:868 u X 0:4946 u Y 0:0405 u Z Þ þ 7148 ð: u X 5: 10 5 u Y þ : u Z Þ ¼1:5 u X :959 u Y þ 0:854 u Z In summary, the position (r) and velocity (r ) vectors of the space object at the epoch of observation, with respect to the geocentric-equatorial system XYZ, are r ¼89:848 u X 178:414 u Y þ 01:47 u Z ðkmþ r 0 ¼1:5 u X :959 u Y þ 0:854 u Z ðkm=sþ Now that the position and velocity vectors are known, the corresponding orbital elements can be computed, as will be shown below. The radius vector (in km) and the square of the velocity (in km /s ) at epoch are h r 0 ¼ ð89:848þ þð178:414þ i1 þ 01:47 ¼ 1705:57 v 0 ¼ ð1:5þ þð:959þ þ 0:854 ¼ 11:01 The vis-viva integral v l ¼ r 1 a makes it possible to compute the major semi-axis a (in km) as follows a ¼ 1 1 ¼ ¼ 801:744 v 11:01 0 r 0 l 1705: :4

48 176 Orbit Determination from Observations The angular momentum vector per unit mass (in km /s) is u X u Y u Z h ¼ r r 0 6 ¼ 4 89: : :47 5 1:5 :959 0:854 ¼4107:11 u X þ 1857:85 u Y þ 69795:191 u Z and its magnitude (in km /s) is h h ¼ ðh hþ 1 ¼ ð4107:11þ i1 þ 1857:85 þ 69795:191 ¼ 75:005 The semi-latus rectum p (in km) results from p ¼ h l ¼ 75: :4 ¼ 146:084 The eccentricity e results from a = p(1 e ), which, solved for e, yields e ¼ 1 p 1 ¼ 1 146:084 1 ¼ 0:79458 a 801:744 The inclination angle i (in degrees) of the orbit with respect to the equatorial plane is i ¼ arccos h Z ¼ arccos 69795:191 ¼ 17 : h 75:005 The eccentricity vector e results from e ¼ r0 h r l r 0 ðr 0 hþ=l ¼ ½ð: : :85 0:854Þu X þð1: : :11 0:854Þ u Y þ ð1:5 1857: :11 :959Þu Z =98600:4 ¼0:564951u X þ 0:074494u Y 0: u Z r=r 0 ¼ð89:848 u X 178:414 u Y þ 01:47 u Z Þ=1705:57 ¼0:91 u X 0: u Y þ 0: u Z

49 .5 Orbital Elements from Angle and Range Measurements 177 e ¼ðr 0 hþ=l r=r 0 ¼ð0: þ 0:91Þ u X þð0: þ 0:401914Þ u Y þð0: :075875Þ u Z ¼ 0:4711 u X þ 0: u Y 0: u Z The node vector n is defined by n u Z h. In the present case, there results n ¼ n X u X þ n Y u Y ¼h Y u X þ h X u Y ¼1857:85 u X 4107:11 u Y The magnitude n of the node vector is n ¼ ðn nþ 1 ¼ 1857:85 þ 4107:11 1 ¼ 40:68 The right ascension X of the ascending node (in degrees) is X ¼ arccos n X n X ¼ 60 arccos n X n ðn Y 0Þ ðn Y \0Þ In the present case, n Y = < 0; thus, X ¼ 60 arccos 1857:85 ¼ 190 : :68 The argument of perigee x results from x ¼ arccos n e ðe Z 0Þ ne x ¼ 60 arccos n e ðe Z \0Þ ne In the present case, e Z = < 0; hence, x ¼ :85 0: :11 0: arccos 40:68 0:79458 ¼ 1 :54665 The true anomaly at epoch / 0 results from / 0 ¼ arccos e r ðr r 0 0Þ er 0 / 0 ¼ 60 e r arccos ðr r 0 \0Þ er 0

50 178 Orbit Determination from Observations In the present case, there results r r 0 ¼89:848 ð1:5þ 178:414 ð:959þ þ 01:47 0:854 ¼ 75511:58 [ 0 e r ¼ 0:4711 ð89:848þ þ 0: ð178:414þ 0: :47 ¼18111:44 er 0 ¼ 0: :57 ¼ 91:41147 Thus, the true anomaly (in degrees) at epoch is / 0 ¼ arccos 18111:44 ¼ 14 : :41147 In summary, the object detected by the given radar station at the given time revolves about the Earth in an elliptic orbit having the following elements a ¼ 801:744 km X ¼ 190 :64185 e ¼ 0:79458 x ¼ 1 :54665 i ¼ 17 : / 0 ¼ 14 : Orbital Elements from Three Measurements of Angles (Method of Gauss) As has been shown in the preceding paragraphs, a set of six independent quantities is required to determine the motion of a celestial body. These six quantities may be the three components of the position vector and the three components of the velocity vector, or the six classical elements. A radar station like that described in Sect..5 provides range and range-rate measurements. Thus, three range measurements (range, declination, and azimuth) plus three range-rate measurements (range rate, declination rate, and azimuth rate) provide the six required quantities. This implies the availability of a Doppler radar. By contrast, the present paragraph and the following one will show how to determine the orbit of a celestial body when only angular observations are possible. This happens when only a telescope can be used as a means of observation. When only angular measurements are possible (e.g. the declination and the right ascension of the body observed), then three distinct observations are required, each of which provides the declination and the right ascension of the body. Following Curtis [0] and Boulet [1], let t 1, t, and t be the three distinct times at which the three single angular observations are performed. Let P 1,P, and P be the three positions of the observed body at, respectively, t 1, t, and t, as shown in the following figure. Let r 1, r, and r be the three position vectors of the observed

51 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 179 body with respect to the geocentric-equatorial reference system XYZ. Let r E1, r E, and r E be the three position vectors of the point of observation with respect to the same system of reference. The relation between r 1, r, and r and r E1, r E, and r E is r 1 ¼ r E1 þ q 1 ¼ r E1 þ q 1 ðq 1 =q 1 Þ ¼ r E1 þ q 1 u 1 r ¼ r E þ q ¼ r E þ q ðq =q Þ ¼ r E þ q u r ¼ r E þ q ¼ r E þ q ðq =q Þ ¼ r E þ q u where q 1, q, and q are the three position vectors of the observed body, with respect to the topocentric-equatorial reference system X t Y t Z t, at, respectively, t 1, t, and t. Likewise, q 1, q, and q are the magnitudes of the vectors q 1, q, and q, and u 1 = q 1 /q 1, u = q /q, and u = q /q are the three unit vectors along, respectively, q 1, q, and q. The three unit vectors u 1, u, and u are determined by measuring the declination d and the right ascension a of the observed body at, respectively, t 1, t, and t, as follows u 1 ¼ðcos d 1 cos a 1 Þ u X þ ðcos d 1 sin a 1 Þu Y þ ðsin d 1 Þu Z u ¼ðcos d cos a Þ u X þ ðcos d sin a Þu Y þ ðsin d Þu Z u ¼ðcos d cos a Þ u X þ ðcos d sin a Þu Y þ ðsin d Þu Z where u X, u Y, and u Z are the three unit vectors along, respectively, X, Y, and Z. The three vector equations written above r 1 ¼ r E1 þ q 1 ¼ r E1 þ q 1 ðq 1 =q 1 Þ ¼ r E1 þ q 1 u 1 r ¼ r E þ q ¼ r E þ q ðq =q Þ ¼ r E þ q u r ¼ r E þ q ¼ r E þ q ðq =q Þ ¼ r E þ q u provide nine scalar equations in twelve unknowns, which are the three components of each of the three position vectors r 1, r, and r plus the three magnitudes q 1, q,

52 180 Orbit Determination from Observations and q ( + = 1 unknowns). Three additional scalar equations are provided by the fact that the motion of the observed body is confined in a plane, because the moment of momentum per unit mass h is a constant vector (see Sect. 1.1). Consequently, the three position vectors r 1, r, and r are coplanar. This means that one of these vectors results from a linear combination of the other two. Without loss of generality, we suppose r to be a linear combination of r 1 and r, so that r ¼ c 1 r 1 þ c r This equation, added to the 9 + = 1 scalar equations written above, introduces two new unknowns (c 1 and c ). Thus, we have 1 scalar equations and 1 + = 14 unknowns. In addition to the constancy of h, the Keplerian unperturbed motion of the observed body implies that the position vector of that body at any time can be expressed in terms of the position (r) and velocity (v) vectors at any other time by means of the Lagrangian coefficients (f and g). Thus, in the present case, the position vectors r 1 and r of the observed body at times, respectively, t 1 and t can be expressed in terms of r and v at time t, as follows r 1 ¼ f 1 r þ g 1 v r ¼ f r þ g v where f 1 and g 1 are the Lagrangian coefficients computed at time t 1 ; likewise, f and g are the Lagrangian coefficients computed at time t. As shown in Sect. 1.1, in case of small intervals of time between two consecutive observations, the Lagrangian coefficients f and g depend only on the distance existing at the initial time between the attracted body and its centre of attraction. In this case, designating the intermediate time t as the initial time and r as the distance existing at time t between the two mutually attracting bodies, the Lagrangian coefficients f 1, g 1, f, and g depend only on the distance r. The two vector equations r 1 ¼ f 1 r þ g 1 v r ¼ f r þ g v correspond to six scalar equations. We have then = 18 equations. On the other hand, the new unknowns introduced by these six equations are four: the three components of the velocity vector v and the distance r. We have then = 18 unknowns. Thus, under the hypothesis made above, there is only one solution to the problem of determining the vectors r and v at the initial time t. To this end, we first solve the equation r ¼ c 1 r 1 þ c r

53 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 181 for c 1 and c. The vector product of all terms of this equation by r yields r r ¼ c 1 ðr 1 r Þþc ðr r Þ Since the vector product of any vector by itself is the zero vector, then c (r r ) vanishes identically. Thus, r r ¼ c 1 ðr 1 r Þ The scalar product of (r 1 r ) by all terms of the preceding expression yields ðr 1 r Þðr r Þ¼c 1 ðr 1 r Þðr 1 r Þ¼c 1 jr 1 r j where jr 1 r j is the squared magnitude of r 1 r. The preceding equation, solved for c 1, yields ð c 1 ¼ r 1 r Þðr r Þ jr 1 r j By operating likewise, we form the vector product of all terms of the equation by r 1. This yields r ¼ c 1 r 1 þ c r r r 1 ¼ c 1 ðr 1 r 1 Þþc ðr r 1 Þ¼c ðr r 1 Þ The scalar product of (r r 1 ) and all terms of the preceding expression yields Hence, ðr r 1 Þðr r 1 Þ¼c ðr r 1 Þðr r 1 Þ¼c jr r 1 j ð c ¼ r r 1 Þðr r 1 Þ jr r 1 j Now, we form the vector product r 1 r and introduce r 1 ¼ f 1 r þ g 1 v r ¼ f r þ g v into the product r 1 r. This yields r 1 r ¼ ðf 1 r þ g 1 v Þðf r þ g v Þ ¼ f 1 f ðr r Þþf 1 g ðr v Þþg 1 f ðv r Þþg 1 g ðv v Þ

54 18 Orbit Determination from Observations Now, since the vector product of any vector by itself yields the zero vector, and r v ¼ h v r ¼ðr v Þ¼h where h is the constant moment of momentum, then It follows that r 1 r ¼ðf 1 g g 1 f Þh r r 1 ¼ðr 1 r Þ¼f ð 1 g g 1 f Þh jr 1 r j ¼ jr r 1 j ¼ðf 1 g g 1 f Þ h Likewise, we form the vector product r r and introduce r = f r + g v into the product r r. This yields r r ¼ r ðf r þ g v Þ ¼ f ðr r Þþg ðr v Þ¼g h Again, we form the vector product r r 1 and introduce r 1 = f 1 r + g 1 v into the product r r 1. This yields r r 1 ¼ r ðf 1 r þ g 1 v Þ ¼ f 1 ðr r Þþg 1 ðr v Þ ¼ g 1 h In summary, we have obtained the following expressions r r 1 ¼ðr 1 r Þ¼f ð 1 g g 1 f Þh jr 1 r j ¼ jr r 1 j ¼ðf 1 g g 1 f Þ h r r 1 ¼ g 1 h which in turn, substituted into yield ð c ¼ r r 1 Þðr r 1 Þ jr r 1 j ½ c ¼ ð f 1g g 1 f Þh½g 1 h ðf 1 g g 1 f Þ h

55 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 18 Since h h = h, then Let us consider now the expression g 1 c ¼ f 1 g g 1 f ð c 1 ¼ r 1 r Þðr r Þ jr 1 r j which has been derived above. Substituting into this expression yields that is, ðr 1 r Þ¼ðf 1 g g 1 f Þh r r ¼ g h jr 1 r j ¼ jr r 1 j ¼ðf 1 g g 1 f Þ h ½ð c 1 ¼ f 1g g 1 f Þh½g h ðf 1 g g 1 f Þ h g c 1 ¼ f 1 g g 1 f By so doing, the coefficients c 1 and c appearing in the equation r ¼ c 1 r 1 þ c r depend on the Lagrange coefficients f and g only. We set s 1 ¼ t 1 t s ¼ t t where s 1 and s are, by hypothesis, small intervals. Therefore, we can take the first two terms of the series expansions for the Lagrangian coefficients f and g f ¼ 1 1 0s þ 1 0k 0 s þ k 0 þ w 0 s 4 þ g ¼ s 1 6 0s þ 1 4 0k 0 s 4 þ (see Sect. 1.1). In other words, we truncate these series expansions after the first two terms, as follows

56 184 Orbit Determination from Observations f l r s 1 f 1 1 l r s g 1 s 1 1 l s 1 g s 1 l s 6 6 r This is because, by definition, e 0 = l/r. Thus, the quantity f 1 g g 1 f can be approximated as follows f 1 g g 1 f ¼ 1 1 l r s 1 s 1 6 ¼ ðs s 1 Þ 1 l 6 r þ 1 l 1 r 6 s 1 s s 1 s ¼ ðs s 1 Þ 1 6 l r r l r s s 1 1 l 6 r s s 1s þ s 1 s s 1 s s 1 ð Þ þ 1 1 l r 6 s 1 s 1 s s 1 s 1 1 l r s Again, since s 1 and s are small intervals, then the term (l /r 6 )(s 1 s s 1 s )/1 can be neglected. Thus, setting s = s s 1 yields f 1 g g 1 f s 1 l s 6 where s is the interval of time elapsed from the first of the three observations to the last. Now, substituting f 1 g g 1 f s 1 l s g s into c 1 = g /(f 1 g g 1 f ) yields s 1 l 6 r s 1 1 l s c 1 ¼ s 1 6 r ¼ l s s s 1 1 ¼ l s 6 6 r Using the well-known binomial expansion r r s s r 1 1 l 6 r l r s s 1 1 l 6 r s ða þ bþ n ¼ a n þ na n1 b þ nn ð 1Þ a n b þ nn ð 1Þðn Þ a n b þ!! 1

57 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 185 with a ¼ 1; b ¼ 1 6 ðl=r Þs and n ¼1; truncated after the term containing the second power of s, yields 1 1 l 1 s 1 1 þð1þ l s ¼ 1 þ 1 l s yields r This in turn substituted into c 1 s s c 1 s s 1 1 l 6 r 1 1 l 6 s r s 1 þ 1 l 6 By operating likewise, there results c s 1 s r r 1 1 l 6 s ¼ s s r s 1 þ 1 l s s 1 6 r 1 r 1 þ 1 l s s 6 We have hitherto obtained approximate expressions of the coefficients (c 1 and c ) appearing in the equation r = c 1 r 1 + c r. These expressions depend only on the (known) time intervals between observations and the (as yet unknown) distance r of the attracted body from its centre of attraction at time t. The next stage of this development is the expression of the ranges q 1, q, and q as functions of c 1 and c. To this effect, we substitute r r 1 ¼ r E1 þ q 1 r 1 r ¼ r E þ q u r ¼ r E þ q u into r = c 1 r 1 + c r. This yields r E þ q u ¼ c 1 ðr E þ q 1 u 1 Þþc ðr E þ q u Þ which can also be written as follows c 1 q 1 u 1 q u þ c q u ¼c 1 r E1 þ r E c r E The scalar product of each term of the preceding expression by (u u ) yields

58 186 Orbit Determination from Observations c 1 q 1 u 1 ðu u Þq u ðu u Þþc q u ðu u Þ ¼c 1 r E1 ðu u Þþr E ðu u Þc r E ðu u Þ Since u (u u )=u (u u ) = 0, then the preceding expression becomes c 1 q 1 u 1 ðu u Þ¼c 1 r E1 ðu u Þþr E ðu u Þc r E ðu u Þ To simplify the notation, we set D 0 ¼ u 1 ðu u Þ D 11 ¼ r E1 ðu u Þ D 1 ¼ r E ðu u Þ D 1 ¼ r E ðu u Þ Assuming D 0 6¼ 0, that is, assuming that u 1, u, and u are not coplanar, we have c 1 q 1 D 0 ¼c 1 D 11 þ D 1 c D 1 which, solved for q 1, yields q 1 ¼ D 11 þ 1 D 1 c 1 D 1 c 1 c 1 D 0 By operating likewise, we take the scalar product of each term of by (u 1 u ). This yields c 1 q 1 u 1 q u þ c q u ¼c 1 r E1 þ r E c r E c 1 q 1 u 1 ðu 1 u Þq u ðu 1 u Þþc q u ðu 1 u Þ ¼c 1 r E1 ðu 1 u Þþr E ðu 1 u Þc r E ðu 1 u Þ Since u 1 (u 1 u )=u (u 1 u ) = 0, then the preceding expression becomes q u ðu 1 u Þ¼c 1 r E1 ðu 1 u Þþr E ðu 1 u Þc r E ðu 1 u Þ Since q u (u 1 u )=q u (u u 1 )=q u 1 (u u )=q D 0, then the preceding expression becomes q D 0 ¼c 1 r E1 ðu 1 u Þþr E ðu 1 u Þc r E ðu 1 u Þ

59 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 187 By setting D 1 ¼ r E1 ðu 1 u Þ D ¼ r E ðu 1 u Þ D ¼ r E ðu 1 u Þ and solving the preceding expression for q, we obtain q ¼ ðc 1 D 1 þ D c D Þ 1 D 0 By operating likewise, we take the scalar product of each term of by (u 1 u ). This yields c 1 q 1 u 1 q u þ c q u ¼c 1 r E1 þ r E c r E c 1 q 1 u 1 ðu 1 u Þq u ðu 1 u Þþc q u ðu 1 u Þ ¼c 1 r E1 ðu 1 u Þþr E ðu 1 u Þc r E ðu 1 u Þ Since u 1 (u 1 u )=u (u 1 u ) = 0, then the preceding expression becomes c q u ðu 1 u Þ¼c 1 r E ðu 1 u Þþr E ðu 1 u Þc r E ðu 1 u Þ By noting that u (u 1 u ) = u 1 (u u ) = D 0, setting D 1 ¼ r E1 ðu 1 u Þ D ¼ r E ðu 1 u Þ D ¼ r E ðu 1 u Þ and solving the preceding expression for q, we obtain q ¼ c 1 D 1 þ 1 1 D D c c D 0 Now, substituting c 1 s s 1 þ 1 l 6 r ðs s Þ c s 1 s 1 þ 1 l 6 r ðs s 1 Þ

60 188 Orbit Determination from Observations into q =( c 1 D 1 + D c D )/D 0 yields l 6 þ r q ¼D 1 s 6D 0 s which, after setting becomes into ðs s Þ þ D 6 þ l r þ D s 1 D 0 s A ¼ D 1 s þ D s 1 1 þ D s D 0 B ¼ D 1s s s þ D s 1 s s 1 6D 0 s q ¼ A þ lb r Operating the same substitution, that is, c 1 s s 1 þ 1 l 6 r ðs s Þ q 1 ¼ D 11 þ 1 c 1 D 1 c c 1 D 1 0! c s 1 s 1 þ 1 l 6 r ðs s 1 Þ 6D 0 s ðs s 1 Þ! 1 q D ¼ c 1 D 1 þ 1 D D 0 c c 1 D 0 leads to s 6ðD 1 s 1 s þ D 1 Þr s þ ld 1ðs s 1 Þ s 1 6 q 1 ¼ s 7 4 6r þ lðs s Þ D D 0 s 6ðD s 1 s 1 D Þr s þ ld 1ðs s Þ s 6 q ¼ 1 s r þ lðs s Þ D 5 1 D 0 The equation written above q = A + lb/r expresses the range q as a function of the radius vector r of the observed object measured from the centre of mass of the Earth at time t. Another relation between q and r is provided by r ¼ r E þ q u

61 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 189 The scalar product of r by itself yields r r ¼ ðr E þ q u Þðr E þ q u Þ ¼ r E þ q ðr E u Þþq ¼ r E þ Eq þ q where E = r E u. Substituting q = A + lb/r into r = r E r ¼ r E þ E lb Aþ r þ A þ lb r ¼ re leb þ EA þ r þ A þ lab r þ l B r 6 Multiplying all terms of the preceding expression by r 6 yields r 8 r E þ EA þ A r 6 lbðeþ AÞr l B ¼ 0 If we set for convenience x ¼ r a ¼ re þ EA þ A b ¼lB ðe þ AÞ c ¼l B the preceding expression becomes x 8 þ ax 6 þ bx þ c ¼ 0 +Eq + q yields which is known as Lagrange s equation. Since this polynomial has four terms, then (according to the Descartes rule of signs, which states that the number of positive roots of a polynomial with real coefficients is equal to the number of changes of sign in the sequence of the coefficients of the polynomial, or is less than this number by a multiple of ) the polynomial may have no more than three positive roots. When the correct root r of this equation has been found, it must be substituted into s 1 s 6 D 1 þ D 1 r s q 1 ¼ s þ ld s 1 s s 1 1 s 6 4 6r þ lðs s Þ D q ¼ A þ lb r s s 6 D 1 D r s q ¼ 1 s þ ld 1ðs s Þ s 1 s r þ lðs s Þ D D 0 D 0

62 190 Orbit Determination from Observations in order to compute the ranges q 1, q, and q. Then, the following equations r 1 ¼ r E1 þ q 1 u 1 r ¼ r E þ q u r ¼ r E þ q u yield the position vectors r 1, r, and r of the observed object at times, respectively, t 1, t, and t. It remains to compute the velocity vector v of the observed object at time t. To this end, the equation r 1 = f 1 r + g 1 v is solved for r. This yields r ¼ r 1 g 1 v f The resulting value of r is substituted into r = f r + g v. This yields r ¼ f ðr 1 g 1 v Þþg v f 1 and the preceding equation is solved for v. This yields f 1 f v ¼ r r 1 f 1 g g 1 f f 1 g g 1 f At first, the values of f 1, f, g 1, and g to be used for computing r and v are those derived previously, which are rewritten below for convenience: f l r g 1 s 1 1 l 6 s 1 f 1 1 r s 1 g s 1 6 l r l s r s Successively, improved values of fsince the condition 1, f, g 1, and g are computed, as will be shown below, and new values of r and v are computed by means of these improved values, until convergence is reached. This method can be illustrated by the following example, which is based on a series of astronomical observations performed by Healy [] and concerning the COBE artificial satellite (USSPACECOM Catalogue No. 0; International Designation code A). On the 6 th of November 000, Healy made seven observations of the COBE satellite, three of which are shown in the following table. Observed time (EST) Right ascension (hh:mm:ss) Declination ( ) 17:1:9 1:48: :4:0 1:14: :7:0 11:0:

63 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 191 Since UTC = EST + 5, then (neglecting the difference between UTC and UT1) the three EST times indicated above correspond, respectively, to UT1 1 ¼ EST 1 þ 5 ¼ h 1 m 9 s UT1 ¼ EST þ 5 ¼ h 4 m 0 s UT1 ¼ EST þ 5 ¼ h 7 m 0 s The three values of the right ascension, expressed in degrees, are as follows ð1 þ 48=60Þ60=4 ¼ 7:00 ð1 þ 14=60Þ60=4 ¼ 18:50 ð11 þ =60Þ60=4 ¼ 165:75 Healy found these values by means of the 14 Schmidt Cassegrain telescope of the observatory of the University of Maryland, located at latitude u = North, longitude k = east and altitude H =5m. As has been shown above, the same values can be written as follows UT1 Right ascension ( ) Declination ( ) :1: :4: :7: By applying the methods shown in Sect..5, the Greenwich sidereal time h G0 corresponding to the 6 November 000, at 00 h :00 m :00 s UT1, results from a ¼ INTðy=100Þ ¼ INTð000=100Þ ¼ 0 b ¼ INTða=4Þ ¼ INTð0=4Þ ¼ 5 c ¼ a þ b ¼ 0 þ 5 ¼1 e ¼ INT½65:5ðy þ 4716Þ ¼ INT½65:5 ð000 þ 4716Þ ¼ f ¼ INT½0:6001ðm þ 1Þ ¼ INT½0:6001 ð11 þ 1Þ ¼ 67 J 0 ¼ c þ d þ e þ f 154:5 ¼1 þ 6 þ þ :5 ¼ :5 T 0 ¼ ðj Þ=655 ¼ ð451854: Þ=655 ¼ 0: h G0 ¼ 100: þ 6000:77005 T 0 þ 0: T 0 : T 0 ¼ 100: þ 6000: : þ 0: : : : ¼ 405 :51847

64 19 Orbit Determination from Observations This value is brought into the range 0 h G0 60 by subtracting 60, as follows h G0 ¼ 405 : ¼ 45 :51847 The Greenwich sidereal times h G1, h G, and h G corresponding to UT1 1 = :1:9, UT1 = :4:0 and UT1 = :7:0 are, respectively, h G1 ¼ h G0 þ 60: ð þ 1=60 þ 9=600Þ=4 ¼ 84 :146 h G ¼ h G0 þ 60: ð þ 4=60 þ 0=600Þ=4 ¼ 85 : h G ¼ h G0 þ 60: ð þ 7=60 þ 0=600Þ=4 ¼ 85 :8648 The local sidereal times of the three observations are, respectively, h 1 ¼ h G1 þ k ¼ 84 : :95667 ¼ 07 :5769 h ¼ h G1 þ k ¼ 85 : :95667 ¼ 08 :1195 h ¼ h G1 þ k ¼ 85 : :95667 ¼ 08 : We want to compute the position and velocity vectors of the COBE satellite at UT1 = h :4 m :0 s, with an accuracy of five significant figures. The angles measured by the station and the local sidereal times are given in the following table. Obs. No. Time (s) Right ascension ( ) Declination ( ) Local sidereal time ( ) First, we compute the three geocentric position vectors (r E1, r E, and r E ) of the observatory at the three given times t 1, t, and t, as shown in Sect..5 (that is, taking the equatorial radius a E and the flattening f of the Earth equal, respectively, to km and ). Assuming the geodetic latitude equal to the geographic latitude, the position vectors r E1, r E, and r E are n x H ¼ a E = 1 f f sin o 1= u þ H cos u ¼ f678:17=½1 ð0:0058 o 0:0058 Þ sin 9: = þ 0:05 cos 9:00167 ¼ 496:7 km n z H ¼ a E ð1 f Þ = 1 f f sin 1= o n u þ H sin u ¼ 678:17 ð1 0:0058Þ = 1 ð0:0058 0:0058 Þ sin 1= o 9:00167 þ 0:05 sin 9:00167 ¼ 99:517 km

65 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 19 r E1 ¼ ðx H cos h 1 Þu X þ ðx H sin h 1 Þu Y þ z H u Z ¼ð496:7 cos 07:5769Þ u X þð496:7 sin 07:5769Þ u Y þ 99:517 u Z ¼ 011:7 u X 945:1 u Y þ 99:5 u Z r E ¼ ðx H cos h Þu X þ ðx H sin h Þu Y þ z H u Z ¼ð496:7 cos 08:1195Þu X þð496:7 sin 08:1195Þu Y þ 99:517 u Z ¼ 06:5 u X 905:0 u Y þ 99:5 u Z r E ¼ ðx H cos h Þu X þ ðx H sin h Þu Y þ z H u Z ¼ð496:7 cos 08:865978Þu X þð496:7 sin 08:865978Þu Y þ 99:517 u Z ¼ 114:5 u X 864:5 u Y þ 99:5 u Z The three unit vectors u 1, u, and u result from u 1 ¼ðcos d 1 cos a 1 Þu X þ ðcos d 1 sin a 1 Þu Y þ ðsin d 1 Þu Z ¼½cosð16:Þ cos 7:0 u X þ ½cosð16:Þ sin 7:0u Y þ ½sin ð16:þu Z ¼ 0:80496 u X 0:575 u Y 0:8067 u Z u ¼ðcos d cos a Þ u X þ ðcos d sin a Þu Y þ ðsin d Þu Z ¼ ðcos 46:9 cos 18:5Þu X þ ðcos 46:9 sin 18:5Þu Y þ ðsin 46:9Þu Z ¼ 0:51174 u X 0:4575 u Y þ 0:7016 u Z u ¼ðcos d cos a Þ u X þ ðcos d sin a Þu Y þ ðsin d Þu Z ¼ ðcos 76:1 cos 165:75Þu X þ ðcos 76:1 sin 165:75Þu Y þ ðsin 76:1Þu Z ¼0:84 u X þ 0:0591 u Y þ 0:9707 u Z The time intervals s 1, s, and s are computed as follows s 1 ¼ t 1 t ¼ ¼181 s s ¼ t t ¼ ¼ 180 s s ¼ t t 1 ¼ 61 0 ¼ 61 s The vector products (u u ), (u 1 u ), and (u 1 u ) result from u X u Y u Z 6 u u ¼ 4 0: : : :84 0:0591 0:9707 ¼0:4867 u X 0:66677 u Y 0: u Z

66 194 Orbit Determination from Observations u X u Y u Z 6 u 1 u ¼ 4 0: : : :84 0:0591 0:9707 ¼0:49085 u X 0:71604 u Y 0: u Z u X u Y u Z 6 u 1 u ¼ 4 0: : : : :4575 0:7016 ¼0:50876 u X 0:718 u Y 0:09694 u Z The scalar products D 0, D 11, D 1, D 1, D 1, D, D, D 1, D, and D are D 0 ¼ u 1 ðu u Þ¼0:80496 ð0:4867þ0:5754 ð0:66677þ 0:8067 ð0:075158þ ¼0: D 11 ¼ r E1 ðu u Þ¼011:7 ð0:4867þ945:1 ð0:66677þ þ 99:5 ð0:075158þ ¼876:75 km D 1 ¼ r E1 ðu 1 u Þ¼011:7 ð0:49085þ945:1 ð0:71604þ þ 99:5 ð0:074117þ ¼ 1050:6 km D 1 ¼ r E1 ðu 1 u Þ¼011:7 ð0:50876þ945:1 ð0:718þ þ 99:5 ð0:09694þ ¼ 966:1 km D 1 ¼ r E ðu u Þ¼06:5 ð0:4867þ905:0 ð0:66677þ þ 99:5 ð0:075158þ ¼85:01 km D ¼ r E ðu 1 u Þ¼06:5 ð0:49085þ905:0 ð0:71604þ þ 99:5 ð0:074117þ ¼ 996:51 km D ¼ r E ðu 1 u Þ¼06:5 ð0:50876þ905:0 ð0:718þ þ 99:5 ð0:09694þ ¼ 910:44 km D 1 ¼ r E ðu u Þ¼114:5 ð0:4867þ864:5 ð0:66677þ þ 99:5 ð0:075158þ ¼77:9 km D ¼ r E ðu 1 u Þ¼114:5 ð0:49085þ864:5 ð0:71604þ þ 99:5 ð0:074117þ ¼ 94:47 km D ¼ r E ðu 1 u Þ¼114:5 ð0:50876þ864:5 ð0:718þ þ 4069:057 ð0:09694þ ¼ 854:88 km

67 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 195 The quantities A and B are computed as follows s =s ¼ 180=61 ¼ 0:49861 s 1 =s ¼181=61 ¼0:5019 s s s ¼ ¼ 1: s 1 s s 1 ¼ ¼1: A ¼ ðd 1 s =s þ D þ D s 1 =sþ=d 0 ¼ ½1050:6 0:49861 þ 996:51 þ 94:47 ð0:5019þ=ð0:018881þ ¼6:66 km B ¼ D 1 s s s þ D s 1 s s 1 = ½ 6D0 s ¼ 1050:6 1: þ 94:47 ð1: Þ = ½6 ð0:018881þ61 ¼8: km s The quantities E and r E result from E ¼ r E u ¼ 06:5 0: :0 ð0:4575þ þ 99:5 0:7016 ¼ 650:9km r E ¼ r E r E ¼ 06:5 þ 905:0 þ 99:5 ¼ 4: km The coefficients a, b, and c of the polynomial x 8 + ax 6 + bx + c result from a ¼ re h i þ EA þ A ¼ 4: þ 650:9 ð6:66þ þð6:66þ ¼4: km b ¼lBðEþAÞ ¼ 98600:4 8: ð650:9 6:66Þ ¼4: km 5 c ¼l B ¼98;600:4 ð Þ 8: ¼0: km 8 We search a value of x such that the following function f ðxþ ¼x 8 þ ax 6 þ bx þ c should be equal to zero in a given interval x min x x max. This search is limited to the values of x (where x r ) which are physically meaningful. Consequently, x cannot be negative, or smaller than the mean radius of the Earth (671 km). In addition, since the values computed above of the three coefficients a, b, and c are, all of them, negative, then the equation f(x) =x 8 + ax 6 + bx + c = 0 has only one positive root. To search this root, we first evaluate f(x) for x ranging from 6400 km

68 196 Orbit Determination from Observations to 9400 km, with a step size of 100 km. The results of this evaluation are shown below. x 10 (km) f(x) 10 0 x 10 (km) f(x) The plot shows that the physically meaningful root of f(x) = 0 is roughly the midpoint of the interval 700 x 700 km. Therefore, the search of the root of interest is confined to this interval. As shown in the preceding table, we find f ð700þ ¼ : : : ¼0: ð\0þ f ð700þ ¼ : : : ¼ 0: ð [ 0Þ

69 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 197 Since the condition f(700)f(700) < 0 is satisfied, we search a zero of f(x) by means of Müller s method of parabolic interpolation, which has been described in Chap. 1, Sects. 5 and 8. At the midpoint x 0 = 750 km of the interval we also find and f ð750þ ¼ : : Then we set and compute the coefficients 0: ¼ 0: x ¼ 700 km f fðx Þ ¼ 0: x 0 ¼ 750 km f 0 fðx 0 Þ ¼ 0: x 1 ¼ 700 km f 1 fðx 1 Þ ¼ 0: h 1 ¼ x 1 x 0 ¼ ¼ 50 km h ¼ x 0 x ¼ ¼ 50 km c ¼ h =h 1 ¼ 50=50 ¼ 1 A ¼ ½cf 1 f 0 ð1 þ cþ þf = ch 1 ð1 þ cþ ¼½10: : ð1 þ 1Þ0: = ½1 50 ð1 þ 1Þ ¼ 0: B ¼ f 1 f 0 Ah 1 =h1 ¼½0: : : =50 ¼ 0: C ¼ f 0 ¼ 0: of the interpolating parabola f(x) =A(x x 0 ) + B(x x 0 )+C. We compute the estimated root of f(x) = 0, as follows C x ¼ x 0 B ðb 4 ACÞ 1 where in the present case the plus sign in front of the square root takes effect, because the value of B is greater than zero. Thus, n x ¼ 750 0: = 0: þ½ð0: Þ 4 0: : =o ¼ 748:4km

70 198 Orbit Determination from Observations This value, substituted into f(x) =x 8 + ax 6 + bx + c, yields f ð748:4þ ¼ 748:4 8 4: :4 6 4: :4 0: ¼ 0: Now, since is less than 750, then we take 700, 750 and for the next step. At the same time, we reset the subscripts 0, 1 and, as follows x ¼ 700:0km f fðx Þ ¼ 0: x 0 ¼ 748:4km f 0 fðx 0 Þ ¼ 0: x 1 ¼ 750:0km f 1 fðx 1 Þ ¼ 0: Now we compute again h 1 ¼ x 1 x 0 ¼ 750:0 748:4 ¼ 1:6km h ¼ x 0 x ¼ 748:4 700:0 ¼ 48:4km c ¼ h =h 1 ¼ 48:4=1:6 ¼ 0:5 and the coefficients A, B, and C of the interpolating parabola, as follows A ¼ ½cf 1 f 0 ð1 þ cþ þf = ch 1 ð1 þ cþ ¼½0:5 0: : ð1þ 0:5Þ0: = ½0:5 1:6 ð1 þ 0:5Þ ¼ 0: B ¼ f 1 f 0 Ah 1 =h1 ¼ð0: : : :6 Þ=1:6 ¼ 0: C ¼ f 0 ¼ 0: Thus, the estimated root is n x ¼ 748:4 0: = 0: þ½ð0: Þ 4 0: : =o ¼ 748:4km Since this value is the same as that computed in the preceding step, then we take km as the correct root, within the chosen limits of accuracy, of f ðxþ ¼x 8 þ ax 6 þ bx þ c ¼ 0 This means that we take x r = km in this preliminary approximation.

71 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 199 Now we compute the ranges q 1, q, and q, as follows s 1 s 6 D 1 þ D 1 r s q 1 ¼ s þ ld 1ðs s 1 Þ s 1 s 6 4 6r þ lðs s Þ D q ¼ A þ lb r s s 6 D 1 D r s q ¼ 1 s þ ld 1ðs s Þ s 1 s r þ lðs s Þ D D 0 D 0 In the present case, there results s 1 =s ¼181=180 ¼1:0056 s=s ¼ 61=180 ¼ :0056 s=s 1 ¼ 61= ð181þ ¼ 1:9945 s s 1 s1 =s ¼ ½61 ð181 Þð181=180Þ ¼ 9810 s s s =s 1 ¼ ð Þ180= ð181þ ¼ 9780 s s ¼ ¼ 9791 and consequently q 1 ¼ ½6 ð77:9 ð1:0056þþ 85:01 :0056Þ748:4 þ 98600:4 77:9 ð9810þ=½6 748:4 þ 98600: :75 =ð0:018881þ ¼ 1460:km q ¼6:66 þ 98600:4 8: =748:4 ¼ 89: km q ¼ ½6 ð966:1= ð1:0056þ 910:44 ð1:9945þþ 748:4 þ 98600:4 966:1 ð9780þ=½6 748:4 þ 98600: :88 =ð0:018881þ ¼ 160:4km We compute r 1, r, and r as follows r 1 ¼ r E1 þ q 1 u 1 ¼ 011:7 u X 945:1 u Y þ 99:5 u Z þ 1460: ð0:80496 u X 0:575 u Y 0:8067 u Z Þ ¼ 4187: u X 4708:5 u Y þ 58:6 u Z r ¼ r E þ q u ¼ 06:5 u X 905:0 u Y þ 99:5 u Z þ 89: ð0:51174 u X 0:4575 u Y þ 0:7016 u Z Þ ¼ 50:6 u X 409:4 u Y þ 4644:7 u Z

72 00 Orbit Determination from Observations r ¼ r E þ q u ¼ 114:5 u X 864:5 u Y þ 99:5 u Z þ 160:4 ð0:84 u X þ 0:0591 u Y þ 0:9707 u Z Þ ¼ 741:4 u X 769:7 u Y þ 5548:0 u Z The Lagrangian coefficients f 1, f, g 1, and g result from f l=r s 1 ¼ 1 1 ð98600:4=748:4 Þð181Þ ¼ 0:9885 f 1 1 l=r s ¼ 1 1 ð98600:4=748:4 Þ180 ¼ 0:9804 g 1 s l=r s 1 ¼ :4 ð181=748:4þ ¼179:97 g s 1 6 l=r s ¼ :4 ð180=748:4þ ¼ 178:98 Now, in order to compute v, we evaluate f 1 = ðf 1 g g 1 f Þ ¼ 0:9885=½0: :98 ð179:97þ0:9804 ¼: f =ðf 1 g g 1 f Þ¼0:9804=½0: :98 ð179:97þ0:9804 ¼: v ¼ ½f 1 = ðf 1 g g 1 f Þr ½f = ðf 1 g g 1 f Þr 1 ¼ 0: ð741:4 u X 769:7 u Y þ 5548:0 u Z Þ0:00786 ð4187: u X 4708:5 u Y þ 58:6 u Z Þ ¼4:099 u X þ :6179 u Y þ 5:477 u Z In summary, the preliminary approximation to the position and velocity vectors of the COBE satellite, observed at time t,is r ¼ 50:6 u X 409:4 u Y þ 4644:7 u Z v ¼4:099 u X þ :6179 u Y þ 5:477 u Z This completes the first part of the computation. The second part is meant to compute more accurate values of the vectors r and v than those computed in the first part, as will be shown below. First iteration. We compute the magnitudes r and v of the vectors r and v, as follows r ¼ ðr r h Þ 1 ¼ 50:6 þð409:4þ i1 þ 4644:7 ¼ 748:4km h v ¼ ðv v Þ 1 ¼ ð4:099þ i1 þ :6179 þ 5:477 ¼ 5:044 1 km=s

73 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 01 Now, we compute a, the inverse of the major semi-axis a of the trajectory of the observed object, by means of the vis-viva integral v /l =/r 1/a, as follows a 1 a ¼ v r l ¼ 5: : :4 ¼ 1: km 1 We compute the radial component of v, as follows v r ¼ r r ¼ ð4:099 Þ50:6 þ :6179 ð409:4þþ5: :7 r 748:4 ¼0: km/s Then we write Kepler s equation in universal variables (see Sect. 1.7) at times t 1 and t respectively, as follows l 1 ð t1 t Þ r v r l 1 l 1 ð t t Þ r v r l 1 v 1 C av 1 v C av þ ð 1 ar Þv 1 S av 1 þ r v 1 þ ð 1 ar Þv S av þ r v where C and S are the Stumpff functions (see Sect. 1.7), which are defined as follows Cz ðþ¼ Sz ðþ¼ 8 >< 1 cos z 1= =z for z [ 0 1= for z ¼ 0 >: cosh z 1= 1 = ðzþ for z\ 0 8 z 1= sin z 1= =z = >< for z [ 0 1=6 for z ¼ 0 >: sinh z 1= z 1= = ðzþ = for z\ 0 z = av, and v 1 and v are the universal variables to be determined. Since t 1 t ¼ s 1 ¼181 t t ¼ s ¼ 180 then Kepler s equations at times t 1 and t become, respectively, h i 98600:4 1= ð181þ ¼ 748:4 ð0:006949þ=98600:4 1= v 1 Cð1: v 1 Þ þð1 1: :4Þ v 1 Sð1: v 1 Þ þ 748:4 v 1

74 0 Orbit Determination from Observations 98600:4 1= 180 ¼½748:4 ð0:006949þ=98600:4 1= v Cð1: v Þ þð1 1: :4Þv Sð1: v Þ þ 748:4 v The preceding equations, after simplification, become, respectively, ¼0: v 1 Cð1: v 1 Þ0:0544 v 1 Sð1: v 1 Þ þ 748:4 vv ¼0: v Cð1: v Þ0:0544 v Sð1: v 1 Þ þ 748:4v We solve iteratively the two equations written above. To this end, an initial estimate of v 1 and v is provided by the following formula suggested by Chobotov [18]: As to the first equation, there results l 1 jajdt l 1 1 jajdt ¼ 98600:4 1: ð181þ ¼ 16:4 km 1 Therefore, an estimate of the unknown value of v 1 is taken tentatively in the interval 18.0 v around We ascertain whether the following function f ðvþ ¼11474 þ 0: v Cð1: v Þþ0:0544 v Sð1: v Þ748:4 v has values of opposite signs at the endpoints of this interval. We find f ð18:0þ ¼11474 þ 0: ð18:0þ C½1: ð18:0þ þ 0:0544 ð18:0þ S½1: ð18:0þ 748:4 ð18:0þ ¼16176 ð[ 0Þ f ð14:0þ ¼11474 þ 0: ð14:0þ C½1: ð14:0þ þ 0:0544 ð14:0þ S½1: ð14:0þ 748:4 ð14:0þ ¼1805 ð\0þ

75 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 0 Since the condition f( 18.0)f( 14.0) < 0 is satisfied, we search a zero of f(v) by means of Müller s method of parabolic interpolation, which has been described in Sects. 1.5 and 1.8. Consequently, we choose arbitrarily a value v 0 falling between the endpoints 18.0 and By choosing v 0 = 16.0, the corresponding f(v 0 )is f ð16:0þ ¼11474 þ 0: ð16:0þ C½1: ð16:0þ þ 0:0544 ð16:0þ S½1: ð16:0þ 748:4 ð16:0þ ¼ 1686:4 Then we set v ¼18:0km 1= v 0 ¼16:0km 1= v 1 ¼14:0km 1= f f ðv Þ¼16176 f 0 f ðv 0 Þ¼1686:4 f 1 f ðv 1 Þ¼1805 It is to be noted that here v and v 1 are the endpoints of the current interval of search for the unknown v and that they have nothing to do with the number of observations. Now we set and compute the coefficients h 1 ¼ v 1 v 0 ¼14:0 ð16:0þ ¼:0km 1= h ¼ v 0 v ¼16:0 ð18:0þ ¼:0km 1= c ¼ h =h 1 ¼ :0=:0 ¼ 1:0 A ¼ ½cf 1 f 0 ð1 þ cþ þf = ch 1 ð1 þ cþ ¼½1:0ð1805Þ 1686:4 ð1 þ 1:0Þ þ16176 = ½1:0 :0 ð1 þ 1:0Þ ¼ 0:5 B ¼ f 1 f 0 Ah 1 =h1 ¼½ :4 ð0:5þ:0 =:0 ¼745: C ¼ f 0 ¼ 1686:4 of the interpolating parabola f(v) =A(v v 0 ) + B(v v 0 )+C. This done, we compute the estimated root of f(v) = 0 as follows C v ¼ v 0 B ðb 4ACÞ 1 where in the present case the minus sign in front of the square root takes effect, because the value of B is less than zero. Thus, there results v ¼16:0 1686:4=½745: ð745: þ 4 0:5 1686:4Þ 1= ¼15:767 km 1=

76 04 Orbit Determination from Observations This value, substituted into f(v), yields f ð15:767þ ¼11474 þ 0: ð15:767þ C ½1: ð15:767þ þ 0:0544 ð15:767þ S ½1: ð15:767þ 748:4 ð15:767þ ¼1:71 Since is greater than 16.0, we take 16.0, 14.0, and for the next step; at the same time, we reset the subscripts as follows Now we set and compute the coefficients v ¼16:0km 1= f f ðv Þ ¼ 1686:4 v 0 ¼15:767 km 1= f 0 f ðv 0 Þ ¼ 1:71 v 1 ¼14:0km 1= f 1 f ðv 1 Þ ¼ 1805 h 1 ¼ v 1 v 0 ¼14:0 ð15:767þ ¼1:767 km 1= h ¼ v 0 v ¼15:767 ð16:0þ ¼0: km 1= c ¼ h =h 1 ¼ 0:=1:767 ¼ 0:1186 A ¼ ½cf 1 f 0 ð1 þ cþ þf = ch 1 ð1 þ cþ ¼½0:1186 ð1805þ ð1:71þ ð1 þ 0:1186Þ þ 1686:4 = ½0:1186 1:767 ð1 þ 0:1186Þ ¼ 0:19 B ¼ f 1 f 0 Ah 1 =h1 ¼½1805 ð1:71þ ð0:19þ1:767 =1:767 ¼745:4 C ¼ f 0 ¼1:71 of the interpolating parabola. The estimated root v of f(v) = 0 results from h i v ¼ v 0 C= B B 1= 4AC ¼15:767 ð1:71þ=½745:4 ð745:4 4 0:19 1:71Þ 1= ¼15:767 km 1= Since this value is the same as that computed previously, within the chosen accuracy, we take it as correct. In other words, v 1 = km ½ is taken as the root of Kepler s equation ¼0:079504v 1 Cð1: v 1 Þ0:0544 v 1 Sð1: v 1 Þ þ 748:4v 1

77 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 05 By using the same iterative method, we search now the root v of the second equation, that is, 1164 ¼0:079504v Cð1: v Þ0:0544v Sð1: v Þ þ 748:4v To this end, it is necessary to find an interval of v where the following function f ðvþ ¼1164 þ 0:079504v Cð1: v Þþ0:0544v Sð1: v Þ 748:4v changes sign. We try 14.0 v 18.0 and find the following values at the endpoints: f ð14:0þ ¼ 1164 þ 0: :0 Cð1: :0 Þþ0: :0 Sð1: :0 Þ748:4 14:0 ¼ 1189 ð[ 0Þ f ð18:0þ ¼ 1164 þ 0: :0 Cð1: :0 Þþ0: :0 Sð1: :0 Þ748:4 18:0 ¼16781 ð\0þ Since the condition f(14.0)f(18.0) < 0 is satisfied, we choose arbitrarily a value v 0 falling between these endpoints. Choosing v 0 = 16.0, the corresponding f(v 0 )is f ð16:0þ ¼ 1164 þ 0: :0 Cð1: :0 Þþ0: :0 Sð1: :0 Þ748:4 16:0 ¼97:1 Now we set v ¼ 14:0km 1= v 0 ¼ 16:0km 1= v 1 ¼ 18:0km 1= f f ðv Þ¼1189 f 0 f ðv 0 Þ¼97:1 f 1 f ðv 1 Þ¼16781 Again, the subscripts used here refer only to the iterations. We also set h 1 ¼ v 1 v 0 ¼ 18:0 16:0 ¼ :0km 1= h ¼ v 0 v ¼ 16:0 14:0 ¼ :0km 1= c ¼ h =h 1 ¼ :0=:0 ¼ 1

78 06 Orbit Determination from Observations and compute the coefficients A ¼ ½cf 1 f 0 ð1 þ cþ þf = ch 1 þ cþ ¼½1ð16781Þ ð97:1þ ð1 þ 1Þ þ1189 ½1 = :0 ð1 þ 1Þ ¼ 0:75 B ¼ f 1 f 0 Ah 1 =h1 ¼½16781 ð97:1þ0:75 :0 =:0 ¼74:5 C ¼ f 0 ¼97:1 of the interpolating parabola f(v) =A(v v 0 ) + B(v v 0 )+C. Now, we compute the estimated root of f(v) = 0 as follows h v ¼ v 0 C= B B i 1= 4AC ¼ 16:0 ð97:1þ=½74:5 ð74:5 þ 4 0:75 97:1Þ 1= ¼15:68 This value, substituted into f(v), yields f ð15:68þ ¼ 1164 þ 0: :68 Cð1: :68 Þþ0: :68 Sð1: :68 Þ748:4 15:68 ¼1:1684 Since is less than 16.0, we take 14.0, 16.0 and for the next step. At the same time, the subscripts are reset as follows Now we set and compute the coefficients v ¼ 14:0km 1= f f ðv Þ ¼ 1189 v 0 ¼ 15:68 km 1= f 0 f ðv 0 Þ ¼ 1:1684 v 1 ¼ 16:0km 1= f 1 f ðv 1 Þ ¼ 97:1 h 1 ¼ v 1 v 0 ¼ 16:0 15:68 ¼ 0:17 km 1= h ¼ v 0 v ¼ 15:68 14:0 ¼ 1:68 km 1= c ¼ h =h 1 ¼ 1:68=0:17 ¼ 5:091 A ¼ ½cf 1 f 0 ð1 þ cþ þf = ch 1 ð1 þ cþ ¼½5:091 ð97:1þ ð1:1684þ ð1 þ 5:091Þ þ1189=½5:091 0:17 ð1 þ 5:091Þ ¼ 0:4895 B ¼ f 1 f 0 Ah 1 =h1 ¼½97:1 ð1:1684þ 0:4895 0:17 =0:17 ¼74:8 C ¼ f 0 ¼1:1684 of the interpolating parabola.

79 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 07 The estimated root v of f(v) = 0 results from h v ¼ v 0 C= B B i 1= 4AC ¼ 15:68 ð1:1684þ=½74:8 ð74:8 þ 4 0:4895 1:1684Þ 1= ¼15:68 km 1= Since this value is the same as that computed previously, within the chosen accuracy, we take it as correct. In other words, v = km ½ is taken as the root of Kepler s equation 1164 ¼0:079504v Cð1: v Þ0:0544v Sð1: v Þ þ 748:4v Now, v 1 = km ½ and v = km ½ are used to compute again the Lagrangian coefficients f 1, f, g 1 and g as follows h i f 1 ¼ 1 ½v 1 =r Cðav 1Þ¼1 ð15:767þ =748:4 C ½1: ð15:767þ ¼ 0:9890 f ¼ 1 ½v =r Cðav Þ¼1 15:68 =748:4 Cð1: :68 Þ ¼ 0:9808 g 1 ¼ s 1 ½v 1 =l1= Sðav 1Þ¼181 ½15:767 ð Þ =98600:4 1= S½1: ð15:767þ ¼179:97 s g ¼ s ½v =l1= Sðav Þ¼180 ð15:68 =98600:4 1= Þ Sð1: :68 Þ¼178:98 s These values of the Lagrangian coefficients are to be compared with those resulting from the preliminary approximation. The two sets are shown below. Preliminary-approximation values First-iteration values f 1 ¼ 0:9885 f 1 ¼ 0:9890 f ¼ 0:9804 f ¼ 0:9808 g 1 ¼179:97 g 1 ¼179:97 g ¼ 178:98 g ¼ 178:98 In order to improve the convergence of the process, Curtis [0] suggests to replace the first-iteration set of values by an arithmetic average of the two sets. This leads to the following values f 1 ¼ ð0:9885 þ 0:9890Þ= ¼ 0:9888 f ¼ ð0:9804 þ 0:9808Þ= ¼ 0:9806 g 1 ¼179:97 g ¼ 178:98

80 08 Orbit Determination from Observations which are used to compute again c 1 and c by means of g c 1 ¼ f 1 g g 1 f g 1 c ¼ f 1 g g 1 f Thus, c 1 ¼ 178:98=½0: :98 ð179:97þ0:9806 ¼0:5076 c ¼179:97 ð Þ=½0: :98 ð179:97þ0:9806 ¼0:51007 These values of c 1 and c are used to compute new values of q 1, q, and q,as follows q 1 ¼ ðd 11 þ D 1 =c 1 c D 1 =c 1 Þ=D 0 ¼ð876:75 þ 85:01=0:5076 0: :9=0:5076Þ= ð0:018881þ ¼ 1484:0km q ¼ ðc 1 D 1 þ D c D Þ=D 0 ¼ð0: :6 þ 996:51 0: :47Þ= ð0:018881þ ¼ 907:95 km q ¼ ðc 1 D 1 =c þ D =c D Þ=D 0 ¼ð0: :1=0:51007 þ 910:44=0: :88Þ= ð0:018881þ ¼ 168:9km These values of q 1, q, and q are used to compute new values of r 1, r, and r, as follows r 1 ¼ r E1 þ q 1 u 1 ¼ 011:7 u X 945:1 u Y þ 99:5 u Z þ 1484:0 ð0:80496 u X 0:575 u Y 0:8067 u Z Þ¼ 406: u X 470:9 u Y þ 576:0 u Z r ¼ r E þ q u ¼ 06:5 u X 905:0 u Y þ 99:5 u Z þ 907:95 ð0:51174 u X 0:4575 u Y þ 0:7016 u Z Þ¼ 58:1 u X 416:1 u Y þ 4655:4 u Z r ¼ r E þ q u ¼ 114:5 u X 864:5 u Y þ 99:5 u Z þ 168:9 ð0:84 u X þ 0:0591 u Y þ 0:9707 u Z Þ¼ 75: u X 768: u Y þ 557:7 u Z The values of r and r obtained above are used to compute a new value of v,as follows f 1 f v ¼ r r 1 f 1 g g 1 f f 1 g g 1 f

81 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 09 Since f 1 = ðf 1 g g 1 f Þ ¼ 0:9888=½0: :98 ð179:97þ0:9806 ¼0: f = ðf 1 g g 1 f Þ ¼ 0:9806=½0: :98 ð179:97þ0:9806 ¼0:00786 then v ¼ ½f 1 = ðf 1 g g 1 f Þr ½f = ðf 1 g g 1 f Þr 1 ¼ 0: ð75: u X 768: u Y þ 557:7 u Z Þ0:00786 ð406: u X 470:9 u Y þ 576:0 u Z Þ ¼4:1004 u X þ :6567 u Y þ 5:566 u Z In summary, the values of the position and velocity vectors of the COBE satellite at time t, at this stage of the iterative procedure, are r ¼ 58:1 u X 416:1 u Y þ 4655:4 u Z v ¼4:1004 u X þ :6567 u Y þ 5:566 u Z The remaining part of the computation is straightforward, but long. Therefore, it is not given here. Suffice it to say that, after the fifth iteration, the values of the four Lagrangian coefficients converge to those given below: f 1 ¼ 0:9897 f ¼ 0:9815 g 1 ¼179:97 g ¼ 178:99 Consequently, the iterative process of refinement terminates after the fifth iteration. The position (r ) and velocity (v ) vectors of the COBE satellite at time t can be computed by introducing these values in the expressions shown above. The computed values of these vectors are given below. r ¼ 55:5 u X 41:7 u Y þ 4651:7 u Z v ¼4:0755 u X þ :645 u Y þ 5:54 u Z The computed values can be compared with the actual values of the position and velocity vectors of the COBE satellite, at the same time t, which are given below (from Healy []). r ¼ 58:0 u X 41:871 u Y þ 4654:98 u Z v ¼4:10 u X þ :658 u Y þ 5:564 u Z

82 10 Orbit Determination from Observations About the difference existing between computed and actual values, it is to be noted that the data used for the computation have been considered as free from errors, which cannot be true in practice; in addition, no account has been taken of the perturbations. The method shown above is due to Gauss [6] and has been refined by Gibbs [7]. Taff [66] has strongly censured this method on the grounds of the small radius of convergence of the f and g series. According to Taff, Moulton [56] has found that the radius of convergence of these series is Ts/p, where T is the orbital period (of the object observed) and s is given by n h s ¼ M0 þ i 1 ln 1 þ 1 o e ln e 1 e 1 where e is the orbital eccentricity and M 0 is the value of the mean anomaly, contained in the interval [ p, p], at the instant t = t 0 [66]. Neither Taff nor Moulton, in the cited papers, define what must be understood by instant t = t 0. It may be presumed either that t 0 is the intermediate (t ) of the three observation times (t 1, t, and t )defined above, which are such that t 1 < t < t, or that t 0 is exactly the midpoint of the interval [t 1, t ], that is, t 0 =(t 1 + t )/. On the other hand, well before Taff and Moulton, Gibbs had noted that The determination of an orbit from three complete observations by the solution of the equations which represent elliptic motion present so great difficulties in the general case, that in the first solution of the problem we must generally limit ourselves to the case in which the intervals between the observations are not very long [7]. In other words, the time intervals s 1 = t 1 t and s = t t must be small fractions of the period T of the orbiting body observed, which period is not known a priori. However, the method of Gauss shown above computes first the components of the position and velocity vectors of the observed object from a preliminary approximation and then refines iteratively the values of such components. The knowledge of the first-approximation set of values makes it possible to compute approximately the major semi-axis a = 1/(/r v /l) of the orbit and hence the orbital period T =p(a /l) ½ of the observed object. In addition, Branham [1] notes that, according some authors, Gauss method is restricted to low-eccentricity orbits, because the radius of convergence of the f and g series becomes smaller and smaller as the orbital eccentricity approaches unity. This is because the value of s in Moulton s [56] formula given above also depends on eccentricity. However, Branham himself admits that this restriction can be removed by using the f and g functions rather than series. On the same line of reasoning is Marsden, who shows that Taff s criticism of Gauss method of initial orbit determination on the grounds of the small radius of convergence of the f and g series is completely unjustified [51]. In the example given above, the preliminary approximation to the position and velocity vectors of the COBE satellite, observed at time t, has given the following results

83 .6 Orbital Elements from Three Measurements of Angles (Method of Gauss) 11 r ¼ 50:6 u X 409:4 u Y þ 4644:7 u Z v ¼4:099 u X þ :6179 u Y þ 5:477 u Z The corresponding magnitudes r and v of r and v have been found to be h r ¼ ðr r Þ 1 ¼ 50:6 þð409:4þ i1 þ 4644:7 ¼ 748:4km h v ¼ ðv v Þ 1 ¼ ð4:099þ i1 þ :6179 þ 5:477 ¼ 5:044 1 km=s The major semi-axis results from the vis-viva integral a ¼ 1 1 ¼ ¼ 7000:5km v 5:044 r l 748: 98600:4 The corresponding orbital period of the COBE satellite results from 1 ¼ T ¼ p a 7000:5 1 ¼ : :1s l 98600:4 and the angular momentum per unit mass is h ¼ Y Z 0 Z Y 0 ux þ Z X 0 X Z 0 uy þ X Y 0 Y X 0 uz ¼ð409:4 5:477 : :7Þ u X þð50:6 5:477 4: :7Þ u Y þ ð50:6 :6179 4: :4Þu Z ¼574 u X 7985 u Y 8149:9 u Z The square of the magnitude of h is h ¼ h h ¼ ð574 The semi-latus rectum results from Þ þ ð7985þ þ ð8149:9þ p ¼ h l ¼ ð574þ þð7985þ þð8149:9þ 98600:4 The orbital eccentricity results from p ¼ a 1 e ¼ 6991:6km

84 1 Orbit Determination from Observations which, solved for e, yields e ¼ 1 p 1 ¼ :6 1 ¼ 0:05656 a 7000:5 Now, since in the present case the time intervals s 1 = 181 s and s = 180 s are equal to a small fraction (about 1/) of the orbital period T = s; and the orbital eccentricity e = is much less than unity; then the iterations converge, as has been shown above..7 Orbital Elements from Three Measurements of Angles (Method of Laplace) Let the topocentric right ascension (a) and declination (d) of a satellite be given at three distinct times t 1, t, and t, that is, let a set of values for a 1, d 1, a, d, a, and d be given. Let u 1, u, and u be the unit vectors along the line-of-sight vectors (respectively, q 1 q 1 u 1, q q u, and q q u ) going from the observation site P placed on the surface of the Earth to the satellite Q observed, as shown in the following figure. Indicating for brevity u i (i = 1,, and ) these unit vectors, Sect.. has shown that the components of u i are u i ¼ðcos d i cos a i Þu X þ ðcos d i sin a i Þu Y þ ðsin d i Þu Z

85 .7 Orbital Elements from Three Measurements of Angles (Method of Laplace) 1 where u X, u Y, and u Z are the unit vectors along the respective axes of the geocentric-equatorial system XYZ. The preceding expression, in matrix terms, is u X cos d cos a u i ¼ 4 u Y 5 ¼ 4 cos d sin a 5 u Z sin d The preceding figure also shows that i i ði ¼ 1; ; Þ r i ¼ r E þ q i ¼ r E þ q i u i ði ¼ 1; ; Þ where q i is the magnitude of the line-of-sight vector q i,andr E is the vector from the centre of the Earth to the observation site. As shown in Sect..5, differentiating two times the preceding expression with respect to time yields r 0 i ¼ r0 E þ q0 i u i þ q i u 0 i r 00 i ¼ r 00 E þ q00 i u i þ q 0 i u0 i þ q0 i u0 i þ q iu 00 i ¼ r 00 E þ q00 i u i þ q 0 i u0 i þ q iu 00 i On the other hand, the equation of motion of the satellite is r 00 i ¼ l ri r i By substituting r i = (l/r i )r i into r i = r E + q i u i +q i u i + q i u i and remembering that r i = r E + q i u i, there results l ðr E þ q i u i Þ ¼ r 00 E þ q00 i u i þ q 0 i u0 i þ q iu 00 i r i that is, h q 00 u i þ q 0 i u0 i þ q i u 00 i þ l r u i i h ¼ r 00 E þ l r r E i (where i = 1,, ). At a given time of observation (e.g. at the intermediate time t ) the preceding vector equation is equivalent to scalar equations in 10 unknowns (u i, u i, q i, q i, q i, and r), where the known quantities are u i, r E, and r E. The line-of-sight vector u i is known at the three times t 1, t, and t. Consequently, when the time intervals t t 1 and t t are small with respect to the orbital period T of the object observed, the values of u and u (that is, the values of u i and u i at the intermediate time t ) can be computed by taking, respectively, the first and second time-derivative of u(t), where u(t) is the Lagrange polynomial which interpolates the three line-of-sight vectors u 1, u, and u. As is well known, this polynomial is

86 14 Orbit Determination from Observations ðt t Þðt t Þ uðtþ ¼ u 1 þ t t ð 1Þðt t Þ u þ t t ð 1Þðt t Þ ðt 1 t Þðt 1 t Þ ðt t 1 Þðt t Þ ðt t 1 Þðt t Þ Differentiating once and twice this expression yields, respectively, u 0 t t t t t 1 t t t 1 t ðtþ ¼ u 1 þ u þ ðt 1 t Þðt 1 t Þ ðt t 1 Þðt t Þ ðt t 1 Þðt t Þ u 00 ðtþ ¼ u 1 þ u þ ðt 1 t Þðt 1 t Þ ðt t 1 Þðt t Þ ðt t 1 Þðt t Þ The value of u results from evaluating u (t ). When more than three observations are available, then u and u can be computed by using either Lagrange polynomials with a higher degree than two or a least-squares polynomial fit, as will be shown below. Thus, at t = t, we have the equation h q 00 u þ q 0 u 0 þ q u 00 þ l i h r u ¼ r 00 E þ l i r r E which, projected onto the axes X, Y, and Z, gives rise to the following scalar equations q 00 u X þ q 0 u 0 X þ q½u00 X þðl=r Þ X ¼½r 00 EX þðl=r Þr EX q 00 u Y þ q 0 u 0 Y þ q½u00 Y þðl=r Þ Y ¼½r 00 EY þðl=r Þr EY q 00 u Z þ q 0 u 0 Z þ q½u00 Z þðl=r Þ Z ¼½r 00 EZ þðl=r Þr EZ for the three unknowns q, q, and q. For the present, we consider r as a parameter. The matrix of the coefficients in the preceding system of equations is u X u 0 X u 00 X þ ð l=r Þu X 4 u Y u 0 Y u 00 Y þ ð l=r Þu Y 5 u Z u 0 Z u 00 Z þ ð l=r Þu Z Let D be the determinant of the preceding matrix. The value of D does not change if the first column of the matrix, multiplied by (l/r ), is subtracted from the third column, as follows u X u 0 X u 00 X þ ð l=r Þu X u X ðl=r Þ 4 u Y u 0 Y u 00 Y þ ð l=r Þu Y u Y ðl=r Þ5 u Z u 0 Z u 00 Z þ ð l=r Þu Z u Z ðl=r Þ u u u

87 .7 Orbital Elements from Three Measurements of Angles (Method of Laplace) 15 Consequently, D is equal to u X u 0 X u 00 X 4 u Y u 0 Y u 00 5 Y ¼ D 0 u Z u 0 Z u 00 Z where D 0 is the determinant of the matrix written above. Solving the preceding system of equations according to Cramer s rule, we have q ¼ D q D ¼ D q D 0 where D q is the determinant of the following matrix u X u 0 X rex 00 þ ð l=r 4 u Y u 0 Y rey 00 þ ð l=r u Z u 0 Z rez 00 þ ð l=r This matrix can be written as follows 4 u X u 0 X r 00 EX u Y u 0 Y r 00 EY u Z u 0 Z r 00 EZ Þr EX Þr EY Þr EZ 5 5 l=r u X u 0 X r EX 4 u Y u 0 Y r EY 5 u Z u 0 Z r EZ Let D 1 and D be the determinants of the two matrices written above. Since q = D q /D, then the following equality holds q ¼ D q D ¼ D 1 D l D r D ¼D 1 l D D 0 r ¼AB l D 0 r where we have set for convenience A ¼ D 1 ¼ u u0 r 00 E D 0 D 0 B ¼ D ¼ u ð u0 r E Þ D 0 D 0 D 0 ¼ u ðu 0 u 00 Þ The equation written above, that is, q ¼AB l r holds, of course, if D 0 6¼ 0.

88 16 Orbit Determination from Observations On the other hand, by remembering that r = r E + q u (where r and u are r i and u i at the intermediate time t ) and taking the scalar product of r with itself, there results r ¼ r r ¼ðr E þ q uþ ðr E þ q uþ ¼r E þ q ðr E uþ þq By setting for convenience E ¼qðr E uþ F ¼ re there results the following system of two algebraic equations q ¼AB l r r ¼ q E q þ F for the two unknowns q and r. Now, substituting q = A B(l/r ) into r = q E q + F leads to r ¼ A þ l AB r þ l B r 6 þ AE þ l BE r 6 þ F Multiplying all terms of the preceding equation by r 6 leads to Lagrange s equation r 8 þ ar 6 þ br þ c ¼ 0 where a ¼A þ AE þ F b ¼lðAB þ BEÞ c ¼l B This equation can be solved numerically by means of one of the methods shown in Sect The result is the value of r relating to the intermediate time t.by introducing this value into q ¼AB l r it is possible to compute q. Thus, the position vector of the satellite at time t results from r ¼ r E þ q u

89 .7 Orbital Elements from Three Measurements of Angles (Method of Laplace) 17 With the view of computing the velocity vector v r of the satellite at time t, let us consider again the system of equations q 00 u X þ q 0 u 0 X þ q u00 X þðl=r Þ X ¼ r 00 EX þðl=r Þr EX q 00 u Y þ q 0 u 0 Y þ q u00 Y þðl=r Þ Y ¼ r 00 EY þðl=r Þr EY q 00 u Z þ q 0 u 0 Z þ q u00 Z þ l=r Z ¼ r 00 l=r rez EZ þ Using again Cramer s rule, we solve now for q and write q 0 ¼ D q 0 D where D 6¼ 0 is the determinant of the matrix of coefficients (see above) and D q is the determinant of the following matrix u X rex 00 þ ð l=r Þr EX u 00 X þ ð l=r Þu X u Y rey 00 þ 4 ð l=r Þr EY u 00 Y þ ð l=r Þu Y 5 u Z rez 00 þ ð l=r Þr EZ u 00 Z þ ð l=r Þu Z This matrix can also be written as follows 4 u X r 00 EX u Y r 00 EY u Z r 00 EZ u 00 X u 00 Y u 00 Z 5 l=r u X r EX u 00 X 4 u Y r EY u 00 5 Y u Z r EZ u 00 Z Let D be the determinant of the first and D 4 be the determinant of the second of the two matrices written above. By so doing, the previous expression q = D q /D can also be written as follows q 0 ¼ D q 0 D ¼D D l D4 r D where D 6¼ 0. Since we have set D =D 0, the previous equation can also be written as follows q 0 ¼ D l D4 D 0 r D 0 By setting for convenience C ¼ D D 0 ¼ u r00 E D 0 u00 G ¼ D 4 ¼ u ð r E u 00 Þ D 0 D 0

90 18 Orbit Determination from Observations the previous equation can be written as follows q 0 ¼CG l r Since r is known, the previous expression makes it possible to compute the value of q. This value, substituted into v r 0 ¼ r 0 E þ q0 u þ q u 0 yields the velocity vector v of the satellite at time t. As mentioned above, this method fails when the determinant D 0 of the matrix u X u 0 X u 00 X 4 u Y u 0 Y u 00 5 Y u Z u 0 Z u 00 Z approaches zero, which happens when the observer lies in the plane of the orbit at the intermediate time t. The following section of this paragraph shows how to fit the observations gathered to an orbit by using the method of least squares, which was invented by Gauss in According to Gauss (Theoria motus, Book, Sect., paragraph 17, page 05 of the cited Ref. [6]), Si observationes astronomicae ceterique numeri, quibus orbitarum computus innititur, absoluta praecisione gauderent, elementa quoque, sive tribus observationibus sive quatuor superstructa fuerint, absolute exacta statim prodirent (quatenus quidem motus secundum leges Kepleri exacte fieri supponitur), adeoque accitis aliis aliisque observationibus confirmari tantum possent, haud corrigi. Verum enim vero quum omnes mensurationes atque observationes nostrae nihil sint nisi approximationes ad veritatem, idemque de omnibus calculis illis innitentibus valere debent, scopum summum omnium computorum circa phaenomena concreta institutorum in eo ponere oportebit, ut ad veritatem quam proxime fieri potest accedamus. Hoc autem aliter fieri nequit, nisi per idoneam combinationem observationum plurium, quam quot ad determinationem quantitatum incognitarum absolute requiruntur. Hoc negotium tunc demum suscipere licebit, quando orbitae cognitio approximata iam innotuit, quae dein ita rectificanda est, ut omnibus observationibus quam exactissime satisfaciat. In plain English, If the astronomical observations and other numbers, on which the computation of orbit is based, were absolutely precise, the elements also, deduced by means of three or four observations, would be absolutely exact (within the limits of validity of Kepler s laws). Hence, the computed elements could only be confirmed, but never corrected, by further observations. In practice, since all our measurements and observations are nothing more than approximations to the truth, the same holds for all calculations based on them. Thus, the principal purpose of all computations concerning concrete phenomena must needs be that of drawing us near the truth to the maximum possible extent. This can be done in no other way

91 .7 Orbital Elements from Three Measurements of Angles (Method of Laplace) 19 than by using a suitable combination of more observations than would be strictly necessary to determine the unknown quantities. This task can only be undertaken, when an approximate knowledge of the orbit has already been reached, which is then to be corrected in such a way as to satisfy all the observations to the maximum extent of exactness. In a few words, since all actual observations are affected by random errors, then more observations than those which are strictly necessary are needed, the usefulness of the redundant measurements being the mutual cancellation of the errors to the maximum possible extent. With reference to the following figure, suppose we have a set of approximate values y i which correspond to discrete values x i (where i =1,,, n) of the independent variable x. If the three points (x 1, y 1 ), (x, y ), and (x, y ) were joined by an interpolating polynomial, as shown in the following figure, then the resulting curve would pass exactly through each of the three points but would also have an oscillating behaviour, which makes the curve unfit to represent the overall trend of our data. If, on the contrary, the same data were represented by a straight line obtained by means of the least-squares method, then the result would be like that shown below.

92 0 Orbit Determination from Observations The least-squares method produces a functional form f(x, c 0, c 1,, c m ) f m (x), which depends not only on x but also on m + 1 parameters c 0, c 1,, c m to be determined in such a way as to minimise the squares of the residuals, that is, the squares of the differences between the functional form f m (x) and each data point. A type of functional form f m (x) frequently used is an algebraic polynomial: f m ðþ¼c x 0 þ c 1 x þ c x þ þc m x m The degree m of this polynomial must be chosen by the solver. However, m must be less than n 1, where n is the number of the given points. Otherwise, if m were equal to n 1, then f m (x) would be just the interpolating polynomial. For the purpose of illustrating the method, suppose that we are given a set of n points (x i, y i ) with i =1,,, n, and that we search the least-squares fit of these points by means of a second-degree polynomial f (x) =c 0 + c 1 x + c x. In this case, according to the least-squares method, the unknown coefficients c 0, c 1, and c must be chosen in such a way as to correspond to the least squares of the residuals, the residuals being the components q 1, q,, q n of the following n 1 residual vector q 1 q q Such components are q n q 1 ¼ y 1 f ðx 1 Þ ¼ y 1 c 0 c 1 x 1 c x 1 q ¼ y f ðx Þ ¼ y c 0 c 1 x c x. q n ¼ y n f ðx n Þ ¼ y n c 0 c 1 x n c x n This is because the Euclidean length q of the residual vector q is just the square root of the sum of the squares of its components: q =(q 1 + q + + q n ) ½. Consequently, in order for the residual vector q to have the minimum Euclidean length, q must have the minimum value, where q ¼ q 1 þ q þ þq n ¼ y 1 c 0 c 1 x 1 c x 1 þ y c 0 c 1 x c x þ þ yn c 0 c 1 x n c x n This condition is satisfied by those values of c 0, c 1,, c m which cause the first derivative of q to vanish. If m =, this leads to the following three conditions:

93 .7 Orbital Elements from Three Measurements of Angles (Method of Laplace) 1 n ¼½ y 1 c 0 c 1 x 1 c x 0 þ y c 0 c 1 x c x þ þ yn c 0 c 1 x n c x n 1 n ¼½x 1 y 1 c 0 c 1 x 1 c x 1 þ x y c 0 c 1 x c x þ þxn y n c 0 c 1 x n c x n 1 n y 1 c 0 c 1 x 1 c x 1 þ x y c 0 c 1 x c x þ þx n y n c 0 c 1 x n c x n ¼0 Since each of the three partial derivatives must be equal to zero, then the coefficient in front of the square brackets (that is, ) can be dropped. Thus, the previous three conditions reduce to nc 0 þ ðx 1 þ x þ... þ x n Þc 1 þ x 1 þ x þ... þ x n c ¼ y 1 þ y þ... þ y n ðx 1 þ x þ... þ x n Þc 0 þ x 1 þ x þ... þ x n c1 þ x 1 þ x þ... þ x n c ¼ x 1 ðy 1 þ y þ... þ y n Þ x 1 þ x þ... þ x n c0 þ x 1 þ x þ... þ x n c1 þ x 4 1 þ x4 þ... þ x4 n c ¼ x 1 ðy 1 þ y þ... þ y n Þ This is a system of three linear algebraic equations for the three unknowns c 0, c 1, and c. The values of c 0, c 1, and c determine the least-squares fit parabola f ðþ¼c x 0 þ c 1 x þ c x Let us consider now the differentiation of a general (m-degree) least-squares fit polynomial. Let f m ðx ¼ c 0 þ c 1 x þ c x þ... þ c n x m be this polynomial. Differentiating once and twice the polynomial with respect to x yields df m dx ¼ c 1 þ c x þ c x þ d f m dx ¼ c þ 6c x þ The expressions shown above make it possible to compute f m (x) and its first and second derivative at any point x of interest. If that point be chosen as the origin of the series expansions, then there results x =0,f m (0) = c 0,(df m /dx) 0 = c 1, and (d f m /dx ) 0 =c.

94 Orbit Determination from Observations.8 Improvement in Orbit Determination by Differential Correction The preceding paragraphs have shown some methods for the preliminary determination of the orbit followed by a space object. These methods use only the minimum number of observations necessary to compute an orbit. Since six independent quantities are strictly necessary to determine the motion of a body, then these methods use a set of six independent quantities. However, as has been shown numerically in an example given in Sect..6, the six orbital elements (or the + components of the position and velocity vectors at a given time t 0 ) computed by means of these methods differ to some extent from the actual orbital elements. For the purpose of reducing, as far as possible, the influence of the errors, the residuals (that is, the differences between the computed and observed quantities) must be determined at each time of observation. In some cases, it is possible to refine a preliminary orbit without taking account of the perturbations, that is, considering a purely Keplerian orbit. As shown in Sect..6, this happens when there are at least three reliable observations which cover a significant part of the orbit but are not too distant in time with respect to the epoch of the preliminary determination of the orbital elements. The differential correction is a numerical procedure based on the principle of least squares, which is meant to correct the computed elements by minimising the residuals. The quantities which concur to determine the observed values of the right ascension (a) and declination (d) of a given body are substantially the orbital elements of that body and the position vector r E which specifies the location of the observer with respect to the geocentric-equatorial system. Of these quantities, r E can be supposed to be known accurately enough to require no improvement in accuracy make it useless to improve r E. Consequently, the uncertainties in the values of a and d can be attributed to the orbital elements, that is, to the components of the position (r) and velocity (r ) vectors of the space object. Let t 0 be an epoch chosen arbitrarily and let x 0, y 0, z 0, x 0, y 0, and z 0 be the components of such vectors at t 0. The dependency of a and d on x 0, y 0, z 0, x 0, y 0, and z 0 can be written as follows a ¼ a x 0 ; y 0 ; z 0 ; x 0 0 ; y0 0 ; z0 0 d ¼ d x 0 ; y 0 ; z 0 ; x 0 0 ; y0 0 ; z0 0 The total differentials of a and d are da dx 0 dy dz dd dx 0 dy dz and express the changes in a and d resulting from the independent changes in x 0, y 0,, z 0. Replacing the differentials by the corresponding finite differences leads to

95 .8 Improvement in Orbit Determination by Differential Correction Da Dx 0 Dy Dz Dd Dx 0 Dy Dz where Da and Dd are the residuals in, respectively, right ascension and declination, and Dx 0, Dy 0,, Dz 0 are the changes to be made in, respectively, x 0, y 0,, z 0 for the purpose of reducing such residuals to zero. Now, Da and Dd are known quantities, which result from measuring the position of the observed body on the celestial sphere. When three couples of values (a, d) are measured at three different times t 1, t, and t, that is, when we have a set of six values {a 1, d 1, a, d, a, d }, then independent equations of conditions can be written as follows Da 1 1 Dx 0 1 Dy 0 þ 0 Dz Dd 1 1 Dx 0 1 Dy 0 þ 0 Dz Da Dx 0 Dy Dz Dd Dx 0 Dy Dz Da Dx 0 Dy Dz Dd Dx 0 Dy Dz When we use more measurements than those strictly necessary, that is, when a set of n measurements {a 1, d 1, a, d,, a n, d n } is available, where n, then n independent equations can be written as follows Da 1 1 Dx 0 1 Dy 0 þ 0 Dz Dd 1 1 Dx 0 1 Dy 0 þ 0 Dz Da Dx 0 Dy Da n n Dx 0 n Dy 0 þ 0 Dz 0 0 Dz Dd n n Dx 0 n Dy 0 þ 0 Dz 0 0 0

96 4 Orbit Determination from Observations Thus, the problem reduces to obtaining suitable numerical values of the partial derivatives, because these values, introduced into the preceding equations, make it possible to obtain the values of the corrections Dx 0, Dy 0,, Dz 0 which best fit the available measurements a 1, d 1, a, d,, a n, d n. To this end, Escobal [] suggests the following procedure. Let e be any one of the six variables x 0, y 0, z 0, x 0, y 0, and z 0. Let De be some small change introduced in that variable. The partial derivatives of a i and d i (i =1,,, n) with respect to e can be approximated by means of the following n a i x 0 ;...; 0 þ D;...; z 0 0 ai x 0 ;...; 0 ;...; z 0 0 d i x 0 ;...; 0 þ D;...; z 0 0 di x 0 ;...; 0 ;...; z 0 0 D where e 0 is the value of e at the epoch chosen (t 0 ). Each of the variables is incremented in turn, while the others maintain their original values. By so doing, the two equations written above approximate the needed 6n partial derivatives. Usually an increment De equal to a few units per cent (Bate et al. [5], suggest 1 or %) suffices to provide a satisfactory approximation of the partial derivatives a i / e and d i / e (i =1,,, n). Now the values of these partial derivatives, obtained as has been shown above, are introduced into Da 1 1 Dx 0 1 Dy 0 þ 0 Dz Dd 1 1 Dx 0 1 Dy 0 þ 0 Dz Da Dx 0 Dy Da n n Dx 0 n Dy 0 þ 0 Dz 0 0 Dz Dd n n Dx 0 n Dy 0 þ 0 Dz and these equations are solved for Dx 0, Dy 0,, Dz 0. Now the variations Dx 0, Dy 0,, Dz 0 are added to the respective quantities x 0, y 0,, z 0. This makes it possible to compute a new couple of position (r) and velocity (r ) vectors x 0 þ Dx 0 x 0 r 0 4 y 0 þ Dy 0 5 r 0 0 þ Dx0 0 4 y 0 þ Dy0 5 0 z 0 þ Dz 0 z 0 0 þ Dz0 0

97 .8 Improvement in Orbit Determination by Differential Correction 5 at the same epoch (t 0 ). Hence, new values of right ascension and declination are computed and compared with the observed values. The differences between the two sets (computed quantities minus observed quantities) are the residuals. If these residuals are greater than a chosen tolerance, then the refinement process is repeated; otherwise it is stopped..9 Improvement in Orbit Determination by Weighted Least-Squares Estimation The preceding paragraph has shown how to determine the components of the position and velocity vectors of an observed object so that the residuals should be as small as possible. In practice, since different measurements have different units and degrees of reliability, then a weighting factor is applied to each residual, and consequently the quantity to be minimised is the square of the weighted residuals. In addition, as shown in Sect..1, the force model used to compute the forces acting upon the observed object at a time t can only provide approximate values of the true forces, because the physical constants are known approximately and also because the mathematical model used to compute the forces can never be exact. The mathematical formulation of this general problem will be shown below following Montenbruck and Gill [5]. Let us consider the m-dimensional column vector x, whose components x 1, x,, x m are the time-dependent components of the position (r) and velocity (r ) vectors of the observed object, and the components of two vectors p and q, which contain the free parameters characterising the force and measurement model, as follows Let r r x p 5 q x 0 ¼ fðt; xþ x 0 ¼ xðt 0 Þ be the differential equation and the initial condition which describe the evolution in time of the augmented state vector x. Let z 1 z z z n

98 6 Orbit Determination from Observations be an n-dimensional column vector, whose components z 1, z,, z n are the observations at times t 1, t,, t n. These observations can be expressed as follows (with i =1,,, n) or in vector terms z i ðt i Þ ¼ g i ðt i ; x i Þ þe i ¼ h i ðt i ; x 0 Þ þe i z ¼ hðx 0 Þþe where g i indicates the computed value of the i th observation as a function of time t i and the instantaneous state vector x i, and h i indicates the same value as a function of the state vector x 0 at the reference epoch t 0. The presence of the quantities e i is due to the difference between computed and actual values because of measurement errors, which are assumed to be randomly distributed with zero mean value. Using the least-squares method, we search the state vector x 0 lsq corresponding to the minimum value of the loss function Jðx 0 Þ¼qq q T q ¼ ½z hðx 0 Þ T ½z hðx 0 Þ that is, corresponding to the minimum value of the squared sums of the residuals q 1, q,, q n, for a given set of observations z 1, z,, z n. The expression of the loss function given above holds for observations of equal type and quality. This assumption will be removed in the following section of this paragraph. In addition, the number (n) of observations is assumed to be greater than or equal to the number (m) of unknowns. The vector-valued function h(x 0 ) appearing in the equation z = h(x 0 )+e is a nonlinear function. Since an approximate value (x 0 appr ) of the actual state vector (x 0 ) at epoch is known, the problem can be simplified by linearising the function h. Let x 0 ref be a reference state vector, which is initially set equal to x 0 appr. The residual vector q = z h(x 0 ) is expressed approximately as follows q ¼ z hx ð 0 Þ z hx 0 ref x 0 x ref 0 ¼ Dz HDx0 where Dx 0 = x 0 x 0 ref is the difference between x 0 and the reference state vector, Dz = z h(x 0 ref ) is the difference between the actual observations and those resulting from the computed reference orbit, and the Jacobian H =( h/ x 0 ) ref is the matrix of the partial derivatives of the computed values with respect to the state vector (here the subscript ref indicates that the partial derivatives are to be evaluated at x 0 = x 0 ref ) at the reference epoch t 0. Thus, by means of q ¼ Dz H D x 0 we obtain the residual vector q after applying a correction Dx 0 to the reference state vector and recomputing the observations.

99 .9 Improvement in Orbit Determination by Weighted Least-Squares Estimation 7 This makes it possible to reduce the original nonlinear least-squares problem of finding the minimum value of the loss function Jðx 0 Þ¼½zhx ð 0 Þ T ½z hx ð 0 Þ to the simpler linear least-squares problem of searching Dx 0 lsq corresponding to the minimum value of the loss function Jðx 0 Þ ¼ ½Dz H Dx 0 T ½Dz H Dx 0 If the Jacobian H has full rank m (the rank of a matrix A is the maximum number of linearly independent rows or columns of A), that is, if the columns of H are linearly independent, then the minimum value of the loss function is attained when the partial derivatives of J(x 0 ) with respect to Dx 0 vanish, that is, when n ½Dz HDx 0 T ½Dz HDx 0 ¼ 0 where the partial derivatives are to be evaluated at Dx 0 = Dx lsq 0. The partial derivatives of the scalar product q q q T q with respect to Dx 0 can be computed by means of the following a T ð bþ ¼ a þ Thus, the general solution of the linear least-squares problem may be obtained by solving the following system of m algebraic equations H T H Dx lsq 0 ¼ H T Dz The solution is 0 ¼ H T 1 H H T Dz Dx lsq where H T H is an m m symmetric matrix. In order to compute the solution, Montenbruck and Gill [5] recommend the use of standard techniques for positive definite linear systems of equations. One of such techniques is the Cholesky decomposition, which is shown below. Let A be an m m symmetric (A = A T, where A T is the transpose of A), positive definite ( Ax, x > 0 for any real vector x other than the zero vector) matrix with real entries a ij, as follows

100 8 Orbit Determination from Observations a 11 a 1... a 1m a 1 a... a m A a m1 a m... a mm Then A has a unique decomposition of the form A = LL T, where L is a lower triangular matrix with positive diagonal entries ( ii > 0), and L T is the transpose of L. The two matrices L and L T have the following entries L m1 m... mm m1 0 L T... m mm For example, we want to decompose the following symmetric matrix A into the product of two matrices L and L T such that L L T By applying the rules of matrix multiplication, we have the following six equations for the six unknowns 11, 1,, 1,, and : 11 ¼ 4 hence 11 ¼ 1 þ ¼ 10 hence ¼ 11 1 ¼ hence 1 ¼ þ ¼ 9 hence ¼ ¼6 hence 1 ¼ 1 þ þ ¼ 6 hence ¼ 1 Consequently, the Cholesky decomposition for A is A = LL T, where L L T

101 .9 Improvement in Orbit Determination by Weighted Least-Squares Estimation 9 For a matrix A, the nonzero entries of L can be computed, as a function of the entries of A, by means of the following sequence of operations: ð1þ 11 ¼ ða 11 Þ 1 ðþ 1 ¼ a 1 ðþ 1 ¼ a ð4þ ¼ a 1 1 ð5þ ¼ a 1 1 ð6þ ¼ a 1 1 In the general case, for the Cholesky decomposition of an m m symmetric, positive definite matrix A, the following procedure can be used. For i =1,,, m and j = i +1,i +,, m: ii ¼ a ii Xi1 ik k¼1!1 ji ¼ 1 ii a ji Xi1 k¼1 jk ik! Since A is symmetric and positive definite, then the expression under square root is always positive and the entries ij of L are all real. Therefore, to solve a system A x ¼ b of linear algebraic equations such that A is a square, symmetric and positive definite matrix, we put the preceding expression in the form This done, the equations LL T x ¼ b Lz ¼ b are solved for z by forward substitution; then the equations L T x ¼ z are solved for x by backward substitution. Another form of the Cholesky decomposition, which eliminates the need to compute square roots, is the following A ¼ LDL T where A is an m m symmetric, positive definite matrix with real entries a ij, D is a diagonal matrix with all positive nonzero entries d jj, and L and L T are, respectively, a unit lower triangular and a unit upper triangular matrix. It is to be noted that the matrix L* { * ij } of the decomposition A = L*L* T and the matrix L { ij } of the decomposition A = LDL T are not the same matrix.

102 0 Orbit Determination from Observations For example, for m =, we have a 11 a 1 a a 1 a a 5 ¼ d d a 1 a a d The entries d jj of D and the entries ij of L (with i > j) result from! d jj ¼ a jj Xj1 jk d kk ij ¼ 1 a ij Xj1 ik jk d kk d jj k¼1 For example, we want to decompose the same symmetric positive definite matrix as that considered previously, that is, A into the product LDL T. There results d 11 ¼ a 11 ¼ 4 1 ¼ a 1 =d 11 ¼ =4 ¼ 1= d ¼ a 1 d 11 ¼ 10 ð1=þ 4 ¼ 9 1 ¼ a 1 =d 11 ¼6=4 ¼1:5 ¼ð1=d Þða 1 1 d 11 Þ¼1=9½9 ð=þ1= 4Þ ¼1: d ¼ a 1 d 11 d ¼ 6 ð6=4þ 4 ð4=þ 9 ¼ 1 Therefore, the LDL T Cholesky decomposition of the given matrix A is ¼ k¼ : :5 1: : 5 1:5 1: The L*L* T and LDL T Cholesky decompositions of the same matrix A are related to each other as shown below A ¼ LDL T ¼ LD 1 1 TL T DL T ¼ LD 1 D 1 T ¼ LD 1 LD 1 where D ½ is also a diagonal matrix, whose nonzero entries are the square roots of the corresponding entries of D, that is, D ½ = diag(d ½ 11, d ½,, d ½ mm ). Hence, A ¼ LDL T ¼ L L T where L* =LD ½. Now, let us come back to the linear least-squares problem. Since h(x 0 ) is actually a nonlinear function, the value Dx 0 lsq computed as shown above (and consequently

103 .9 Improvement in Orbit Determination by Weighted Least-Squares Estimation 1 x lsq 0 = x ref 0 + Dx lsq 0 ) is only an approximate solution of the orbit determination problem. A better solution than this can be obtained by replacing x ref 0 with the value of x lsq 0 computed previously and performing a new iteration. Let the superscripts [i] and [i + 1] denote the value of x lsq 0 resulting from, respectively, the i th and the (i +1) th iteration. The iterative process of refinement can be expressed as follows ½i þ 1 1H h i x0 ¼ x ½i 0 þ H½iT H ½i ½iT z h x ½i 0 which formula has x 0 appr as its starting point, that is, x 0 [0] = x 0 appr, and continues till the difference x 0 [i+1] x 0 [i] is greater in magnitude than a desired tolerance. For the best convergence of the iterative process, the Jacobian H [i] must be computed at each iteration; however, Montenbruck and Gill [5] observe that H [i] can be replaced by a constant H [0]. This replacement implies a higher number of iterations, but results often in less computational work than would be required if H [i] were computed at each step. The method shown above does not take into account the different errors by which the measurements may be affected. However, this method can be made general by weighting each observation z i with the inverse of its mean measurement error r i (where i =1,,, n). By so doing, each residual q i is replaced by the corresponding normalised residual q* i defined as follows q i ¼ q i ½ ¼ z hx ð 0Þ r i The mean measurement error r i includes both the random and systematic errors. A typical example of the latter type of error is provided by the refraction of light rays due to the Earth atmosphere. By using the normalised residuals q* i instead of the ordinary residuals q i, the expression Dx lsq 0 =(H T H) 1 (H T Dz) shown above remains formally identical, the only difference being that in the present case (weighted observations) the Jacobian H and the difference vector Dz must be replaced by their respective counterparts H* and Dz*, as follows Dx lsq 0 ¼ H T H 1 H T Dz where H* =R H and Dz* =R Dz; R is the n n diagonal matrix filled with zeros, except the elements placed along its main diagonal, which are r 1 1,, r n 1. In other words, R diag(r 1 1,, r n 1 ) or, which is the same, r i 1=r =r... 0 R =r n i

104 Orbit Determination from Observations The solution given above Dx lsq 0 ¼ H T H 1 H T Dz of the problem of weighted least-squares estimation may also be expressed in terms of the original Jacobian matrix H and the original difference vector Dz (instead of H* =R H and Dz* =R Dz) as follows 0 ¼ H T 1 WH H T W Dz Dx lsq where W is the weighting matrix defined as W R diag(r 1,, r n ), or, which is the same, 1=r =r W R =r n The preceding definition of W as a diagonal matrix holds in case of uncorrelated errors of measurement. Should such errors be correlated, then W would become a non-diagonal matrix. In order to understand the correlation of errors, it is necessary to have some concepts of probability theory, which are given below. As has been shown above, if x 0 and e designate, respectively, the actual (augmented) state vector at epoch and the vector containing the measurement errors, then the observation vector z results from z ¼ hx ð 0 Þþe The preceding expression, after linearisation, becomes Dz ¼ H x 0 x ref 0 þ e where x ref 0 is a reference state vector sufficiently close to x 0. The solution of this least-squares problem has been shown to be x lsq 0 ¼ x ref 0 þ 1 HT WH H T WDz ¼ x 0 þ H T 1 WH H T We The preceding expression shows that, when measurement errors are committed, the computed state vector x 0 lsq differs from the actual state vector x 0. In the event of systematic errors being negligible, the components of e are only random errors. Let X be a discrete random variable which takes values in S ={x 1, x,, x n }. The expected value (or mean) of X is defined as follows

105 .9 Improvement in Orbit Determination by Weighted Least-Squares Estimation EðXÞ ¼ Xn i¼1 x i pðx i Þ where p(x i ) is the probability of x i. The mean of X is also denoted by X or by l X. The preceding definition holds if the sum x 1 p(x 1 )+x p(x )+ + x n p(x n ) converges to a finite value. Otherwise, the expected value of X is undefined. For example, when a die is cast, the probability of each of the six possible events x i (with i =1,,, 6)isp(x i ) = 1/6. Thus, according to the preceding definition, the expected value is E(X) =( ) (1/6) = 1/6 = 7/ =.5. Let X be a continuous random variable in S =[, ] with probability density function u(x). In this case, the expected value of X is defined as follows EðXÞ ¼ Z 1 1 x uðxþ dx Here, too, the value of the integral must be finite, in order for the expected value of X to be defined. Let X and E(X) be, respectively, a random variable and its expected value. The variance of X is defined as follows n o VarðXÞ ¼E ½X EðXÞ In other words, the variance of X is the expected squared distance of X from its mean E(X). Let X be a discrete random variable which takes values in S ={x 1, x,, x n }. Then the variance of X is defined as follows VarðXÞ ¼ Xn i¼1 " # x i Xn x j px j px ð i Þ In the example of the die proposed above, p(x i ) = 1/6 and E(X) = 7/. Thus, according to the preceding definition, the variance of X is " VarðXÞ ¼ 1 7 þ 7 þ 7 þ 4 7 þ 5 7 þ 6 7 # 1 6 ¼ 5 4 þ 9 4 þ 1 4 þ 1 4 þ 9 4 þ ¼ 5 1 Likewise, let X be a continuous random variable in S =[, ] with probability density function u(x). Then the variance of X is j¼1

106 4 Orbit Determination from Observations VarðXÞ ¼ Z 1 1 4x Z 1 1 nu ðnþ dn5 u ðxþ dx In case of the expected value E(X) of a random variable X being zero, that is, in case of E(X) = 0, then the variance of X is n o VarðXÞ ¼E ½X EðXÞ h i ¼ E ðx 0Þ h i ¼ E ðxþ The variance of a random variable X is often denoted by r X, that is, Var(X) r X, and its standard deviation [Var(X)] ½ by r X. Let X and Y be two independent random variables. Let E(X) and E(Y) be their respective expected values. In this case, there results VarðX þ YÞ ¼ VarðXÞþ VarðYÞ Instead, in case of X and Y being dependent random variables, then there results VarðX þ YÞ ¼VarðXÞþVarðYÞþ Ef½X EðXÞ½Y EðYÞg where the term E{[X E(X)][Y E(Y)]} is called the covariance of X and Y, that is, Cov ðx; YÞ ¼Ef½X EðXÞ½Y EðYÞg The covariance of two random variables X and Y is the average value of the deviation of X from its mean E(X) and the deviation of Y from its mean E(Y). The zero value of the covariance Cov(X, Y) of two random variables X and Y does not imply their independence; it implies only the linear independence of X and Y. The correlation coefficient q XY Corr(X, Y) of two random variables X and Y is defined as follows CorrðX; YÞ ¼ CovðX; YÞ ½VarðXÞVarðYÞ 1 Corr(X, Y) is a dimensionless quantity having the following properties: 1 CorrðX; YÞ1 CorrðaX þ b; cy þ dþ ¼ CorrðX; YÞ where a, b, c, and d are any real constants. This implies that Corr(X, Y) is equal to unity only when a relation of a linear dependence exists between X and Y.

107 .9 Improvement in Orbit Determination by Weighted Least-Squares Estimation 5 Let us see now the practical application of the concepts given above. If the error vector e has only random errors as its components, then the expected values of, respectively, e and ee T are such that EðeÞ ¼0 Eðee T Þ¼ diag ðr 1 ;...; r n Þ In other words, the expected value E(e i ) of each component e i (i =1,,, n) of the error vector e is equal to zero; all these components are uncorrelated, in other words e i e j = 0 for i 6¼ j; and the standard deviation r i of e i is equal to [E(e i )] ½. The least-squares solution x 0 lsq of the orbit determination problem is a random variable, because it depends on e, which is a random variable by the hypothesis made above. Then, E(e) =0 implies that the expected value of x 0 lsq is equal to the actual value of x 0, because E x lsq 0 h ¼ E x 0 þ H T 1 i WH H T W e ¼ x 0 þ H T WH 1 H T W EðeÞ ¼ x0 In addition, since by definition Cov(X, Y) = E{[X E(X)][Y E(Y)]}, then h ih i T Cov x lsq 0 ; xlsq 0 ¼ E x lsq 0 E x lsq 0 x lsq 0 E x lsq 0 h ih i T ¼ E x lsq 0 x 0 x lsq 0 x 0 Substituting x 0 þ H T WH 1 H T W e for x lsq 0 into the preceding expression yields h Cov x lsq 0 ; xlsq 0 ¼ E x 0 þ H T 1 ih WH H T W e x0 x 0 þ H T 1 WH H T W e x0 h ¼ E H T 1 WH H T ih W e H T 1 WH H T i T W e ¼ H T 1 WH H T W E ee T ðwhþ H T WH 1 i T The expression written above can be simplified, if the weighting matrix W is chosen in accordance with the standard deviation r i (where i =1,,, n) of the measurement. In this case, W = diag(r 1,, r n ) is the inverse of E(ee T ), and the preceding expression reduces to Cov x lsq 0 ; xlsq 0 ¼ H T 1 WH

108 6 Orbit Determination from Observations Those elements of the covariance matrix which are placed along its main diagonal are the standard deviation r x lsq 0i ¼ Cov x lsq 0i ; xlsq 0i h i1 (i =1,,, n) of the components of the x lsq 0 vector. Likewise, the off-diagonal elements of the covariance matrix provide a measure of the correlation existing between errors of individual components. The expected value and the covariance of x lsq 0 define the distribution of values of x lsq 0 which would result in an experiment of repeated orbit determinations for the same trajectory, if the measurement errors were only of the random type. If these lsq errors have a normal distribution, then there is a probability of 67% that x 0 (resulting from the actual measurements) deviates from x 0 by less than 1r, and a probability of 99.7% that x lsq 0 deviates from x 0 by less than r. In the presence of systematic errors e*, there is a further deviation d x lsq 0 ¼ H T 1 WH H T W e of x 0 lsq from x 0. Montenbruck and Gill [5] point out that the measurement standard deviation r(e) must be known in order to construct the weighting matrix W. The analysis performed so far is based on the assumption of a Gaussian distribution of errors in the observations. However, this analysis (based only on data noise errors) does not take into account the effect of model errors. It is therefore necessary to take into account the effect due to systematic errors. To this end, the measurement equation can be rewritten as follows z ¼ hx ð 0 ; cþþe where z is the n-dimensional vector containing the observations at times t 1, t,, t n ; x 0 is the vector of the estimated parameters; c is the vector of the so-called consider parameters (which affect the estimated parameters); h is a vector-valued function of x 0 and c; and e is the vector containing the measurement noise errors. The vector c contains the force and measurement parameters which are uncertain but are not modified as a result of the least-squares estimation process. The expected value of these parameters can be assumed, without loss of generality, equal to zero. Previously both the estimated and the consider parameters have been taken together and the expression z ¼ hx ð 0 Þþe ref of the observation vector z has been linearised around the reference state vector x 0 to obtain Dz ¼ H x 0 x ref 0 þ e

109 .9 Improvement in Orbit Determination by Weighted Least-Squares Estimation 7 Just in the same way, now the expression z ¼ hx ð 0 ; cþþe is linearised around x ref 0 to obtain Dz ¼ H x x 0 x ref 0 þ Hc c þ e where H x is the Jacobian matrix containing the partial derivatives of the vector-valued function h with respect to x 0, and H c is the Jacobian matrix containing the partial derivatives of h with respect to c. Now the expression of the least-squares solution becomes x lsq 0 ¼ x 0 þ H T x WH 1H T x x WH ð c c þ eþ This solution differs from the true value x 0 of the estimation parameters by a quantity which depends on the consider parameters (components of c) and the measurement noise (components of e). The consider parameters are assumed to be random variables with zero mean and covariance matrix C, which are uncorrelated with the measurement noise, so that the equality E(ce T )=0holds. In this case, the expected value of the least-squares solution E x lsq 0 ¼ x 0 þ H T x WH 1H T x x WH ½ c EðcÞþEðeÞ ¼ x 0 is equal to the true state vector x 0. The consider covariance matrix P c is larger than the noise-only covariance matrix P =(H T x WH x ) 1, which is also called formal or computed covariance matrix. The consider covariance matrix has the following expression P c ¼ HH T x W Hc CH T c þ EðeeT Þ PH T T x W ¼ P þðph T x WÞðH cch T c Þ PHT x W T where the weighting matrix W has been taken as the inverse of the measurement covariance matrix. As has been shown above, an approximate value (x 0 appr ) of the actual state vector (x 0 ) at epoch must be known to start the least-squares orbit determination process. Some information on the accuracy of this value is often available. We want to incorporate the a priori covariance matrix P 0 apr into the least-squares estimation. To his end, the loss function shown above Jðx 0 Þ ¼ q q q T q ¼ ðdz H Dx 0 Þ T ðdz H Dx 0 Þ is represented in another way. Remembering that Dx lsq 0 ¼ H T 1 H H T Dz

110 8 Orbit Determination from Observations the loss function may be written as follows TðH Jðx 0 Þ ¼ Dx 0 Dx lsq 0 T HÞ Dx 0 Dx lsq 0 þ Dz T Dz Dx lsq T 0 H T H Dx lsq 0 TP0 ¼ x 0 x lsq 0 1 x 0 x lsq 0 þ constant Since this expression of the loss function J(x 0 ) is a quadratic form of (x 0 x 0 lsq ) defined by the inverse covariance matrix P 0 1 =(H T H)of(x 0 x 0 lsq ), then the loss function minimum and the covariance matrix provide the same information on the least-squares estimation as that which comes from the measurement vector Dz and the Jacobian matrix H. Thus, an a priori estimate x apr 0 ¼ x ref 0 þ Dxapr 0 of the state vector x 0 may come from a modified loss function J ¼ x 0 x apr 0 TK x0 x apr 0 þ q T q where K =(P 0 apr ) 1, called information matrix, is used to attribute a contribution to the loss function to each deviation from x 0 apr, and q is the vector containing the residuals. The information matrix K is always positive semi-definite, because it is the inverse of the covariance matrix. By the way, let A be a matrix with real entries a ij. A is said to be positive semi-definite if, for any vector x with real components x i, the dot product Ax, x of Ax and x is non-negative, that is, if ha x; xi0 Consequently, K can be factored to form a product K = S T S, which makes it possible to determine the minimum of the loss function J. If J is written as follows J ¼ðDx 0 Dx apr 0 ÞT KðDx 0 Dx apr 0 ÞþðDz H Dx 0Þ T ðdz H Dx 0 Þ¼A T A A ¼ SDxapr 0 Dz S H Dx 0 then the information matrix K can be considered as the result of additional observations. Consequently, the minimum of the combined loss function results, after simplification, from Dx 0 lsq ¼ K þ H T H 1 K Dx apr 0 þ H T Dz

111 .9 Improvement in Orbit Determination by Weighted Least-Squares Estimation 9 In case of weighted observations, the preceding expression becomes 0 ¼ K þ H T 1 WH K Dx apr 0 þ H T W Dz Dx lsq where W is the weighting matrix. In order for the preceding expression to be computable, the sum K + H T WH must have a nonzero determinant, without the need for K or H T WH to be non-singular (by the way, an n n matrix A is said to be non-singular or invertible when there exists an n n matrix B A 1 such that AB = BA = I, where I is the n n identity matrix). However, the non-singularity of the information matrix K is sufficient to ensure the resolvability of the equations independently of H T WH. In order to take advantage of this fact, a singularity in least-squares problems can be avoided by giving a small a priori weight to each estimation parameter and adding the corresponding diagonal matrix K to the normal equations matrix. The expected value Dx 0 lsq =(K + H T WH) 1 (K Dx 0 apr + H T W Dz) of the estimated state is equal to the actual state x 0, if the a priori information x 0 apr is also a random variable having x 0 as its mean value. The covariance matrix P 0 of the estimate is related to the a priori covariance and the measurement information matrix by the following expression ðp 0 Þ 1 ¼ðP apr 0 Þ1 þðh T WHÞ.10 Numerical Solution of the Least-Squares Estimation Problem The purpose of the present paragraph is to provide the reader of this book with the basic concepts which are necessary to solve numerically the least-squares problem described in the preceding paragraph. Those among the readers who are fully conversant with such concepts can skip this paragraph and go directly to the next. For the sake of generality, the system of m normal equations of Sect..9, that is, (H T H)Dx lsq 0 =(H T Dz), is written here in the following form A T A x ¼ A T b where A, x, and b take the places of, respectively, H, Dx 0 lsq, and Dz. The numerical procedures to be described below are based on the decomposition (also called factorisation) of a given non-singular n m matrix A (with n m) into an orthogonal n n matrix Q and an upper triangular m m matrix R. As the

112 40 Orbit Determination from Observations bottom (n m) rows of an n m upper triangular matrix contain only zeroes, as will be shown below, it is customary to write R mm A nm ¼ Q nn 0 ðnmþm where 0 is an (n m) m partition, containing only zeroes, of the given matrix A. According to the definition given by Olver and Shakiban [60], an orthogonal (or orthonormal) matrix Q is a square matrix which satisfies the condition Q T Q = I, where Q T is the transpose of the given matrix and I is the identity matrix. Since the inverse matrix Q 1 must satisfy the condition Q 1 Q = I, then a square matrix Q is orthogonal if and only if its transpose Q T is equal to its inverse Q 1. In other words, a square matrix is orthogonal if and only if its columns form an orthonormal basis with respect to the Euclidean scalar (or inner or dot) product on an n-dimensional Euclidean space. This is because, if q 1, q,, q n are the columns of Q, then q 1 T, q T,, q n T will be the rows of the transpose matrix Q T. Now, the (i, j) entry of the product Q T Q results from the product between the i th row of Q T and the j th column of Q. Since Q is, by hypothesis, an orthogonal matrix (such that Q T Q = I), then the following relations must hold q i q j ¼ q T i q j ¼ 1 ði ¼ jþ 0 ði 6¼ jþ which are just the conditions for q 1, q,, q n to form an orthonormal basis. Guillemin [1] gives the following numerical example of an orthogonal matrix of order three 0:5 0:5 0:707 Q 4 0:707 0: :5 0:5 0:707 where approximates ½() ½. As is easy to verify, the three columns 0:5 0:5 0:707 q 1 4 0:707 5 q 4 0:707 5 q :5 0:5 0:707 of the matrix Q indicated above satisfy the conditions q i q j = 1 (for i = j) and q i q j = 0 (for i 6¼ j). An orthogonal matrix Q has determinant det(q) =± 1. This is because Q, as an orthogonal matrix, must satisfy the condition Q T Q = I; taking the determinant of this equality yields 1 ¼ det ðiþ ¼detðQ T QÞ¼det Q T det ðqþ ¼½det ðqþ If the determinant of an n n orthogonal matrix is equal to +1, then that matrix is called proper and the corresponding orthonormal basis is a right-handed basis on

113 .10 Numerical Solution of the Least-Squares Estimation Problem 41 an n-dimensional Euclidean space; whereas, if the determinant of an orthogonal matrix is equal to 1, then that matrix is called improper and the corresponding orthonormal basis is a left-handed basis on the same space. The product of two orthogonal matrices is also an orthogonal matrix. This is because, if Q 1 and Q are two orthogonal matrices, that is, two matrices such that Q T 1 Q 1 ¼ I Q T Q ¼ I then there results (Q 1 Q ) T (Q 1 Q )=Q T Q 1 T Q 1 Q = Q T Q = I. Thus, the product matrix Q 1 Q is also orthogonal. An m m matrix R is said to be upper triangular, if its entries r ij below the main diagonal are zero (r ij 6¼ 0 for i j; r ij = 0 for i > j), as shown below: r 11 r 1... r 1m 0 r... r m R r mm The QR decomposition applied to a given non-singular matrix A makes the product A T A less sensitive to small errors affecting the values of the entries a ij of A. The QR decomposition is unique, if all the diagonal elements r ii (i =1,,, m) of R are required to be real and positive [8]. There are several methods for computing the QR decomposition. Among them are the Gram Schmidt procedure, the Householder transformations, and the Givens rotations. These three methods will be described below. The interested reader can find more on the matter in several books or articles, for example in Refs. [5, 9]. The numerically stable Gram Schmidt procedure is a method to make a set of vectors orthogonal in an inner product space. An inner product space is a vector space with an operation, which associates each pair v and w of vectors of the space with a scalar quantity known as the inner product v, w of the considered pair of vectors. A vector space V is the set of all real (column) vectors v with n components v 1, v,, v n, this set being closed under the two operations of addition (if v and w are two vectors of V, then v + w is also a vector of V) and multiplication by a scalar (if v is a vector of V and s is a real scalar, then sv is also a vector of V). Two vectors v and w of a vector space V are said to be orthogonal, if their inner product v, w is zero. Vectors v 1, v,, v n of an n- dimensional vector space V are linearly independent, if the equality c 1 v 1 + c v + + c n v n = 0 can only hold if the coefficients c 1, c,, c n are all equal to zero. Vectors v 1, v,, v n of an n-dimensional vector space V are a basis, if any vector v of V can be expressed as a linear combination v = c 1 v 1 + c v + + c n v n with a unique choice of the coefficients c 1, c,, c n. For an n-dimensional vector space, any nonzero linearly independent vectors form a basis. Let V and W be two sets of vectors. If each vector v contained in V can be expressed as a linear

114 4 Orbit Determination from Observations combination of the vectors contained in W, then W is said to be the basis set (or the generating set or the spanning set) for V. The Gram Schmidt procedure takes a finite set {a 1, a,, a m } of linearly independent vectors and generates an orthogonal set {q 1, q,, q m } which spans the same subspace as the previous set. Let a 11 a 1... a 1m 0 a... a m A a n1 a n... a nm be a given n m matrix with m linearly independent columns. The same matrix can also be indicated briefly A [a 1 a a m ], where a 1, a,, a m are the column vectors of A. We want to construct an orthogonal n n matrix Q {q 1, q,, q n } and an upper triangular m m matrix R {r 1, r,, r m } such that R mm A nm ¼ Q nn 0 ðnmþm For example, let A be a two-column matrix A [a 1 a ] and Q be a two-column matrix Q [q 1 q ]. Since R must be a triangular matrix, then r A ½a 1 a ¼ ½q 1 q 11 r 1 ¼ QR 0 r In this case, the two column vectors q 1 and q of Q and the three nonzero entries r 11, r 1, and r of R are the quantities to be determined as a function of the two column vectors a 1 and a of the given matrix A. Likewise, when A is a three-column matrix A [a 1 a a ], then r 11 r 1 r 1 A ½a 1 a a ¼ ½q 1 q q 4 0 r r 5 ¼ QR 0 0 r In this case, the three column vectors q 1, q,andq of Q and the six nonzero entries r 11, r 1, r 1, r, r, and r of R are the quantities to be determined as a function of the three column vectors a 1, a, and a of the given matrix A. The numerically stable Gram Schmidt procedure is described below by means of an example, due to Wikipedia [8]. Let us consider the following matrix A The three column vectors a 1, a, and a of the given matrix A will, by degrees, be replaced by the three column vectors q 1, q, and q of an orthogonal matrix Q.

115 .10 Numerical Solution of the Least-Squares Estimation Problem 4 At the same time, the six nonzero entries r 11, r 1, r 1, r, r, and r of an upper triangular matrix R will be computed so that A = QR. To this end, we first compute the norm (that is, the Euclidean length) of the first (or leftmost) column vector a 1 of A. The first entry r 11 of R is set equal to the norm a 1. This yields h r 11 ¼ ka 1 k ¼ a 11 þ a 1 þ 1 a 1 ¼ 1 þ 6 þð4þ i1 ¼ 14 Now, we normalise a 1 by dividing all its components by its norm a 1. This yields the first column q 1 of Q, as follows q 11 ¼ a 11 ka 1 k ¼ 1 14 ¼ 6 7 q 1 ¼ a 1 ka 1 k ¼ 6 14 ¼ 7 q 1 ¼ a 1 ka 1 k ¼ 4 14 ¼ 7 At this stage of the procedure, the given matrix A becomes 6= = = Now, we compute the next entries (r 1, r 1,, r 1m )ofr by taking the scalar products of the first column of the matrix shown above with each of the other columns of the same matrix. In the example considered above, this yields r 1 ¼ ð6=7þð51þ þð=7þ167 þð=7þ4 ¼ 1 r 1 ¼ ð6=7þ4 þ ð=7þð68þ þð=7þð41þ ¼14 Now, for the purpose of making the second, the third,, the m th column orthogonal to the first column, we subtract r 1 times the first column from the second column, r 1 times the first column from the third column,, r 1m times the first column from the m th column. This yields 51 ð6=7þ1 ¼69 4 ð6=7þð14þ ¼ ð=7þ1 ¼ ð=7þð14þ ¼ 6 4 ð=7þ1 ¼ 0 41 ð=7þð14þ ¼ 45

116 44 Orbit Determination from Observations and the resulting matrix is 6= = = At this stage of the procedure, the first column vector of the preceding matrix has been normalised to have unit length; in addition, the second column and the third column have been made orthogonal to the first. Now, in order to compute q, we normalise the second column, by dividing all its components by its norm. At the same time, the r entry of R is set equal to the norm (or Euclidean length) of the second column. In the example considered above, this yields r ¼ h ð69þ i1 þ 158 þ 0 ¼ 14 The resulting matrix is 6=7 69= =7 158= =7 6=5 45 Now r is the scalar product of the second column and the third column: r ¼ þ ð Þþ 6 5 ð45þ ¼ 70 By subtracting r times the second column from the third, we obtain 16 ð70þð69=175þ ¼ 406=5 6 ð70þð158=175þ ¼ 4=5 45 ð70þð6=5þ ¼ The resulting matrix is 6=7 69= =5 4 =7 158=175 4=5 5 =7 6=5 Now we compute r as the norm of the third vector: " r ¼ 406 þ 4 #1 þðþ 5 5 ¼ 5

117 .10 Numerical Solution of the Least-Squares Estimation Problem 45 Finally, in order to compute q, the last column is normalised by dividing all its components by its norm. This yields ¼ ¼ ðþ 1 5 ¼ 5 The two matrices Q and R resulting from the procedure described above are then Q 6=7 69=175 58=175 4 =7 158=175 6=175 5 =7 6=5 = R where A = QR. The example shown above has described the numerically stable Gram Schmidt procedure for a given matrix A. The same procedure can be extended to any non-singular m m matrix A, by means of the algorithm given below, which is taken from Olver and Shakiban [60]. This algorithm takes the entries a ij of A and replaces them with the entries q ij of Q; at the same time, it computes the nonzero entries r ij of R, which must be stored in a separate matrix. start for j = 1 to m set r jj = (a 1j + a j + + a mj ) ½ if r jj = 0, stop; print A has linearly dependent columns else for i = 1 to m set a ij = a ij /r jj next i for k = j +1to m set r jk = a 1j a 1k + a j a k + + a mj a mk for i = 1 to m end set a ik next i next k next j = a ik a ij r jk

118 46 Orbit Determination from Observations The QR decomposition based on the Householder transformations is shown below. Let x and y be any two linearly independent vectors, having the same Euclidean length x = y, of an n-dimensional vector space V. Consider the unit vector u =(y x)/ y x and the matrix H = I (u u T ). Then H is the reflection matrix such that H x = y. It is to be noted that uu T is an outer (or tensor) product, which yields a matrix, not a scalar. For example, let 1 14 x y be two column vectors of a three-dimensional vector space V. It is easy to verify that the two vectors x and y given above are linearly independent and have the same Euclidean length x = y = 14. In this case, there results y x ¼ h ky xk ¼ þð6þ i1 þ 4 ¼ ð14þ 1 u ¼ y x 1= ð14þ 1 ky xk ¼ 6 = ð14þ = ð14þ 1 1= ð14þ 1 h i uu T ¼ 6 = ð14þ = ð14þ 1 = ð14þ 1 = ð14þ 1 = ð14þ 1 1=14 =14 = ¼ 4 =14 9=14 6=14 5 =14 6=14 4=14 uu T 1=7 =7 =7 6 7 ¼ 4 =7 9=7 6=7 5 =7 6=7 4=7

119 .10 Numerical Solution of the Least-Squares Estimation Problem 47 H ¼ I uu T =7 =7 = ¼ þ 4 =7 9=7 6= =7 6=7 4=7 6=7 =7 =7 6 7 ¼ 4 =7 =7 6=7 5 =7 6=7 =7 As is easy to verify, the reflection matrix H, found as shown above, satisfies the condition H x = y. In addition, H is an orthogonal matrix (that is, H T H = I) and its determinant det(h) is equal to 1. Since we know how to compute the reflection matrix H in an n-dimensional vector space, we can apply these concepts to produce QR decompositions. To this end, we intend to introduce sub-diagonal zeros into the given matrix A to be decomposed, in order to gradually change A into an upper triangular matrix R. In other words, we multiply the given matrix A nm on the left by a sequence of reflection matrices H 1, H,, H k, so that the product H k H H 1 A should be equal to an upper triangular matrix R. The first step operates on the matrix A itself. We choose the first (that is, the leftmost) column vector a 1 of the given matrix A nm [a 1 a a m ]. We compute the Euclidean length ka 1 k ¼ a 11 þ a 1 þ 1 þa n1 of the first column vector a 1 and then find a reflection matrix H 1 such that the first column of the product H 1 A should be a multiple of the vector [ ] T.In other words, we form v ¼ a 1 ka k u ¼ v kk v H 1 ¼ I ðuu T Þ By so doing, H 1 is a reflection matrix such that the first column of H 1 A is a multiple of the vector [ ] T. In other words, the first column of H 1 A is a column vector having all zeros in its rows except the first. Each of the column vectors a 1, a,, a i,, a m of A nm can be reflected onto a multiple of [ ] T in two ways: it can be reflected onto a i [ ] T or onto a i [ ] T. The choice of the sign in front of a i is important, because when we form the unit vector u =(y x)/ y x we divide by y x. Consequently, a choice which makes this denominator small must be avoided. To this end, the sign in front of a i should be the opposite of the sign in front of the

120 48 Orbit Determination from Observations entry placed in the k th row of a i, where a ki is the pivot co-ordinate after which all entries are zero in the final upper triangular form of A. Therefore, it is advisable to choose u ¼sgn ða ki Þka i k½ where the signum function sgn(x) is such that 8 < 1 ðif x\ 0Þ sgnðxþ ¼ 0 ðif x ¼ 0Þ : 1 ðif x [ 0Þ The second step operates on the matrix A, which results from cancelling the first row and the first column of H 1 A and retaining the rest of H 1 A. We repeat for A the same operations performed in the first step, and compute the reflection matrix H. Since H is of smaller rank than H 1 and we want to operate with H 1 A (and not with A ), then we expand A, byfilling in a 1 in the upper left entry of H 1 A. This means that the second reflection matrix H results from H ¼ H 0 The third step operates on A, which results from cancelling the first row and the first column of H A and retaining the rest of H A. We repeat for A the same operations performed in the second step, and compute the reflection matrix H. The third reflection matrix H results from H ¼ H 0 In general, at the p th step, the p th reflection matrix H p results from H p ¼ I p1 0 0 H 0 p After a number k = min(m 1, n) of steps, the result of this process will be H k H H 1 A ¼ R where R is an upper triangular matrix, and each of the reflection matrices (H 1, H,, H k ) is an orthogonal matrix. Thus, with T Q ¼ H 1 T H T H T k

121 .10 Numerical Solution of the Least-Squares Estimation Problem 49 the desired result A = QR will be reached. For example, we consider the matrix A which was decomposed previously by means of the Gram Schmidt procedure. We consider the first column vector a 1 [1 6 4] T of the matrix A and compute the Euclidean length a 1 of a 1, as follows h k k ¼ 1 þ 6 þð4þ i1 ¼ 14 a 1 Then we compute ka 1 k½ T ¼ ½ v ¼ a 1 sgnða 11 Þka 1 k½ T ¼ ½ 6 4 kk¼ v h ðþ þ 6 þð4þ i1 ¼ 14 ð Þ 1 u ¼ v 1= ð14þ 1 kk v ¼ 6 = ð14þ = ð14þ 1 1= ð14þ 1 h i uu T 6 ¼ = ð14þ = ð14þ 1 = ð14þ 1 = ð14þ 1 = ð14þ 1 1=14 =14 =14 ¼ 4 =14 9=14 6=14 5 =14 6=14 4=14 H 1 ¼ I uu T =7 =7 = ¼ þ 4 =7 9=7 6= =7 6=7 4=7 6=7 =7 =7 6 7 ¼ 4 =7 =7 6=7 5 =7 6=7 =7 T T

122 50 Orbit Determination from Observations Now we consider the matrix resulting from the product H 1 A, that is, H 1 A ¼ 6=7 =7 = =7 =7 6= ¼ =7 6=7 = This matrix, due to the nonzero value of its (H 1 A) entry (because 168 6¼ 0), is not an upper triangular matrix. Therefore, we cancel the first row and the first column of H 1 A, and consider the matrix A 0 ¼ The Euclidean length of the first column vector a 1 of A [a 1 a ] results from a h 0 1 ¼ ð49þ i1 þ 168 ¼ 175 v ¼ ¼ uu T ¼ h kk¼ v ð4þ i1 þ 168 ¼ 80 u ¼ v k v k ¼ 4=5 =5 4=5 16=5 1=5 ½4=5 =5 ¼ =5 1=5 9=5 H 0 ¼ I ðuuþ T ¼ 1 0 =5 4=5 þ 0 1 4=5 18=5 Since the product H H 1 A yields H ¼ 4 0 7=5 4= =5 7=5 ¼ 7=5 4=5 4=5 7= H H 1 A ¼ 4 0 7=5 4= ¼ =5 7= then H H 1 A is the desired upper triangular matrix R and there is no necessity of further steps. The desired matrix Q results from Q = H 1 T H T. Therefore

123 .10 Numerical Solution of the Least-Squares Estimation Problem 51 6=7 =7 = Q ¼ H T 1 H T ¼ 4 =7 =7 6= =5 4=5 5 =7 6=7 =7 0 4=5 7=5 6=7 69=175 58= ¼ 4 =7 158=175 6=175 5 =7 6=5 =5 It is easy to verify that the product QR of the two matrices Q and R determined above is equal to the given matrix A. The QR decomposition based on the Givens rotations is shown below. A Givens rotation is the rotation of a column vector in the plane spanned by two co-ordinate axes. In order to perform a QR decomposition A = QR, each Givens rotation is a matrix G which, multiplied on the left by the matrix A, introduces a zero in the sub-diagonal entries of A, for the purpose of gradually transforming A into an upper triangular matrix R. The product of the transposes of all these Givens rotation matrices produces the orthogonal matrix Q. A Givens rotation matrix is represented by the matrix shown below cos h sin h 0. G sin h cos h In other words, a Givens rotation matrix G with entries g ij results from the identity matrix I after operating the following substitutions: g ii ¼ cos h g jj ¼ cos h g ij ¼ sin h g ji ¼sin h This method applies to the case of a QR decomposition as follows. Let us consider again the matrix A which was decomposed previously. We want to construct a Givens rotation matrix G 1 for the purpose of replacing the a 1 entry of A (which entry is at present a 1 = 4) with a zero. By multiplying on the left the first column vector

124 5 Orbit Determination from Observations 1 a of the matrix A [a 1 a a ] by the following rotation matrix G G cos h sin h 5 0 sin h cos h there results the following column vector cos h 4 sin h 5 6 sin h 4 cos h In order for the third entry ( 6 sin h 4 cos h) of this vector to be zero, h must be h ¼ arctan ¼ :690 and consequently cos(.690) = and sin(.690) = Therefore, G 1 is G :805 0: : :805 The product G 1 A yields a matrix A, whose entry a 1 is equal to zero, as follows A 0 ¼ 4 0 0:805 0: : : ¼ 4 7:111 15:64 : :60 71:84 Now, starting from A indicated above, we find a rotation matrix G cos h sin h 0 G 4 sin h cos h

125 .10 Numerical Solution of the Least-Squares Estimation Problem 5 in order to transform a 1 = into zero. By operating as has been shown above, we find the following condition ð sin hþ1 þðcos hþ7:111 ¼ 0 hence h = 1.00 and consequently cos h = and sin h = Then 0: : G 4 0: : The product A = G A yields a matrix such that its entries a 1 and a 1 are both of them equal to zero: A :96 1: :60 71:84 Now, starting from A indicated above, let us find a rotation matrix G G 4 0 cos h sin h 5 0 sin h cos h which transforms a = into zero. By operating as has been shown above, we find the following condition ð sin hþ1:96 þðcos hþ11:60 ¼ 0 hence h = , and consequently cos h = and sin h = Therefore, G is G 4 0 0: : :6444 0:76550 The product G A yields a matrix R having all of its sub-diagonal entries equal to zero, as follows R

126 54 Orbit Determination from Observations As has been shown above, R results from R = G G G 1 A. Since G 1, G and G are, each of them, orthogonal matrices, then the desired matrix Q of the decomposition A = QR results from Q ¼ G T 1 GT GT In the example considered above, the product indicated above yields 6=7 69=175 58=175 Q 4 =7 158=175 6=175 5 =7 6=5 =5 It is easy to verify that the product QR of the two matrices Q and R determined above is equal to the given matrix A. Generally speaking, the QR decomposition A = QR can be used to solve systems of linear algebraic equations. This is because, by using this decomposition, the system can be written as follows Ax ¼ b QRx ¼ b Now, since Q is an orthogonal matrix (such that Q T Q = I), then multiplying on the left the two members of the preceding equality by Q T yields R x ¼ Q T b Since R is an upper triangular matrix, then the preceding expression can be solved for x by back-substitution, as will be shown below. In the specific case of the least-squares estimation, the QR decomposition A = QR can be used to write the loss function J described in the preceding paragraph, that is, J ¼ ðb Ax Þ T ðb AxÞ in the way shown below. By multiplying on the left the two members of the preceding equality by Q T and remembering that Q T Q = QQ T = I, there results J ¼ðQ T b Q T AxÞ T ðq T b Q T AxÞ ¼ðd Rx Þ T ðd RxÞ þr T r where the two vectors d and r are partitions of the matrix Q T b. The number of their rows is m for d and n m for r. The minimum value (r T r) of the loss function J is reached for Rx = d.

127 .10 Numerical Solution of the Least-Squares Estimation Problem 55 If m is the rank of A, m will also be the rank of R. Consequently, the following system of linear algebraic equations r 11 r 1... r 1m x 1 d 1 0 r... r m x ¼ d r mm x m d m has a unique solution. Since R is an upper triangular matrix, the components x 1, x,, x m of x can be computed by back-substitution, that is, beginning from the last (x m ) and going towards the first (x 1 ), as follows x m ¼ d m r mm x m1 ¼ d m1r ðm1þm x m r ðm1þðm1þ..! x i ¼ 1 r ii d i Pm r ij x j j¼i þ 1 ði ¼ m 1; m ;...; 1Þ In addition to the QR decomposition methods shown above, a singular value decomposition may be used to solve a least-squares problem. A singular value decomposition is particularly suited in case of systems of linear algebraic equations Ax = b having ill-conditioned coefficient matrices A, that is, in case of matrices A such that small changes in their entries a ij result in large changes in the solution x of the system, as will be shown below. Singular value decomposition is a means of decomposing a matrix into the product of three simpler matrices, as the sequel will show. Any nonzero n m matrix A of rank r > 0 can be decomposed as follows A ¼ PDQ T that is, decomposed into the product of an n r matrix P with orthonormal columns (such that P T P = I), an r r diagonal matrix D diag(d 1, d,, d r 1, d r ), whose nonzero entries are supposed to be d 1 d d r 0, and an r m matrix Q T with orthonormal columns (such that Q T Q = I). Let d d... 0 D d r

128 56 Orbit Determination from Observations be the diagonal matrix of the singular value decomposition A = PDQ T. Its nonzero entries d 1, d,, d r, called the singular values of A, are the positive square roots of the nonzero eigenvalues k i of the Gram matrix K = A T A associated with A, that is, d i ¼ ðk i Þ 1 [ 0 (i =1,,, r). The columns of P, called the left singular vectors, are the normalised eigenvectors of AA T. The columns of Q, called the right singular vectors, are the normalised eigenvectors of A T A. As has been shown by Abdi [], singular vectors come in pairs of one left and one right singular vector corresponding to the same singular value. They can be computed either separately or as a pair. When they are computed as a pair (by rewriting equation A = PDQ T in another form, as will be shown below), then it is possible to: compute only one (instead of two) decomposition in eigenvectors; and prevent a problem which may arise because the normalised eigenvectors of a matrix are determined up to a factor equal to 1. Since singular vectors being pairs of vectors must have compatible parities, then care must be taken of the signs in front of the components of the eigenvectors; otherwise, the matrices P and Q, if computed separately, might fail to reconstruct the given matrix A. Since P and Q are orthogonal matrices, the preceding expression can be rewritten as follows A ¼ PDQ T P ¼ AQD 1 Consequently, P results from the product of A, Q and D 1. In addition, D 1 is a diagonal matrix having its nonzero entries equal to the reciprocals of the corresponding entries of D, because D is a diagonal matrix. It is to be remembered that the eigenvalues of any n n matrix M are the scalars k 1, k,, k n which satisfy the following characteristic equation detðm kiþ ¼0 and that the eigenvectors of M are the corresponding nonzero vectors v such that ðm kiþv ¼ 0 If v is an eigenvector of M, then any other nonzero vector w, resulting from the product between v and a scalar, is also an eigenvector of M.

129 .10 Numerical Solution of the Least-Squares Estimation Problem 57 By the way, the number of the singular values of a matrix M is always equal to the rank of the matrix itself. Therefore, if an n n matrix M has less than n singular values, then M is singular. The condition number k(m) of a non-singular n n matrix M is defined as the ratio between the largest (d 1 ) and the smallest (d n ) of the singular values of M, that is, k(m) =d 1 /d n. A singular matrix M is said to have condition number k(m) equal to infinity; a matrix M having a very large condition number k(m) is said to be ill-conditioned. In practice, this condition occurs when the condition number k(m) of the given matrix M is greater than the reciprocal of the precision of the machine used. In case of single-precision computations, this occurs typically when k(m) is greater than An example, taken from Leach [47], of a single value decomposition is given below. Let A 0:96 1:7 :8 0:96 A T 0:96 :8 1:7 0:96 be, respectively, the given matrix to be decomposed and its transpose. The Gram matrix K = A T A associated with A is A T 0:96 :8 0:96 1:7 6:1 :84 A ¼ ¼ 1:7 0:96 :8 0:96 :84 :88 The eigenvalues k 1 and k of K = A T A satisfy the equation detðk kiþ ¼ 0 that is, 6:1 k :84 det ¼ 0 :84 :88 k By expanding the determinant, there results ð6:1 kþð:88 kþ :84 :84 ¼ 0 k 10k þ 9 ¼ 0 The preceding equation has two roots (sorted in descending order, in the absolute sense), which are k 1 ¼ 5 þ ð5 9Þ 1 ¼ 9 k ¼ 5 ð5 9Þ 1 ¼ 1

130 58 Orbit Determination from Observations Therefore, the singular values of A are d 1 ¼ ðk 1 Þ 1 ¼ d ¼ ðk Þ 1 ¼ 1 Thus, d 1 = and d = 1 are the nonzero entries of the required matrix D, that is, D ¼ D 1, the inverse of D, results immediately from D 1 ¼ 1= Now, for k 1 = 9 and k = 1, let us compute the eigenvectors of K = A T A.As shown above, these eigenvectors become column vectors in a matrix (Q) ordered by the size of the corresponding eigenvalues. In other words, the eigenvector of the largest eigenvalue, in the absolute sense, becomes the first (that is, the leftmost) column, the eigenvector of the next largest eigenvalue becomes the second column, and so on; the eigenvector of the smallest eigenvalue becomes the last (that is, the rightmost) column of Q. (For k = k 1 =9) A T A ki ¼ 6:1 9 :84 :88 :84 ¼ :84 :88 9 :84 5:1 Let a and b be the components of the eigenvector q 1 corresponding to k 1 =9. The components of q 1 result from (A T A ki)q 1 = 0. Therefore :88 :84 a ¼ 0 :84 5:1 b 0 :88a þ :84b ¼ 0 :84a 5:1b ¼ 0 Solving either equation for b yields b = 0.75 a. Therefore q 1 a 0:75 a

131 .10 Numerical Solution of the Least-Squares Estimation Problem 59 Dividing the two components by the Euclidean length q 1 = 1.5 a of q 1 yields the normalised eigenvector q 1, that is, q 1 0:8 0:6 (For k = k =1) A T A ki ¼ 6:1 1 :84 5:1 :84 ¼ :84 :88 1 :84 :88 The components a and b of the corresponding eigenvector q result from A T A ki q ¼ 0 Therefore, 5:1 :84 a ¼ 0 :84 :88 b 0 5:1 a þ :84 b ¼ 0 :84 a þ :88 b ¼ 0 Solving either equation for b yields b = 4a/. Therefore q a 4 a Dividing the two components by the Euclidean length q = 5a/ of q yields the normalised eigenvector q, that is, q 0:6 0:8 Therefore, the required matrix Q is given by Q =[q 1 q ], that is, Q ¼ 0:8 0:6 0:6 0:8 As is easy to verify, the product Q T Q is equal to the identity matrix I. Therefore, Q is an orthogonal matrix.

132 60 Orbit Determination from Observations Now, as has been shown above, we compute P = AQD 1. The product QD 1 is QD 1 0:8 0:6 1= 0 0:8= 0:6 ¼ ¼ 0:6 0: : 0:8 The product AQD 1 yields P, as follows 0:96 1:7 0:8= 0:6 0:6 0:8 P ¼ ¼ :8 0:96 0: 0:8 0:8 0:6 It is easy to verify that P is an orthogonal matrix and that the product PDQ T yields the given matrix A. The singular value decomposition can be used for computing the pseudo-inverse of a matrix, which in turn provides a general method for solving the least-squares problem in the form Ax = b, as will be shown below. Following Golub and Reinsch [9], let A be a real n m matrix. An m n matrix X is said to be the pseudo-inverse of A, if X satisfies the following four properties: AXA ¼ A XAX ¼ X ðaxþ T ¼ AX ðxaþ T ¼ XA Let A + denote the unique solution X. IfA = PDQ T, then the pseudo-inverse A + is such that A þ ¼ QD þ P T where D þ ¼ diagðdi þ Þ di þ ¼ 8 < 1 ðfor d i [ 0Þ d : i 0 ðfor d i ¼ 0Þ Consequently, the pseudo-inverse A + of a given matrix A can easily be computed as a result of the singular value decomposition of A. This concept applies to the search for the least-squares solution of an over-determined system of n algebraic equations in m unknowns, where n > m. Let be such a system in matrix form. Let Ax ¼ b q ¼ b Ax

133 .10 Numerical Solution of the Least-Squares Estimation Problem 61 be the residual vector for some vector x. We want to determine the vector x lsq corresponding to the least-squares solution of Ax = b, that is, the vector corresponding to the minimum possible residual vector q. To this end, let us consider the squared Euclidean length kqk ¼ q T q ¼ ðb AxÞ T ðb AxÞ ¼ b T b x T A T b þ x T A T Ax of the residual vector q = Ax b. To determine the minimum value of q,we take the derivative with respect to x and set it to zero. This yields A T b þ A T A x ¼ 0 and hence the following m m system of normal equations A T A x ¼ A T b Multiplying both terms of the preceding expression on the left by (A T A) 1 yields x ¼ A T 1A A T b ¼ A þ b Therefore, the least-squares solution (x) ofax = b corresponding to the minimum value of q is given by x = A + b. Weights can be applied to the original system of equations Ax = b by using a diagonal matrix W R diag(r 1,, r n ). In this case, the least-squares solution to be found is x ¼ A T 1A WA T W b Bebis [6] gives the following (over-determined) system Ax = b of three equations for the two unknowns x 1 and x : 11x 1 þ x ¼ 0 x 1 þ x ¼ 7 x 1 x ¼ 5 In matrix form, the preceding equations are as follows x 0 1 ¼ x 1 5 As has been shown above, the least-squares solution (x) results from x ¼ A þ b

134 6 Orbit Determination from Observations Therefore, the problem reduces to determining the pseudo-inverse A + of the given matrix A. Using the expression written above A þ ¼ A T 1A T A we compute A T A as follows A T A ¼ ¼ Since A T A is a matrix, it is easy to compute its inverse (A T A) 1 as follows. Let M be the following matrix M a b c d Then, the inverse matrix M 1 is given by M 1 ¼ d=det ðmþ b=det ðmþ c=det ðmþ a=det ðmþ where det(m) =ad bc is the determinant of the given matrix M. In this case, det A T A ¼ ¼ 148 Therefore A T 1¼ 14=148 18=148 A 18=148 19=148 The pseudo-inverse A + =(A T A) 1 A T results from the product 14=148 18= =148 19= :0796 0:0551 0: ¼ 0: :854 0:0675 Consequently, the least-squares solution x = A + b of the given system Ax = b is 0 0:0796 0:0551 0: : :854 0: ¼ 0:4106 1:684

135 .10 Numerical Solution of the Least-Squares Estimation Problem 6 The same result can also be obtained by computing the singular value decomposition of the given matrix A, as will be shown below. The matrix A T A has been computed above. Its eigenvalues result from det A T A k I ¼ ð19 kþð14 kþ18 18 ¼ 0 The preceding equation, solved for k, yields k 1 ¼ 11:75 ðk 1 Þ 1 ¼ 11:478 1 ðk 1 Þ 1 k ¼ 11:48 ðk Þ 1 ¼ :59 1 ðk Þ 1 ¼ 0:08711 ¼ 0:9816 For k = k 1 = 11.75, the corresponding normalised eigenvector q 1 is q 1 0:9885 0:15111 For k = k = 11.48, the corresponding normalised eigenvector q is q 0: :9885 The orthogonal matrix Q, having q 1 and q as its columns, is then Q ¼ The matrix D 1 results from D 1 ¼ diag 1 k 1 1 ; 1 k 1! The product QD 1 is equal to QD 1 ¼ 0:9885 0: : :9885 ¼ 0: :9816 0: : : :9474 The orthogonal matrix P results from the product AQD 1 and is then 0:9766 0:09875 P ¼ 4 0:175 0: : :046

136 64 Orbit Determination from Observations Finally, A + results from A þ ¼ QD þ P T where D + = diag(d + i ) = diag[1/(k 1 ) ½, 1/(k ) ½ ] has the same entries as those of D 1 computed above. The product QD + P T yields 0:0796 0:0551 0: : :854 0:0675 This matrix is to be compared with the matrix which has been computed above by means of the expression A + =(A T A) 1 A T. The singular value decomposition applies to the least-squares problem as will be shown below. Let J ¼ ðb Ax Þ T ðb AxÞ be the loss function. Let A = PDQ T be the matrix of the least-squares problem Ax = b. Bydefining the vectors s and t as follows s ¼ Q T x t ¼ P T b the condition (A T A)x = A T b for the minimum value of the loss function can be written as follows D s ¼ D t If the matrix of the normal equations is non-singular, then the inverse D 1 of the D matrix exists and the solution of the least-squares problem Ax = b is s ¼ D 1 t Remembering the definitions of the vectors s and t and the property of orthogonality (Q T = Q 1 ) of the matrix Q, the preceding expression becomes x ¼ QD 1 P T b or, using the column vectors p i and q i of the matrices, respectively, P and Q, x ¼ Xm i¼1 1 d i p T i bq i

137 .10 Numerical Solution of the Least-Squares Estimation Problem 65 In order to apply the mathematical methods shown above to a problem of orbit determination, let us consider again the seven observations of the COBE artificial satellite performed by Healy [] on the 6 th of November 000. They are: Observed time (EST) Right ascension (hh:mm:ss) Declination ( ) Offset 17:1:9 1:48: ::0 1:41: ::0 1:1: :4:0 1:14: :5:9 0:8: :6:0 15:01: :7:0 11:0: As the satellite came nearest the cross hairs, time was recorded. The offset in the last column is an estimate of how close to the cross hairs the satellite came. We want to construct a function which best fits, in the least-squares sense and in the time interval given above, the seven observations tabulated above. Since UTC = EST + 5, then (neglecting the difference between UTC and UT1) the seven EST times indicated above correspond, respectively, to UT1 1 ¼ EST 1 þ 5 ¼ h 1 m 9 s UT1 ¼ EST þ 5 ¼ h m 0 s UT1 ¼ EST þ 5 ¼ h m 0 s UT1 4 ¼ EST 4 þ 5 ¼ h 4 m 0 s UT1 5 ¼ EST 5 þ 5 ¼ h 5 m 9 s UT1 6 ¼ EST 6 þ 5 ¼ h 6 m 0 s UT1 7 ¼ EST 7 þ 5 ¼ h 7 m 0 s The seven values of right ascension, expressed in degrees, are given below. ð1 þ 48=60Þ60=4 ¼ 7:00 ð1 þ 41=60Þ60=4 ¼ 5:5 ð1 þ 1=60 þ 45=600Þ60=4 ¼ :975 ð1 þ 14=60Þ60=4 ¼ 18:50 ð0 þ 8=60Þ60=4 ¼ 07:00 ð15 þ 1=60Þ60=4 ¼ 5:5 ð11 þ =60Þ60=4 ¼ 165:75 In the hypothesis of uncorrelated errors of measurement, the weighting matrix is W ¼ diagðw 11 ; w ;...; w 77 Þ

138 66 Orbit Determination from Observations where the values of the weights w 11, w,, w 77 are preliminarily set, all of them, to unity. In other words, the starting value of the weighting matrix W is taken equal to I, where I is the 7 7 identity matrix. Using a method described by several authors (see, e.g., Refs. [14, ]), called iteratively reweighted least squares, we intend to update W iteratively, in such a way as to give less weight to the more uncertain data points. To this end, we use a mathematical model to predict the law of variation of the right ascension and declination with time. Then we use the residuals q 1, q,, q 7 as a measure of uncertainty. These residuals are the differences between the computed (on the basis of the mathematical model) and the observed data points. The weights chosen in each iteration are related to the magnitudes of the residuals computed in the previous iteration, so that a large residual gives rise to a small weight, as will be shown below. The preliminary values of the weights along the main diagonal of W are indicated in the following table. Obs. No. x Right ascension ( ) Declination ( ) Preliminary weight The right ascension (a) and declination (d) of the observed satellite vary with time t according to functions a(t) and d(t), which are not known a priori. We approximate these unknown functions, within the time interval t 1 t t 7, by means of Chebyshev polynomials T 0, T 1,, T n as follows aðxþ ¼a 0 T 0 ðxþ þa 1 T 1 ðxþ þ þa n T n ðxþ dðxþ ¼d 0 T 0 ðxþ þd 1 T 1 ðxþ þ þd n T n ðxþ where a 0, a 1,, a n, d 0, d 1,, d n are constant coefficients to be determined, and T 0 ðxþ ¼1 T 1 ðxþ ¼x T n þ 1 ðxþ ¼xT n ðxþt n1 ðxþ ðn Þ are Chebyshev polynomials of the first kind. Supposing that a and d are subject to errors independently of each other, we want to determine the best (in the weighted least-squares sense) polynomial approximation for a and d. In other words, we want to determine the unknown values of the coefficients a 0, a 1,, a n, d 0, d 1,, d n corresponding to the minimum value of some norm of the residual vector (q). In the case of the classical least-squares method, this norm is the so-called norm, that is,

139 .10 Numerical Solution of the Least-Squares Estimation Problem 67 we search the minimum value of the sum of the squares of the residuals. However, according to Bube and Langan [14], solutions found by using this method tend to be very sensitive to data points affected by large errors. By contrast, solutions coming from searching the minimum value of the p norm (where 1 p < ) are less sensitive to errors. The method described in Ref. [14] is a hybrid 1 / technique, by means of which a fit (or minimum value of the sum of the squared residuals) is used for small residuals, and a 1 fit (or minimum value of the sum of the absolute residuals) is used for large residuals. A smooth transition from the search for the minimum norm to the search for the minimum 1 norm is obtained by choosing an appropriate value for a positive parameter e which results from an estimate of the standard deviation of the residuals, as will be shown below. In the general case of n-degree polynomials, the model matrix A of the system of linear algebraic equations Ax = b is T 01 T T n1 T 0 T 1... T n A T 07 T T n7 where the first subscript in the entries T ij of A indicates the degree of the Chebyshev polynomial, and the second subscript indicates the number of observation. In order for the system Ax = b to be over-determined with our set of seven observations, the value of n cannot be greater than five. We set n = 5, which corresponds to a fifth-degree Chebyshev polynomial approximation to the true unknown functions a(x) andd(x). In order for the argument x of the Chebyshev polynomials to be within the interval 1 x 1, we have operated a change of variable (from t to x) in the second column of the preceding table, as follows x ¼ t ð t 7 þ t 1 Þ t 7 t 1 where the subscript indicates the number of observation. As to declination (d), we compute the entries T ij of the 7 6 matrix A by computing the Chebyshev polynomials up to the fifth degree for the seven observations indicated above. This yields 1:0000 1:0000 1:0000 1:0000 1:0000 1:0000 1:0000 0:6605 0:18 0:854 0: :4587 1:0000 0:964 0:7868 0: :516 0:99409 A 1:0000 0: : : : : :0000 0:964 0:7868 0: :516 0: :0000 0: : :8165 0:9769 0: :0000 1:0000 1:0000 1:0000 1:0000 1:0000

140 68 Orbit Determination from Observations The column vectors x and b contain, respectively, the six unknown coefficients d 0, d 1,, d 5 of the fifth-degree approximating polynomial d ðxþ ¼d 0 T 0 ðxþþd 1 T 1 ðxþ þ þd 5 T 5 ðxþ and the seven observed values of declination. They are d 0 16:00 d 1 :0000 x 6. 7 b :100 d 5 As shown above, the weighting matrix W is preliminarily set equal to the 7 7 identity matrix I. Now we form the products A T WA and A T Wb. Then, the values of the coefficients d 0, d 1,, d 5 (contained in the vector x) of the fifth-degree approximating polynomial indicated above result from the product of the inverse of A T WA and A T Wb, as follows x ¼ A T 1A WA T W b By so doing, according to our preliminary evaluation (with W = I) of the weights w 11, w,, w 77, the polynomial approximation which best fits, in the weighted least-squares sense, the observed declinations of the satellite is d ðxþ ¼9:74 T 0 ðxþþ54:564 T 1 ðxþ8:776 T ðxþ9:085 T ðxþ 0:99597 T 4 ðxþþ0:67411 T 5 ðxþ The approximation shown above makes it possible to compute the values of declination at the given arguments x 1, x,, x 7. These values are compared with the observed values of declination; then the residuals q 1, q,, q 7 (computed minus observed values) are also computed, as will be shown below. Then, the weighting matrix W is updated by means of some non-negative weighting function f(q i ) of the residuals q i. A possible way to do this is to compute the new values (which will be placed in the rightmost column of the following table) of the weights w ii, as suggested by Bube and Langan [14]: w ii ¼ 1 1 þ q i 1 4 where i = 1,,, 7, and e is a positive constant, called damping parameter, whose value must be chosen by the solver. By so doing, we search the minimum value of the following hybrid loss function

141 .10 Numerical Solution of the Least-Squares Estimation Problem 69 JðxÞ ¼ X7 i¼1 1 þ q 1 4 i 1 Bube and Langan suggest to take e roughly equal to 1/ times the standard deviation r of the residuals. In the present case, with reference to the residuals q 1, q,, q 7 given in the following table (which are computed with W = I), we have l ¼ q 1 þ q þ þq 7 ¼0: " ð r ¼ q 1 lþ þ ðq lþ þ þðq 7 lþ #1 ¼ 0: ¼ r 1:164 ¼ 0: and use this value of e to compute new values of the weights, as shown below. Obs. No. x Decl. comp. ( ) Decl. obs. ( ) Residual Weight When the new values of the weights w 11, w,, w 77 have been computed, the weighting matrix W is updated as follows W = diag(w 11, w,, w 77 ), and then the least-squares problem is solved again with the new weighting matrix, updated as indicated above. In the present case, the rightmost column of the preceding table, containing the weights to be used for the next iteration, has been filled as follows 1 w 11 ¼ " #1 ¼ 0: þ 0: : w ¼ " #1 ¼ 0: þ 0: :000961

142 70 Orbit Determination from Observations 1 w ¼ " #1 ¼ 0: þ 0: : w 44 ¼ " #1 ¼ 0: þ 0: : w 55 ¼ " #1 ¼ 0: þ 0: : w 66 ¼ " #1 ¼ 0: þ 0: : w 77 ¼ " #1 ¼ 0: þ 0: : and consequently the weighting matrix W is updated as follows W ¼ diagð0:9950; 0:8740; 0:65140; 0:5789; 0:64987; 0:87551; 0:9951Þ The iterative process described above comes to an end when the set of weights computed in a given iteration does not differ, within a specified tolerance, from the set computed in the preceding iteration. With an accuracy of five significant figures, two further iterations lead to the following results. Since the weights shown below are the same, within the accuracy of five significant figures, as those computed in the previous iteration, we stop here the iterative process. The final approximating polynomial for declination is then dðxþ ¼ 9:74 T 0 ðxþþ54:564 T 1 ðxþ8:77 T ðxþ9:085 T ðxþ 0:9966 T 4 ðxþþ0:67411 T 5 ðxþ Obs. No. x Decl. comp. ( ) Decl. obs. ( ) Residual Weight (continued)

143 .10 Numerical Solution of the Least-Squares Estimation Problem 71 (continued) Obs. No. x Decl. comp. ( ) Decl. obs. ( ) Residual Weight The final set of weights is given in the last column of the table shown above. Now, we use again the 7 7 identity matrix I as the first estimate of the weighting matrix W, in order to compute the weighted least-squares best fit for the observed values of right ascension. The 7 6 matrix A is the same as that shown above for the case of declination. The column vectors x and b contain now the six unknown coefficients a 0, a 1,, a 5 of the approximating polynomial aðxþ ¼a 0 T 0 ðxþþa 1 T 1 ðxþþ þa 5 T 5 ðxþ and the seven observed values of right ascension. These vectors are a 0 7:00 a 1 5:5 x 6. 7 b :75 a 5 The results found iteratively for right ascension are given below. First iteration (W = I): Obs. No. x R.A. computed ( ) R.A. observed ( ) Residual Weight where, again, the rightmost column of the table shown above contains the weights computed for the next iteration. The value of the damping parameter e results from the residuals q 1, q,, q 7 (computed minus observed values of right ascension) given above, as shown below:

144 7 Orbit Determination from Observations l ¼ q 1 þ q þ þq 7 ¼0: " ð r ¼ q 1 lþ þ ðq lþ þ þðq 7 lþ #1 ¼ : ¼ r 1:164 ¼ 1:9474 By updating the weighting matrix W (set previously equal to I) by means of the values contained in the rightmost column of the table shown above, the values of the coefficients a 0, a 1, a, a, a 4, and a 5 of the fifth-degree approximating polynomial are determined, as follows aðxþ ¼78:87 T 0 ðþ80:9 x T 1 ðxþ 9:000 T ðxþ 10:9 T ðxþ þ 6:09 T 4 ðxþ þ9:898 T 5 ðxþ These values make it possible to compute new values of right ascension; these, in turn, are compared with the observed values of right ascension; then the residuals q 1, q,, q 7 (computed minus observed values) are computed again. Third (and last) iteration: Obs. No. x R.A. comp. ( ) R.A. obs. ( ) Residual Weight Since the weights are the same, within the accuracy of five significant figures, as those computed in the previous iteration, we stop here the iterative procedure. The final approximating polynomial for right ascension is then aðxþ ¼78:78 T 0 ðxþ 80:89 T 1 ðþ9:09 x T ðxþ 10:8 T ðxþ þ 6:661 T 4 ðxþ þ9:889 T 5 ðxþ The final set of weights is given in the last column of the table shown above.

145 .11 The Kalman Filter 7.11 The Kalman Filter The batch estimation method described in the preceding paragraphs is based on the least-squares principle. It provides an estimate of the state vector x for a spacecraft at a given epoch by processing the whole amount of observations in each iteration. To this end, all the observations which are necessary to determine the spacecraft orbit must be available before the process of their improvement can take place. This requirement makes the method of batch estimation less desirable than other methods in real-time or near-real-time applications, which require a quasi-continuous update of information to produce an estimate of the state vector x. As has been shown by Vergez et al. [80], one of these applications is the tracking of an Earth-orbiting satellite by means of ground stations placed on the surface of the Earth. Since the number of such stations is limited, an orbiting satellite cannot be tracked continuously. For example, the following figure, due to the courtesy of NASA [57], shows the orbit, the US ground stations and their acquisition circles used to track the Landsat 7 satellite. A ground station can receive data from a satellite, when the satellite is within the acquisition circle of the station. When the link between the satellite and a ground station is lost, then the satellite position at some later time must be predicted, in order for another ground station to be able to establish a new link. The tracking data result from measurements (range, azimuth angle, altitude angle, and their rates of change with time) made on an orbiting spacecraft by established ground stations. Alternative data are range measurements between two spacecraft and positions determined by using the Global Positioning System (GPS).