Spaceflight Dynamics - PDF Free Download

Spaceflight Dynamics Prof. S. P. Bhat Department of Aerospace Engineering Indian Institute of Technology, Bombay April 21, 2006 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 1 / 97

Outline 1 Orbital Mechanics 2 Attitude Dynamics Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 2 / 97

Introduction Space engineering Supports astronomy, astrophysics, space sciences, telecommunications, military, meteorology Through spacecraft such as Interplanetary spacecraft Earth satellites Unmanned satellites Manned space stations Reusable space vehicles This course earth satellites Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 3 / 97

Where to put them? Orbit dictated by mission Orbit described in terms of shape, size and orientation Orbit depends on position and velocity at the start of orbital motion Orbital mechanics Description and prediction of orbital motion Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 4 / 97

Aristotle 384BC-322BC Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 5 / 97

Ptolemy 85AD-165AD Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 6 / 97

Copernicus 1473-1543 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 7 / 97

Galileo 1564-1642 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 8 / 97

Kepler 1571-1630 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 9 / 97

Newton 1643-1727 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 10 / 97

What do they do up there? Attitude dynamics (rotational motion) Description Variables Equations of motion Solutions Attitude control Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 11 / 97

How to put them there? Satellites injected by launch vehicles Initial conditions for orbital motion decided by burnout position and velocity Rocket performance Limited by structural mass Leads to staging Rocket trajectories Predict burnout conditions Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 12 / 97

Two-Body Problem Motion of two bodies moving under mutual gravitational acceleration m 1 Primary C.M. r 1 r c Secondary m 2 r 2 m 1 r 1 = Gm 1m 2 r 1 r 2 3 (r 1 r 2 ) m 2 r 2 = Gm 1m 2 r 1 r 2 3 (r 2 r 1 ) Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 13 / 97

Translation of Center of Mass Six degrees of freedom Three for motion of center of mass Three for relative motion (m 1 + m 2 ) r c = m 1 r 1 + m 2 r 2 = 0 Center of mass moves along a straight line with uniform velocity Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 14 / 97

Relative Motion In terms of displacement vector r = r 2 r 1 of secondary relative to primary m 1 second equation m 2 first equation = r = µr r 3, µ = G(m 1 + m 2 ) Central force motion with inverse square attraction Cannot be derived by using Newton s law directly Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 15 / 97

Energy Integral ṙ r = µ (r ṙ), r3 r = r = r r LHS = d ( ) 1 (ṙ ṙ) = d ( ) 1 dt 2 dt 2 v2 RHS = µ ( ) d 1 r 3 (r r) = 1 µ d dt 2 2 r 3 dt (r2 ) = µ r 3 rṙ = d ( µ ) dt r d ( 1 dt 2 v2 µ ) = 0 r }{{} E Specific energy E = 1 2 v2 µr 1 = constant Note: E total mechanical energy of the two-body system Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 16 / 97

Conclusions from the Energy Integral v = 2E + 2 µ r If E < 0, then v = 0 at r = µ 1 E Satellite falls back, orbit is bounded If E 0, satellite can be in motion at any distance Satellite escapes?? Escape speed at distance r def 2µ v esc = r Note: Escape verified as possible but not guaranteed Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 17 / 97

Angular Momentum Specific angular momentum H = r ṙ Ḣ = ṙ ṙ + r r = 0 H = Constant along the orbit r and ṙ lie in a fixed plane perpendicular to the constant H Orbit lies in a plane Only uses the fact that H has a constant direction Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 18 / 97

Areal Rate Area swept out by the radius vector in a small time increment t A = 1 2 r(t) r(t + t) = 1 2 t r(t) ṙ(t) = 1 2 t H da dt = 1 2 H = 1 2 H = constant Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 19 / 97

Areal Rate Kepler s law of areas: Radius vector sweeps out equal areas in equal interval of time So far: Speed as function of radius Planar nature of orbit Motion along the orbit Next: Shape Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 20 / 97

Eccentricity Vector Define eccentricity vector e def = µ 1 (ṙ H) r 1 r Lies in the plane of motion ė = 0 along motion Define true anomaly ν to be angle between r and e Eccentricity e def = e e µre cos ν = µr e = r (ṙ H)µr = H 2 µr r = H2 /µ 1 + e cos ν ν r e Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 21 / 97

Nature of the Orbit Polar equation of the orbit with r = H2 /µ 1 + e cos ν e as the positive x-axis Primary body as the origin Orbit bounded if and only if e < 1 Also the polar equation of a conic section of eccentricity e with origin at its focus Kepler s law of orbits: Orbit is a conic section with focus at its primary Conic section: curve of intersection between a right circular cone and a plane Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 22 / 97

Conic Sections Circle Ellipse Parabola Hyperbola Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 23 / 97

Shape and Size of the Orbit Shape determined by the eccentricity e 2H e = 2 E µ 2 + 1 e = 0 circular orbit 0 < e < 1 elliptic orbit e = 1 parabolic orbit e > 1 hyperbolic orbit Size determined by the semilatus rectum H 2 /µ Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 24 / 97

Circular Orbits Zero eccentricity r = H 2 /µ = constant For a circular orbit, H = rv = orbital speed at radius r v = E = 1 2 v2 µ r = 1 µ 2 r < 0 µ r Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 25 / 97

Elliptic Orbits 0 < e < 1, orbit is elliptical with one focus at the primary Periapsis (perigee/perihelion) point of closest approach at ν = 0 r p = H2 /µ 1 + e Apoapsis (apogee/aphelion) farthest point from the primary at ν = π r a = H2 /µ 1 e Semimajor axis a = r p + r a 2 = H2 /µ 1 e 2 H = µa(1 e 2 ) Speed at periapsis v p = H µ (1 + e) = r p a (1 e) Total specific energy E = 1 2 v2 p µ r p = µ 2a Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 26 / 97

Geometrical and Mechanical Description H = µa(1 e 2 ) a = µ 2E E = µ 2H 2a e = 2 E µ + 1 2 r p = a(1 e) r a = a(1 + e) v p = µ 1+e a 1 e v a = µ a (1 e) (1+e) Semiminor axis b = a 1 e 2 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 27 / 97

Parabolic Orbits e = 1, r = H2 /µ 1 + cos ν Periapsis distance r p = H2 2µ, v p = H r p = 2µ H v = 2µ r 0 as r E = 1 2 v2 p µ r p = 0 Just enough energy to reach at rest v esc = 2µ r is sufficient for escape. v v esc guarantees escape Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 28 / 97

Hyperbolic Orbits e > 1, orbit is one branch of a hyperbola with its focus at the primary speed v = def r as v v = π cos 1 1/e 2(E + µ r ) hyperbolic excess velocity v def = 2E as r Periapsis distance r p = H2 /µ 1 + e v p = H r p = µ(1 + e) H E = µ2 (e 2 1) 2H 2 v = µ H e2 1 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 29 / 97

8ν Geometric Description of Hyperbolic Orbits r ν H = v c sin ν = c µ H e2 1 c = H2 µ e e 2 1 Semimajor axis a = c r p = r p c a H 2 µ(e 2 1) = c e ( ) e2 1 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 30 / 97 e

Geometric versus Mechanical Description H = µa(e 2 1) a = µ 2E E = µ 2H 2 E e = 2a µ 2 + 1 r p = a(e 1) µ (e + 1) v p = a (e 1) µ v = a Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 31 / 97

Motion Along an Elliptic Orbit: Orbital Period Total area of orbit = πab = πa 2 1 e 2 Areal rate da dt = H 2 = 1 µa(1 e2 ) 2 a 3 Orbital period T = 2π µ Kepler s law of periods (period) 2 (semimajor axis) 3 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 32 / 97

Motion along an Elliptic Orbit: Kepler s Equation Need position along the orbit as function of time Use Kepler s law of areas need area of a sector of an ellipse Auxiliary circle A A O E ae r ν O A" a B Area(OAB) = b a Area(OA B) = b a [Area(O A B) Area(O A O)] Area(O A B) = 1 2 a2 E, Area(O A O) = 1 2 (ae)a sin E = 1 2 a2 e sin E Area(OAB) = 1 ab(e e sin E) 2 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 33 / 97

Kepler s Equation Let t p be the instant of periapsis passage Area(OAB) = 1 } t t p 2 H Define mean motion E e sin E = H(t t p) ab = Law of areas µ a 3 (t t p) Kepler s equation: n def = 2π T = µ a 3 E e sin E = n(t t p ) }{{} Mean anomaly M Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 34 / 97

True and Eccentric Anomalies Need to relate E and ν Use polar equation of the orbit O A = OO + OA a cos E = ae + r cos ν cos E = (e + cos ν) (1 + e cos ν) 2 sin 2 (E/2) = 1 cos E = (1 e)2 sin2 (ν/2) (1 + e cos ν) 2 cos 2 (E/2) = 1 + cos E = (1 + e)2 cos2 (ν/2) (1 + e cos ν) (1 e) tan (E/2) = tan (ν/2) (1 + e) Use along with Kepler s equation to find ν as function of time Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 35 / 97

Geocentric Frame Need to describe orientation of the orbit or position of the satellite With respect to an earth/inertial frame Using quantities that help to visualize the orbit Define a non-rotating geocentric frame with Origin at earth s center Axes directions fixed with respect to solar system Z axis along earth s axis of rotation pointing north Precesses with a period of 25, 800 years Nutates with an amplitude 9 and period 18.6 years X axis along the line of intersection of earth s orbital plane (ecliptic) and earth s equatorial plane Along line joining the equinoxes, pointing along the vernal equinox Along the first point of Aries Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 36 / 97

The Ecliptic Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 37 / 97

The Ecliptic: View from Earth Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 38 / 97

First Point of Aries Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 39 / 97

Orientation of the Orbit: Right Ascension H Z e Descending Node X Orbit Ω ω Y i Ascending Node Line of Nodes Ascending node point where the orbit crosses equatorial plane from S to N Right ascension of ascending node Ω Eastward from the X axis to the ascending node 0 Ω < 2π Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 40 / 97

Orientation of the Orbit: Inclination Inclination of the orbit i Measured at the ascending node between east and direction of motion 0 i < π i < 90 prograde orbit, orbital motion in the same direction as earth s rotation i > 90 retrograde orbit i 90 polar orbit i = 0 equatorial orbit Inclination determines north and south limits of visibility Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 41 / 97

Orientation of the Orbit: Argument of Perigee H Z e Descending Node X Orbit Ω ω Y i Ascending Node Line of Nodes Argument of perigee ω measured in the orbital plane from the ascending node along the motion 0 ω < 2π Ω, i, ω describe the orientation of the orbit with respect to the geocentric frame In addition, a determines the size, e the shape Six classical orbital elements a, e, i, ω, Ω, t p Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 42 / 97

Determination of Classical Elements from Initial Conditions Given r and v = ṙ Compute E, H, H, e, e Line of nodes is perpendicular to H and k Unit vector along the line of nodes (pointing to the ascending node) n = k H 1 (k H) Ω [0, 2π) from n = cos(ω)i + sin(ω)j Compute i [0, π] from cos i = k H H Compute ω [0, 2π), the angle between n and e, by cos ω = ω = cos 1 n e e, e k 0 = 2π cos 1 n e e, e k < 0 n e e Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 43 / 97

Determination of Classical Elements (continued) To find t p, first find initial anomaly ν from cos ν = e r er ν = cos 1 e r er, = 2π cos 1 e r er, Compute initial eccentric anomaly v e 0 (satellite traveling from perigee to apogee) v e > 0 (satellite traveling from apogee to perigee) tan(e/2) = Compute t p from Kepler s equation (1 e) (1 + e) tan(ν/2) µ a 3 (t 0 t p ) = E e sin E Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 44 / 97

Determination of Position and Velocity Given t, a, e, i, Ω, ω, t p Compute eccentric anomaly at t E e sin E = µ a 3 (t t p) Compute true anomaly at t (1 e) tan(ν/2) = tan(e/2) (1 + e) Compute geocentric distance at t r = a(1 e2 ) 1 + e cos ν Position in the orbital plane determined Need to transform to geocentric coordinates Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 45 / 97

Position in Perifocal Frame Introduce a perifocal coordinate system, origin at earth s center and unit vectors p pointing to the perigee q along the position ν = 90 w orthogonal to the orbital frame such that p q = w q w Z p X Ω ω i Y r = r cos ν p + r sin ν q Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 46 / 97

Velocity in Perifocal Frame ṙ = (ṙ cos ν r ν sin ν)p + (ṙ sin ν + r ν cos ν)q To find r ν, note H = r ṙ = r 2 νw r ν = H µ r = a(1 e 2 (1 + e cos ν) ) To find ṙ, differentiate polar equation µ ṙ = a(1 e 2 ) e sin ν Perifocal components of velocity µ ṙ = a(1 e 2 [ sin ν p + (e + cos ν) q] ) Need to transform to geocentric frame Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 47 / 97

Transformation to Geocentric Frame To obtain the transformation, perform a sequence of 3 rotations on G to get P Rotate G about Z through Ω to get G1 Rotate G1 about X through i to get G 2 Rotate G2 about Z through ω to get P For any vector r r G = R 1 (Ω)(r G1 ) r G1 = R 2 (i)(r G2 ) r G2 = R 3 (ω)(r P ) r G = R 1 (Ω)R 2 (i)r 3 (ω)r P Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 48 / 97

Transformation Matrices G X Ω G 1 Y R 1 (Ω) = cos Ω sin Ω 0 sin Ω cos Ω 0 0 0 1 G 2 Z i G 1 Y R 1 (i) = 1 0 0 0 cos i sin i 0 sin i cos i G 2 X ω P Y R 1 (ω) = cos ω sin ω 0 sin ω cos ω 0 0 0 1 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 49 / 97

Geocentric Components of Position and Velocity r cos ν r G = R 1 (Ω)R 2 (i)r 3 (ω) r sin ν 0 sin ν µ ṙ G = a(1 e 2 ) R 1(Ω)R 2 (i)r 3 (ω) e + cos ν 0 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 50 / 97

Complete Solution of the Two-Body Problem m 1 Primary C.M. r 1 r c Secondary m 2 r 2 r 2 r c = m 1 m 1 + m 2 r r 1 r c = m 2 m 1 + m 2 r Each body moves along a conic section with focus at the center of mass Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 51 / 97

Satellite Tracking and Orbit Determination Predicting orbit from measured position and velocity data Position and velocity known only at injection point from launch vehicle INS Optical tracking Each observation yields right ascension and declination, no range information Three observations required to determine orbit Observations made from rotating, translating earth Approximate method by Laplace, exact method by Gauss Radar tracking for low-earth satellites Azimuth, elevation, range in each observation Some method interpolate between closely spaced observations, differentiate to get velocity Other use two position measurement with elapsed time Range range-rate tracking for deep space craft Range-rate measured by using Doppler shift No angular information available Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 52 / 97

Errors in Orbit Determination Measurement errors lead to errors in estimated orbital parameters Errors between actual position and estimated position grows with time Example: error in period Need for improving accuracy by making new observations and updating the orbit Need for correcting the orbit Body of observations increases with time Use all data rather than the minimum amount required Best fit method of least squares Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 53 / 97

Carl Friedrich Gauss Number theory Astronomy Statistics Analysis Differential geometry Geodesy Geomagnetism Carl Friedrich Gauss 1777-1855 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 54 / 97

Orbital Maneuvers Needed to transfer a geostationary satellite from its low earth parking orbit to its final high altitude geostationary orbit Needed to correct changes in orbital elements due to perturbing forces Impulsive thrust maneuvers Velocity changes instantaneously without change in position Thrust duration (burn times) small compared to orbital period (coast time) Hohmann transfer between two coplanar circular orbits Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 55 / 97

Hohmann Transfer Two impulsive maneuvers Apogee boost: increase speed from vc1 to v 1 so that the satellite enters an elliptical transfer orbit with apogee on the final orbit Circularization: increase speed at the apogee of the transfer orbit to enter the final circular orbit Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 57 / 97

Hohmann Transfer (cont d) Circular orbits Transfer orbit a = a 1 + a 2, v 1 = 2 Impulse magnitudes v c1 = 2 ( µ a 1 µ µ, v c2 = a 1 a 2 µ a 1 + a 2 ) ( µ, v 2 = 2 a 2 µ a 1 + a 2 ) v 1 = v 1 v c1 v 2 = v c2 v 2 Minimum duration between maneuvers = T 2 = π a 3 µ Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 58 / 97

Inclination Change Maneuver Needed if geostationary satellite is not launched from the equator Combined with one of the maneuvers (usually second) of the Hohmann transfer To calculate magnitude and direction of the impulse required v 2 2 = v c2 2 + v 2 2 2v c2 v 2 cos i Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 59 / 97

Coordinate Transformation View classical elements as new coordinates x def = [ r x r y r z v x v y v z ] T = Φ (a, e, i, Ω, w, M) Inverse transformation known [ a e i Ω w M ] T = Φ 1 (x) Equations of motion in the two-body problem ẋ = [ r x r y r z v x v y ] T v z = f (rx, r y, r z, v x, v y, v z ) = f(x) Use transformation to write equation of motion in terms of classical elements [ȧ ė i Ω ] 1 T Φ ẇ Ṁ = f (Φ (a, e, i, Ω, w, M)) x In the two-body problem [ȧ ė i Ω ẇ Ṁ ] T = [ 0 0 0 0 0 n ] T Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 60 / 97

Perturbation Forces Inhomogeneity and oblateness of earth Third body gravitational influence, eg. sun, moon Solar wind Solar radiation pressure Atmospheric drag in low earth orbit ẋ = f(x) + p(x, t) }{{} perturbation [ȧ ė i Ω ẇ Ṁ ] T = [ 0 0 0 0 0 n ] T + perturbation Trajectory no longer a conic section Can be thought of as path traced by a point on an ellipse that is osculating, that is, changing shape, size and orientation x(t) = Φ (a(t), e(t), i(t), Ω(t), w(t), M(t)) Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 61 / 97

Gauss s Planetary Equation Resolve perturbation force along perifocal frame Perturbation force = P p + Qq + W w ȧ = ė = i = Ω = ω = Ṁ = 2 n [ep sin ν + (1 + e cos ν) Q] 1 e2 1 e 2 [P sin ν + (cos E + cos ν)q] na 1 na rw cos (ν + ω) 1 e 2 a 1 na rw sin(ν + ω) 1 e 2 a sin i Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 62 / 97

Earth Inhomogeneity and Oblateness Gravitational potential due to earth in spherical coordinates U(r, λ, φ) = µ r + B(r, λ, φ) }{{} perturbation B(r, λ, φ) = µ ( ) n Re J n P n (sin λ) + r r n=2 }{{} oblateness n ( ) n Re J mn (C nm cos mφ + S nm sin mφ) P nm (sin λ) r }{{} m=1 asymmetry Re = mean equatorial radius Pn = Legendre polynomials Pnm = Legendre functions of the first kind Jn, C nm, S nm = coefficients Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 63 / 97

Effect of J 2 Perturbation J 2 is two orders of magnitude larger than others Arises from the first-order deviation of the oblate earth from a sphere Small periodic changes in a, e, i with Secular changes in Ω, w, M Regression of nodes: dω dt = 3 2 n ȧ 0, ė 0, i 0 J2 cos i (1 e 2 ) 2 ( ) 2 Re a dw Advance of perigee: dt = 3 4 n (1 5 cos 2 i) J2 (1 e 2 ) 2 Change in mean anomaly: dm dt ( ) 2 Re a = n + 3nJ2(3 cos2 i 1) 4(1 e 2 ) 3/2 ( ) 2 Re a Superimposed periodic variations + secular and periodic variations due to higher order terms Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 64 / 97

Application: Sun Synchronous Orbits Orbits that have a nodal regression rate of 360 per year Orbital plane makes a constant angle with respect to sun Sun Satellite revisits any point at the same local time Useful for earth observation satellites Solar illumination the same in pictures takes at different times Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 65 / 97

Launch to Rendezvous Launch a spacecraft to rendezvous with a space station already in the orbit Problem: Find time of launch so that both orbits are coplanar Orbital plane changes after injection is expensive Turning the launch vehicle into the required plane is also expensive Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 66 / 97

Launch to Rendezvous Launch a spacecraft to rendezvous with a space station already in the orbit Problem: Find time of launch so that both orbits are coplanar Orbital plane changes after injection is expensive Turning the launch vehicle into the required plane is also expensive Solution: Launch when launch site lies in the space station orbital plane For a given latitude, this occurs at most twice in every sidereal day Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 67 / 97

Projection of the Orbit 100 80 90 N 60 Geographical Latitude (deg) 40 20 0 O E 20 40 180 E 180 W 0 W 60 80 100 0 50 100 150 200 250 300 350 Geographical Longitude (deg) 90 S Ground Trace 100 80 90 N 60 Geographical Latitude (deg) 40 20 0 O E 20 40 180 E 180 W 0 W 60 80 100 0 50 100 150 200 250 300 350 Geographical Longitude (deg) 90 S Orbital Projection Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 68 / 97

Geometry of Coplanar Launch to Rendezvous α φ Ω i δ Ascending Node λ A 13.75 deg N 0 deg γ 0 deg 80.25 deg E Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 69 / 97

Launch Times for Rendezvous Two solutions which add up to 180 Right ascension of launch site sin δ = tan λ cot i Right ascension of Greenwich meridian Launch time α = α 0 + = α + φ = Ω + δ 2π T sidereal (t t 0 ) t = t 0 + T sidereal (Ω + δ φ α 0 ) 2π Two solutions Launch azimuth sin A = cos i cos λ Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 70 / 97

Rotational Motion of Satellites Orbital dynamics: satellites treated as point masses Rotational motion as extended bodies has to be considered Attitude maneuvering Pointing requirements of optical, communications, imaging payload Solar panel orientation Thruster orientation for orbital maneuvers and station keeping Treat satellite as a rigid body Collection of particles such that the distance between any two remains fixed Six degrees of freedom, 3 translational + 3 rotational Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 71 / 97

Translational Dynamics of a Rigid Body Consider a rigid arrangement of a finite number of particles For the i th particle Fext def = F i ext + i=1 N F ij = m i a i j i N N N F i ext + i=1 j i F ij }{{} =0 Fext = Macm, acm = = N m i a i i=1 mi a i mi Translates as a point particle of mass M located at center of mass under Fext Rotational and translational motions decoupled? Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 72 / 97

Rotational Dynamics of a Rigid Body Take moments of Newton s law about some convenient point R O d ρ dm r CM Inertial Frame df = dm r = dm ( R + d + ρ) dm O = (d + ρ) df = (d + ρ) ( R + d + ρ)dm Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 73 / 97

Rotational Dynamics of a Rigid Body (cont d) M O = = [ (d + ρ) ( d + ρ)] dm + (d R)dm + d [ (d + ρ) ( ] d + ρ) dm +(d R) dm dt }{{}}{{} dh M O( ) + ρdm R } {{ } =0 = d dt H O + m(d R) (ρ R)dm If O is inertially fixed ( R = 0) or the center of mass (d = 0) then M O = d dt H O Attitude dynamics equation HO = moment about O of linear momentum relative to O Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 74 / 97

Attitude Representation Consider two right handed orthonormal frames I with unit vectors l, m, n, B with unit vectors i, j, k Components of any vector v along I and B (v) I = (v) I = [ v.l v.m v.n ] T, (v)b = [ v.i v.j v.k ] T vb 1 i.l + v2 B j.l + v3 B k.l i.l j.l k.l vb 1 i.m + v2 B j.m + v3 B k.m = i.m j.m k.m vb 1 i.n + v2 B j.n + v3 B k.n i.n j.n k.n vb 1 vb 2 vb 3 = [ ] (i) I (j) I (k) I (v)b There exists a unique matrix R such that R(v) B = (v) I for every v R determined solely by orientation of B relative to I R special orthogonal matrix, rotation matrix, direction cosine matrix R T R = I detr = 1 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 75 / 97

An Alternative Situation Rotate a frame to go from I to B What are the new I-components of a vector fixed to the moving frame? I B Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 76 / 97

An Alternative Situation Rotate a frame to go from I to B What are the new I-components of a vector fixed to the moving frame? I B old I components = new B components new I components = R (new B components) = R (old I components) R relates Components of a given vector in two frames Components of a rotated vector to its original components in the same frame Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 76 / 97

Rotation Matrix for an Elementary Rotation Rotate I about a unit vector v through an angle θ to obtain B Transformation matrix (v) I = (v) B def = v R 3 R(v, θ) = I + (1 cos θ)(v ) 2 + sin θ(v ) 1 0 0 0 v 3 v 2 I = 0 1 0, (v ) = v 3 0 v 1 0 0 1 v 2 v 1 0 If (u)b = u R 3, then (v u) B = (v )u Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 77 / 97

Elementary Rotation (cont d) Check: Rv = v Let u v, u u = 1. Let w be obtained from u by rotation about v v u θ w u v (w) B = (u) I def = u w = cos θu + sin θ(v u) = u + (1 cos θ) (v (v u)) + sin θ(v u) (w) I = [I + (1 cos θ)(v ) 2 + sin θ(v )]u = R(w) B Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 78 / 97

Composite Rotations α ψ/ u θ/ v φ/ w C A B (x) I (x) C (x) A (x) R 1 ( u, ψ) R 2 ( v, θ) R 3 ( w, φ) B R composite = R 1 (u, ψ)r 2 (v, θ)r 3 (w, φ) Composition of rotations matrix multiplication on the right Noncommutativity of matrix multiplication non commutativity of rotations Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 79 / 97

Relative Motion Between Frames Frame B rotates relative to I (not necessarily about a fixed axis) r 1, r 2, r 3 vectors fixed in B, r 1, r 2 linearly independent ṙ 1, ṙ 2, ṙ 3 instantaneous derivatives with respect to I r1 r 1 = constant ṙ 1 r 1 = 0 r1 r 2 = constant ṙ 1 r 2 + r 1 ṙ 2 = 0 r 3 = α 1r 1 + α 2r 2 + α 3(r 1 r 2) ṙ 3 = α 1ṙ 1 + α 2ṙ 2 + α 3(ṙ 1 r 2 + r 1 ṙ 2) = (ṙ 1 ṙ 2) ṙ 3 = 0 Instantaneous derivative of every vector fixed in B lies in the plane perpendicular to (ṙ 1 ṙ 2 ) ṙ = α(r)(e r) e = unit vector along (ṙ1 ṙ 2) Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 80 / 97

Relative Motion (cont d) r 1, r 2 vectors fixed in B, linearly independent from e ṙ 1 = α(r 1 )(e r 1 ) ṙ 2 = α(r 2 )(e r 2 ) r 2 ṙ 1 + ṙ 2 r 1 = 0 = α(r 1 ) = α(r 2 ) = constant There exists a vector ω such that instantaneous derivatives of vectors in B are given by ṙ = ω r ω = instantaneous angular velocity of B relative to I Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 81 / 97

Attitude Kinematics B rotates relative to I Instantaneous relative orientation described by rotation matrix R(t) How does R vary? For any vector fixed in B, (ṙ) I = d dt (r) I = d dt R(r) B = Ṙ(r) B + R d dt (r) B }{{} =0 (ṙ) I = (ω r) I = R(ω r) B = R(ω )(r) B Ṙ = R(ω ) Attitude kinematics equation 0 ω 3 ω 2 (ω ) = ω 3 0 ω 1 ω 2 ω 1 0 ω =column vector of B components of instantaneous angular velocity of B relative to I Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 82 / 97

Attitude Kinematics Equation Used for Navigation Control design Simulation Solution involves integrating 9 differential equations R contains 9 elements subject to 6 constraints, only 3 free parameters Question: Is it possible to parametrize rotation matrices with fewer parameters? If yes, rewrite attitude kinematics in terms of fewer parameters Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 83 / 97

Euler Angles Fact: Given a sequence of unit vectors v 1, v 2, v 3 fixed in I such that no two consecutive vectors are linearly dependent, I can be rotated to any desired orientation by a sequence of three rotations, one each about v 1, v 2, v 3 For every rotation matrix R, there exist ψ, θ, φ such that R = R 1(v 1, ψ)r 2(v 2, θ)r 3(v 3, φ) Examples: 3-2-1 Euler angles used in aircraft; v1 = n, v 2 = m, v 3 = l 3-1-3 Euler angles; v1 = n, v 2 = l, v 3 = n Problem: Euler angles do not combine well for successive rotations C ψ 1 θ 1 φ 1 ψ 2 θ 2 φ 2 I??? B Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 84 / 97

Axis-Angle Variables Euler s Theorem: Given two frames I and B, there exists an axis-angle pair (v, θ) such that I coincides with B when rotated about v through θ For every rotation matrix R, there exist v R 3, θ [0, 2π) such that R = R(v, θ) = I + (1 cos θ)(v ) 2 + sin θ(v ) q 1 1 3 3 3 2 q 2 2 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 85 / 97

Axis-Angle Variables Euler s Theorem: Given two frames I and B, there exists an axis-angle pair (v, θ) such that I coincides with B when rotated about v through θ For every rotation matrix R, there exist v R 3, θ [0, 2π) such that R = R(v, θ) = I + (1 cos θ)(v ) 2 + sin θ(v ) θ 1 1 3 3 3 2 q 2 2 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 86 / 97

Axis-Angle Variables Euler s Theorem: Given two frames I and B, there exists an axis-angle pair (v, θ) such that I coincides with B when rotated about v through θ For every rotation matrix R, there exist v R 3, θ [0, 2π) such that R = R(v, θ) = I + (1 cos θ)(v ) 2 + sin θ(v ) θ 1 1 3 3 2 3 θ 2 θ 1 θ 1/2 a b θ 2 /2 2 1 θ 2 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 87 / 97

Axis-Angle Variables Euler s Theorem: Given two frames I and B, there exists an axis-angle pair (v, θ) such that I coincides with B when rotated about v through θ For every rotation matrix R, there exist v R 3, θ [0, 2π) such that R = R(v, θ) = I + (1 cos θ)(v ) 2 + sin θ(v ) Problem: Axis angle variables do not combine well for successive rotations v 1 θ 1 C v 2 θ 2 I??? B Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 88 / 97

Leonhard Euler Leonhard Euler 1707-1783 Rigid body motion Fluid mechanics Solid mechanics Number theory Real and complex analysis Calculus of variations Differential geometry and topology Differential equations Mathematical notation Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 89 / 97

Attitude Kinematics with Euler Angle To obtain attitude kinematics in terms of ψ, θ, φ, substitute R = R 1 (v 1, ψ)r 2 (v 2, θ)r 3 (v 3, φ) in Ṙ = R(w ) Solve for ψ, θ, φ Example: 3-2-1 Euler angles φ θ ψ = 1 0 sin θ 0 cos φ sin φ cos θ 0 sin φ cos φ cos θ 1 } {{ } singular at θ=±90 Fact: Every three-parameter representation of attitude possesses a kinematic singularity ω 3 ω 2 ω 1 Euler angles suitable only for simulating limited angular motion Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 90 / 97

Quaternions To multiply quaternions, use q = q }{{} 0 + q 1 i + q 2 j + q 3 k }{{} Real part Imaginary part ii = jj = kk = 1, ij = ji = k, jk = kj = i, ki = ik = j Multiplication noncommutative Conjugate q = q 0 q 1 i q 2 j q 3 k Magnitude = qq = q 2 1 + q2 2 + q2 3 + q2 4 If v = [v 1 v 2 v 3 ] T, then define v = v 1 i + v 2 j + v 3 k Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 91 / 97

Quaternion Representation of Rotations If I is rotated through θ about unit vector v to obtain B, set q = cos θ 2 + sin θ 2 v Unit quaternion If x B = (x) B and x I = (x) I then x I = q x B q No trigonometric formulae q 1 B I = q B I = q I B q B I = q C I q B C Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 92 / 97

William Rowan Hamilton Algebra Optics Mechanics William Rowan Hamilton 1805-1865 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 93 / 97

Attitude Dynamics Revisited M = Ḣ M = moment of external forces about center of mass H = angular momentum of body about center of mass Derivative with respect to inertial frame Let B be a body fixed frame with unit vectors i, j, k H = H 1 i + H 2 j + H 3 k, ω = ω 1 î + ω 2 ĵ + ω 3 k, M = M 1 î + M 2 ĵ + M 3 k, H B = [H 1 H 2 H 3 ] T ωb = [ω 1 ω 2 ω 3 ] T MB = [M 1 M 2 M 3 ] T Ḣ = Ḣ1î + Ḣ2ĵ + Ḣ3 k + (ω H) (M) B = (Ḣ) B = d dt H B + (ω H) B = d dt H B + (ω B )H B d dt H B = (ω B )H B + M B Attitude dynamics equation in terms of body components Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 94 / 97

Body Component of Angular Momentum R O d ρ dm r CM Inertial Frame Let H B (ρ ρ)dm = ρ (ω ρ)dm = (ρ (ρ ω))dm 0 z y ρ= xî + yĵ + z k, ρ B = [x y z ] T, (ρ B ) = z 0 x y x 0 = (ρ (ρ ω)) B dm = (ρ B )(ρ ω) B dm = ((ρ B ) 2 ω B )dm (y 2 + z 2 )dm xydm xzdm I B = (ρ B ) 2 dm = xydm (x 2 + z 2 )dm yzdm xzdm yzdm (x 2 + y 2 )dm H B = I B ω B I B = moment-of-inertia matrix about body axes Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 95 / 97

Principal Axes of Inertia Moment-of-inertia matrix about inertial axes H I = RH B = RI B ω B = RI B R T ω I = I I = RI B R T II varies as body rotates Two body frames, B amd B, related by rotation matix R H B = I B ω B, H B = I B ω B H B = RH B = RI B ω B = RI B R T ω B, I B = RI B R T I 1 0 0 Fact: There exsists a rotation matrix R such that RI B R T = 0 I 2 0 0 0 I 3 There exists a body frame B such that I B = digonal Corresponding axes are principal axes of inertia, I1, I 2, I 3 are principal moment of inertia Principal axes are along eigenvectors of IB Principal moments of inertia are eigenvalues of IB Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 96 / 97

Euler s Equation for Rotational Dynamics d dt H B = (ω B )H B + M B Put H B = I B ω B, I B constant I B ω B = (ω B )I B ω B + M B Determine evolution of angular velocity component Determine rotational motion together with attitude kinematics equation Euler s equation written for principal axes of inertia ω 1 = (I 3 I 2 ) I 1 ω 2 ω 3 + 1 I 1 M 1 ω 2 = (I 1 I 3 ) I 1 ω 1 ω 3 + 1 I 2 M 2 ω 3 = (I 2 I 1 ) I 3 ω 1 ω 2 + 1 I 3 M 3 Prof. S. P. Bhat (IITB) Spaceflight Dynamics April 21, 2006 97 / 97