Session 5: Review of Classical Astrodynamics

Transcription

1 Session 5: Review of Classical Astrodynamics In revious lectures we described in detail the rocess to find the otimal secific imulse for a articular situation. Among the mission requirements that serve as constraints, is the change in velocity, Δv. The urose of the roulsion system is to rovide this Δv, which deends on the articular trajectory followed by the sacecraft and is determined by the orbital mechanics at lay. Because of this, the use of concets in astrodynamics becomes very relevant to sace roulsion. In this Lecture, we resent a short review of the fundamental laws in astrodynamics (Further reading: R. Battin, An Introduction to the Mathematics and Methods of Astrodynamics, AIAA Education Series, 1999). We start by analyzing the two-body roblem with central forces. Such analysis rovides the fundamentals of most orbital mechanics roblems. Consider the situation deicted in the figure below: x z m 1 r 1 r = r r 1 r y m We have two oint masses interacting through central (gravitational) forces. Under no external forces, the equations of motion for these articles are, d r 1 r r 1 m 1 = Gm 1 m dt r r 1 3 (1) d r r r 1 m = Gm 1 m dt r r 1 3 Defining the relative osition vector r = r r 1, we subtract Eqs. (1) and obtain a single equation of motion, d r r d v r + µ = 0 or + µ = 0 () dt r 3 dt r 3 where µ = G(m 1 + m ) and v = d r/dt. In most cases of interest in astrodynamics, one of the masses dominates over the other. For the articular situation in the figure, we could assume the motion of m 1 is negligible for m 1» m, therefore µ = Gm 1. This is a erfectly accetable aroximation for orbital motion about lanetary bodies. In the case of earth, µ E = 398,601 km 3 /s. This innocently looking equation () ket occuied several of the most rodigious minds in mathematics, like Gauss and Euler, ever since Newton roclaimed his gravitational law and in arallel with Leibniz, develoed infinitesimal calculus. In general, there are two ways 1

2 to tackle this equation: i) through numerical integration, by taking advantage of today s comutational ower (a luxury, that neither Gauss of Euler ever hoed to ta into) and ii) analytically, by uncovering a number of integrals of motion related to the structure of this non-linear differential equation. The analytical method, as exected, is much more enlightening and will be exlored in some detail in this lecture. We begin by taking the cross roduct of vector r with Eq. (), ( ) d v r d v d( r v) d r d( r v) r + µ = r = v = = 0 (3) dt r 3 dt dt dt dt where r r = v v = 0. From here, we see that the resulting vector r v must be constant. We define this constant as, h = r v (4) which reresents a massless angular momentum, which is conserved when no external forces are alied to the masses. The angular momentum is our first integral of motion. We now take Eq. () and calculate its cross roduct with h, d ( v ) r h + µ h = 0 (5) dt r 3 We use the vector identity r h = r ( r v) = r( r v) r v and notice that r v = rv r, where v r = dr/dt is the radial comonent of the velocity vector. The second term in Eq. (5) can be written as (excluding µ), Using this result, we can write Eq. (5) as, r ( ) h v r v d r = r = (6) r 3 r r dt r d dt This equation can be directly integrated, ( d r v ) ( ) h µ = 0 (7) dt r r v h µ = µ e (8) r where the constant (vector) of integration was selected as µ e for convenience, since later on will be identified as the eccentricity of a geometrical orbit (times µ). The eccentricity is our second integral of motion. We calculate the magnitude of the eccentricity vector, e e = ( v ) h r µ r (9)

3 Exanding the square on the right hand side, v ) ( ) h r v h = µ r µ ( ( ) v h r + 1 µr (10) Here we note that ( v h) = v h since v h. For the last term in Eq. (10), we use the vector identity v h = v ( r v) = rv rv r v, and therefore, ( v h) r = r (v v r) = r v θ = h The magnitude of the eccentricity vector becomes, e v h h 1 µ µ ( ) = + 1 E v = e (11) µ T 1 µr r h which is known as the vis-viva integral and reresents the total energy ET (er unit mass), our third and last integral of motion. In this way, without exlicitly solving the differential equation Eq. (), it is ossible to write exressions for the two-body roblem in terms of integrals of motion. To see this, use Eq. (8) to calculate the dot roduct e r, ( v h) r h e r = r = (1) µ µ r Define f as the angle between the eccentricity and radial vectors (known as the true anomaly) and solve for r, h /µ r = or r = (13) 1 + e cos f 1 + e cos f Eq. (13), known as the equation of orbit, is the solution in sace (not in time) of the differential Eq. (). It describes the relative lanar motion of the two oint masses considered in the roblem. A closer look at Eqs. (11) and (13) reveals three motion regimes that deend on the magnitude of the eccentricity vector. 0 e < 1 E T < 0 bounded ellitic e = 1 E T = 0 unbounded (zero velocity at ) arabolic e > 1 E T > 0 unbounded hyerbolic Unbounded trajectories are interesting, esecially in sace roulsion where in some instances the objective is to achieve an escae trajectory for exloration missions. However, for the moment, we focus on bounded (ellitical) orbits. Eventually, we will look at the result of erturbing these orbits through small forces rovided by electric roulsion. The lanar motion in the bounded case (0 e < 1) defines an ellitical trajectory (circular, if e = 0), as shown in the figure below. From Eq. (13), we see that when f = π/ the radius 3

4 ! r f! e a becomes h /µ. This quantity is tyically reresented by and is known as the arameter of the orbit. The eriaxis and aoaxis can be found directly from Eq. (13) with f = 0 and f = π, resectively, r = and r a = (14) 1 + e 1 e Since (twice) the semi-major axis of the orbit is a = r a + r, we find, Consequently, the orbital energy in Eq. (11) can be written as, = a(1 e ) (15) µ E T = (16) a Given the eccentricity and semi-major axis, we use Eq. (13) to find the orbital radius as a function of the true anomaly. Then, we use Eq. (11) to find the orbital velocity at that articular location. What remains is orbital timing. In other words, given an initial time and osition, find where the orbiting object will be some time afterwards. Use the conservation of angular momentum, h = r (df/dt), and substitute in the equation of orbit, Aroriately, this equation is known as Keler s equation. df µ = dt (17) (1 + e cos f) 3 The eriod of the orbit could be easily found by integrating Keler s law (orbit swees equal areas in equal u times) since the area of the ellise is πab and the semi-minor axis is b = a 1 e = h a/µ, the familiar result is, a T = π 3 (18) µ We mentioned earlier that a number of brilliant mathematicians worked intensively in solving the two-body roblem. In fact, most of that work was devoted to solve Keler s equation (17). Once the orbital timing is found, the solution of the two-body roblem is comlete. There are a number of different ways in which this equation could be written, each one with its own rocedure to find an aroximate solution. In general, Eq. (17) could be solved 4

5 numerically, but of course, little is gained that way since, after all, the original equation of motion Eq. () could also be solved numerically. At the end, aroximate solutions using iterative methods, are extremely accurate. It is of course, more interesting for our articular alication to solve the equations of motion Eq. (1) when there are forces acting on the masses (from thrusters, for instance). In articular, if we assume a force F is alied to m, the resulting equation of motion will be, d v r + µ = a c ( r, t) (19) dt r 3 where a c = F /m is the acceleration due to external forces (other than gravity) acting on the moving mass. This new equation brings no additional difficulty if the roblem is solved numerically. However, our analytical aroach is criled since it is no longer ossible to find integrals of motion (neither the angular momentum, eccentricity or energy are longer constant). It is said that the resulting trajectories are non-kelerian. Desite this, if the acceleration is a small quantity (which is tyical for the tye of maneuvers using electric roulsion), the constants of motion will drift slowly. This allows us to follow these changes and rovide accurate redictions of the actual trajectory. Not surrisingly, this lies in the realm of the method of variation of arameters. For now, let us finalize this review by writing the vectorial equation of motion, Eq. (19), as two differential equations in olar coordinates. This will be useful later on to find analytical solutions to some simle roblems where small accelerations are imarted and motion is confined to a constant orbital lane: ( ) d r dθ µ d θ dr dθ a θ r + = a r and + = (0) dt dt r dt r dt dt r 5

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