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Orbital Dynamics: Formulary 1 Introduction

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Orbital Dynamics: Formulary 1 Introduction Orbital Dynamics: Formulary Prof. Dr. D. Stoffer Department of Mathematics, ETH Zurich 1 Introduction Newton's law of motion: The net force on an object is equal to the mass of the object multiplied by its acceleration. (1) F(t) = ma(t) where F(t) : the net force acting on the object at time t. m : the mass of the object. x(t) : position of the object at time t, in an inertial frame. v(t) = _x(t) : velocity of the object at time t. a(t) = ¨x(t) : acceleration of the object at time t. Newton's law of gravity: The attractive force F between two bodies is proportional to the product of their masses m1 and m2, and inversely proportional to the square of the distance r between them: m m (2) F = G 1 2 : r2 The constant of proportionality, G, is the gravitational constant. G = (6:67428 ± 0:00067) × 10−11 m3 kg−1 s−2 = (6:67428 ± 0:00067) × 10−11 N m2 kg−2 : The n{body problem n point masses mi with positions Ri (with respect to an inertial frame) move under the influence of gravity. Let rij := Rj − Ri be the position of the point mass mj relative to the position of the point mass mi. The equations of motion are n X mimj (3) m R¨ = G e ; (i = 1; 2; : : : ; n) i i r2 ij j=1;(j6=i) ij 1 P P where eij := rij. Let R := miRi= mi be the centre of mass. rij i i The 10 classical first integrals (conserved quantities) (4) R(t) = C1t + C2 n X mi(Ri × R_ i) = C3 (angular momentum)(5) i=1 T + V = C4 (total energy)(6) 1 Pn _ 2 Pn P mimj T := miR is the kinetic energy and V := −G is a potential 2 i=1 i i=1 j6=i rij of the force field F. The components of C1; C2; C3 and C4 are the 10 classical first integrals. 1 2 The Two{Body Problem: Orbits Consider two point masses m1 and m2 under the influence of gravity. Let R :(m1 + m2)R = m1R1 + m2R2 be the centre of mass, r := R2 − R1 be the relative position of m2 with respect to m1. Then µ µ (7) ¨r = − e = − r r2 r r3 with µ := G(m1 + m2). The angular momentum (8) h := r × r_ = constant is the specific angular momentum (angular momentum per mass). It is a constant of motion. Let v = r_ = vrer + v?e? then h := rv?eh where eh is the unit vector in the direction of h. The orbit equation h2 1 p (9) r = = µ 1 + e cos # 1 + e cos # where # is the true anomaly , p is the semilatus rectum and e is the excentricity of the conic. e 2 [0; 1) : ellipse (e = 0: circle) e = 1 : parabola e > 1 : hyperbola The energy integral More precisely: the specific energy, energy per mass. 1 2 Kinetic energy: 2 v µ Potential energy: − r According to (6) the (specific) energy 1 µ (10) E = v2 − = constant 2 r is constant along orbits of the two{body problem. Equation (10) relates the velocity to the radius along an orbit. Computing the energy at the periapsis yields 1 µ µ (11) E = v2 − = − (1 − e2) 2 r 2p Formulae for the radial and the perpendicular components of the velocity h (12) v = e sin # r p h v = (1 + e cos #)(13) ? p 2 Circular orbits (e = 0) µ E = − (energy)(14) 2r rµ v = (velocity)(15) r 2π T = p r3=2 (period)(16) µ Elliptic orbits (e 2 (0; 1)) The following formulae hold. a(1 − e2) r = (radius)(17) 1 + e cos # 1 µ µ E = v2 − = − (energy)(18) 2 r 2a Kepler's laws of planetary motion 1. The orbit of every planet is an ellipse with the sun at one of the foci. Thus, Kepler rejected the ancient Aristotelean, Ptolemaic and Copernican belief in circular motion. 2. A line joining a planet and the sun sweeps out equal areas during equal intervals of time as the planet travels along its orbit. This means that the planet travels faster while close to the sun and slows down when it is farther from the sun. With his law, Kepler destroyed the Aristotelean astronomical theory that planets have uniform velocity. 3. The squares of the orbital periods of planets are directly proportional to the cubes of the semi{major axes of their orbits. This means not only that larger orbits have longer periods, but also that the speed of a planet in a larger orbit is lower than in a smaller orbit. More precisely: 4π2 (19) T 2 = a3 µ or, for the sun, planet1, planet2 of mass ms, m1 and m2: T 2 m + m a 3 (20) 1 s 1 = 1 T2 ms + m2 a2 Parabolic orbits (e = 1) 1 µ E = v2 − = 0 (energy)(21) 2 r r2µ v = (escape velocity)(22) esc r The following holds: If v < vesc then the orbit is an ellipse (circles included). 3 If v = vesc then the orbit is a parabola. If v > vesc then the orbit is a hyperbola. Hyperbolic orbits (e > 1) The semi{major axis is negative! 1 µ µ E = v2 − = > 0 (energy)(23) 2 r −2a 2 1 v2 = µ − (vis{viva equation)(24) r a 3 The Two{Body Problem: Position as a Function of Time 3.1 Elliptic orbits # : True anomaly E : Excentric anomaly M : Mean anomaly The Mean anomaly is the rescaled time; the period T is rescaled to 2π; passage through pericentre at time t0 corresponds to M = 0. Kepler's equation: ( relationship between excentric and mean anomaly) (25) E − e sin E = M: Relationship between true and excentric anomaly # r1 + e E (26) tan = tan : 2 1 − e 2 3.2 Hyperbolic orbits h Again, #, E and M := ab (t − t0) are the true anomaly, the excentric anomaly and the mean anomaly. Kepler's equation for hyperbolic orbits: (27) e sinh E − E = M: Relationship between true and excentric anomaly # re + 1 E (28) tan = tanh : 2 e − 1 2 3.3 The orbit in space Consider an inertial system with x; y; z{coordinates. For arbitrary initial conditions r(t0) = r0 6= 0 and r_(t0) = v0 6= 0 there exists a unique solution of (7) µ (29) ¨r = − r r3 4 • The vectors r0 and v0 are not very descriptive. The orbit is easy to describe in a (ξ; η; ζ){frame with periapsis on the ξ{axis and the ξ; η{plane containing the orbit. Three parameters are needed to describe the position of the (ξ; η; ζ){frame with respect to the (x; y; z){frame, for instance the tree Euler angles. Ω: longitude of the ascending node ( (e ; e ) where e = 1 r ). \ x k k rasc asc i: inclination (\(ez; eζ ) = \(ez; eh)). !: argument of the pericentre (\(ek; ep)). • To describe the conic section (with periapsis on the positive ξ{axis) two param- eters are needed. e: describes the shape of the orbit. p: describes the size of the orbit. Alternatively, e and a could be used. • To describe a point on the orbit one parameter is needed. #: the true anomaly. Alternatively, the excentric anomaly E, the mean anomaly M or the elapsed time t − t0 since passage through periapsis may be used. Orbit elements The parameters Ω; i; !; p; e; # are called elements of the orbit. They may be determined as follows from given r and v. 1. Inclination i from he ; hi h (30) cos i = z = z h h where h = r × v. If i < π=2 = 90o the orbit is prograde, if i > π=2 the orbit is retrograde. 2. Define k := ez × h. k points to the direction of the ascending node. If 0 1 0 1 hx −hy (31) h = @ hy A ; then k = @ hx A : hz 0 Ω = \(ex; k) may be determined from he ; ki −h (32) cos Ω = x = y p 2 2 k hx + hy If hx > 0, i.e., ky > 0, then Ω 2 (0; π). If hx < 0, i.e., ky < 0, then Ω 2 (π; 2π). 3. From the vis{viva equation (24) r (33) a = : v2r 2 − µ 5 4. From v2 1 1 (34) e = − r − hr; viv : µ r µ determin e = jej. If necessary p = a(1 − e2). 5. ! = \(e; k) may be determined from he; ki (35) cos ! = : e k If e3 > 0 then ! 2 (0; π)(e3 being the third component of e). If e3 < 0 then ! 2 (π; 2π). 6. # = \(e; r) may be determined from he; ri (36) cos # = : e r If the distance to the pericentre is • increasing, i.e., if hr; vi > 0 then # 2 (0; π) • decreasing, i.e., if hr; vi < 0 then # 2 (−π; 0) or # 2 (π; 2π). 4 Rocket dynamics There are a lot of different rocket engines, usually categorised as either high{ or low{ thrust engines. High{thrust engines can provide thrust accelerations significantly larger than the local gravitational acceleration, while for low{thrust engines the thrust accel- eration is much smaller than the local gravitational acceleration. To provide thrust, mass is expelled out of the rocket nozzle. Thus the rocket mass is decreasing. 4.1 The thrust The thrust of a rocket is (37) S = −m_ c wherem _ describes the loss of mass (it is negative) and where (p − p )A (38) c = v + e a e −m_ is the effective exhaust velocity.
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