Orbital Dynamics: Formulary


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1 Orbital Dynamics: Formulary 1 Introduction Prof. Dr. D. Stoffer Department of Mathematics, ETH Zurich 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) = ẋ(t) : velocity of the object at time t. a(t) = ẍ(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 m 1 and m, and inversely proportional to the square of the distance r between them: () F = G m 1m r. The constant of proportionality, G, is the gravitational constant. G = ( ± ) m 3 kg 1 s = ( ± ) N m kg. The n body problem n point masses m i with positions R i (with respect to an inertial frame) move under the influence of gravity. Let r ij := R j R i be the position of the point mass m j relative to the position of the point mass m i. The equations of motion are (3) m i Ri = G n j=1,(j i) m i m j e rij ij, (i = 1,,..., n) where e ij := 1 r ij r ij. Let R := i m ir i / i m i be the centre of mass. The 10 classical first integrals (conserved quantities) (4) (5) R(t) = C 1 t + C n m i (R i Ṙ i ) = C 3 (angular momentum) i=1 (6) T + V = C 4 (total energy) T := 1 n i=1 miṙ i is the kinetic energy and V := G n m i m j i=1 j i r ij is a potential of the force field F. The components of C 1, C, C 3 and C 4 are the 10 classical first integrals. 1
2 The Two Body Problem: Orbits Consider two point masses m 1 and m under the influence of gravity. Let R : (m 1 + m )R = m 1 R 1 + m R be the centre of mass, r := R R 1 be the relative position of m with respect to m 1. Then (7) r = r e r = r 3 r with := G(m 1 + m ). The angular momentum (8) h := r ṙ = constant is the specific angular momentum (angular momentum per mass). It is a constant of motion. Let v = ṙ = v r e r + v e then h := rv e h where e h is the unit vector in the direction of h. The orbit equation (9) r = h e cos ϑ = p 1 + e cos ϑ where ϑ is the true anomaly, p is the semilatus rectum and e is the excentricity of the conic. e [0, 1) : ellipse (e = 0: circle) e = 1 : parabola e > 1 : hyperbola The energy integral More precisely: the specific energy, energy per mass. Kinetic energy: 1 v Potential energy: r According to (6) the (specific) energy (10) E = 1 v r = constant 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 (11) E = 1 v r = p (1 e ) Formulae for the radial and the perpendicular components of the velocity (1) (13) v r = h p e sin ϑ v = h (1 + e cos ϑ) p
3 Circular orbits (e = 0) (14) (15) (16) E = r v = r T = π r 3/ (energy) (velocity) (period) Elliptic orbits (e (0, 1)) The following formulae hold. (17) (18) r = a(1 e ) 1 + e cos ϑ E = 1 v r = a (radius) (energy) 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.. 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: (19) T = 4π a3 or, for the sun, planet 1, planet of mass m s, m 1 and m : (0) ( T1 T ) m s + m 1 m s + m = ( a1 a ) 3 Parabolic orbits (e = 1) (1) E = 1 v r = 0 (energy) () v esc = r (escape velocity) The following holds: If v < v esc then the orbit is an ellipse (circles included). 3
4 If v = v esc then the orbit is a parabola. If v > v esc then the orbit is a hyperbola. Hyperbolic orbits (e > 1) The semi major axis is negative! (3) (4) E = 1 v r = a > 0 (energy) v = ( r 1 ) (vis viva equation) a 3 The Two Body Problem: Position as a Function of Time 3.1 Elliptic orbits ϑ : E : M : True anomaly Excentric anomaly Mean anomaly The Mean anomaly is the rescaled time; the period T is rescaled to π; passage through pericentre at time t 0 corresponds to M = 0. Kepler s equation: ( relationship between excentric and mean anomaly) (5) E e sin E = M. Relationship between true and excentric anomaly (6) tan ϑ = 1 + e 1 e tan E. 3. Hyperbolic orbits Again, ϑ, E and M := h ab (t t 0) are the true anomaly, the excentric anomaly and the mean anomaly. Kepler s equation for hyperbolic orbits: (7) e sinh E E = M. Relationship between true and excentric anomaly (8) tan ϑ = e + 1 e 1 tanh E. 3.3 The orbit in space Consider an inertial system with x, y, z coordinates. For arbitrary initial conditions r(t 0 ) = r 0 0 and ṙ(t 0 ) = v 0 0 there exists a unique solution of (7) (9) r = r 3 r 4
5 The vectors r 0 and v 0 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 x, e k ) where e k = 1 r asc r asc ). i: inclination ( (e z, e ζ ) = (e z, e h )). ω: argument of the pericentre ( (e k, e p )). To describe the conic section (with periapsis on the positive ξ axis) two parameters 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. ϑ: the true anomaly. Alternatively, the excentric anomaly E, the mean anomaly M or the elapsed time t t 0 since passage through periapsis may be used. Orbit elements The parameters Ω, i, ω, p, e, ϑ are called elements of the orbit. They may be determined as follows from given r and v. 1. Inclination i from (30) cos i = e z, h h = h z h where h = r v. If i < π/ = 90 o the orbit is prograde, if i > π/ the orbit is retrograde.. Define k := e z h. k points to the direction of the ascending node. If h x h y (31) h = h y h z, then k = h x 0. Ω = (e x, k) may be determined from (3) cos Ω = e x, k k = h y h x + h y If h x > 0, i.e., k y > 0, then Ω (0, π). If h x < 0, i.e., k y < 0, then Ω (π, π). 3. From the vis viva equation (4) (33) a = r v r. 5
6 4. From (34) e = ( v determin e = e. If necessary p = a(1 e ). 1 r 5. ω = (e, k) may be determined from ) 1 r r, v v. e, k (35) cos ω = e k. If e 3 > 0 then ω (0, π) (e 3 being the third component of e). If e 3 < 0 then ω (π, π). 6. ϑ = (e, r) may be determined from e, r (36) cos ϑ =. e r If the distance to the pericentre is increasing, i.e., if r, v > 0 then ϑ (0, π) decreasing, i.e., if r, v < 0 then ϑ ( π, 0) or ϑ (π, π). 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 acceleration 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 = ṁ c where ṁ describes the loss of mass (it is negative) and where (38) c = v e + (p e p a )A ṁ is the effective exhaust velocity. v e is the velocity of the expelled particles relative to the rocket. p e is the pressure of the exhaust at the nozzle exit, p a is the outside ambient pressure (atmospheric pressure, which has value 0 in vacuum), and A is the nozzle exit area. Assumptions: Rocket in force free space, c is constant, one dimensional motion. (39) m v = ṁ c or, integrating from t 0 to t 1 (40) v 1 v 0 = c log m 0 m 1, m 1 = e v c. m 0 This equation (in either form) is referred to as the rocket equation. 6
7 4. The equations of motion Let be Then γ : the flight path angle ϕ, r : the polar coordinates v : the tangential velocity of the rocket a : the tangential acceleration of the rocket a : the normal acceleration of the rocket ρ : the radius of curvature u = (S, v) : controll variable (41) a = v γ + v r cos γ (4) a = v From these equations one derives the equations of motion of a rocket in a central gravitational field. Thrust S, atmospheric drag R = 1 σv Ac w (σ: density, A: cross sectional area, c w : drag coefficient). (43) v = S m cos u R m g sin γ, (g = r ) (44) v γ = (g v r ) cos γ + S m sin u (45) ϕ = v r cos γ (46) ṙ = v sin γ There is no lifting force as for aircrafts u is a controll variable in order to inject the rocket into the desired orbit. u = 0 corresponds to the motion when S v. 4.3 Injection into orbit Velocity in a circular LEO with altitude of 300km: v r = km/s From (43) one gets τ τ τ (47) v = g sin γdt S m dt 0 } {{ } idealer Antriebsbedarf R m dt 0 } {{ } Widerstandsverlust 0 } {{ } gravity loss The total loss for injection into a LEO is about 14% ( 4% air drag, 10% gravity loss). The required v is km/s 4.4 Multistage rockets Let (48) m 0 = m t + m s + m L 7
8 be the total mass of a rocket at the start. m t is the mass of propellant, m s the structural mass and m L the payload mass. From the mass ratio at burnout m 0 m 0 (49) Z = = m s + m L m 0 m t one immediately gets the characteristic velocity, cf. (40) (50) v = c log Z. Moreover, define the structural coefficient σ = m L m t+m s = m L m 0 m L. ms m t+m s and the payload ratio ν = Optimal staging Let m ti, m si, m Li, respectively, be the propellant mass, the structural mass, the payload mass of the ith stage, respectively, i = 1,..., n. Note that the total mass m 0i = m ti + m si + m Li of the i stage is the payload mass of the (i 1)th stage. Define m i = m ti + m si to be the sum of the propellant mass and the structural mass of the ith stage. Problem: For given effective exhaust velocity c i and stuctural coefficient σ i of the ith stage, given payload mass m L and given v tot := i v i, i = 1,..., n, minimise M := n i=1 m i. Solution: Solve (51) v tot for λ, then determine n i=1 c i log c i + 1/λ c i σ i = 0 (5) Z i = c i + 1/λ c i σ i and (53) M + m L m L = 5 Orbital Manoeuvres n i=1 (1 σ i )Z i 1 σ i Z i Part A: Impulsive Orbit Transfer The limiting case of finite characteristic velocities v during a short time t 0 is considered. At times t k, k = 1,,... the vehicle undergoes velocity changes v k = v + k v k 5.1 Hohmann transfer The Hohman transfer is the minimum fuel two impulse transfer between circular orbits. From the visviva equation (4) one gets for 0 < r 1 < r (54) v tot = [ ( 1 ) ( 1 + )] r 1 r 1 + r r 1 r r r 1 + r 8
9 5. Bi elliptic transfer For 1 < α < β the bi elliptic transfer from a circular orbit of radius r 1 over a point B with r B = βr 1 to a circular orbit of radius r = αr 1 the total characteristic velocity satisfies (55) v bi v 1 = (β 1) + β(β + 1) 5.3 Change of the orbit plane ( 1 α + 1 ) β 1 α 1 Oneimpuls changes of the orbit plane of circular orbits are very costly. v (56) = sin(ϕ/) v circle For large changes of the orbit plane bi elliptic transfers are more efficient. 5.4 Rendezvous Synodic period for two objects on coplanar circular orbits (57) S = π 1 = n 1 n with angular velocities n 1, n. 1 T 1 1 T Initial phase angle for rendezvous with Hohmann transfer: ( ( 1 + r1 /r ) 3/ ) (58) β = π 1 Part B: Low Thrust Manoeuvres The equations of motion are d (59) dt v = d dt x = S m + g where S is the thrust and g is the gravitational acceleration. For almost circular orbits one gets by integrating (60) v low thrust = r 0 r = v circle(r 0 ) v circle (r) i.e., v is equal to the difference of the orbital velocities on the circles. 6 Interplanetary Mission Analysis 6.1 Domain of influence of a planet Inspect the three body problem spacecraft sun planet. Considering this problem as a perturbed vehicle planet two body problem one gets (61) r pv = G(m p + m v ) ( rsv Gm s r ) sp r 3 pv r pv } {{ } A p 9 r 3 sv r 3 sp } {{ } S s
10 where A p is the acceleration due to the planet and S s is the perturbation due to the sun. For short, (6) r pv A p = S s. Analogously, (63) r sv A s = S p. with (64) A s = G(m s + m v ) r rsv 3 sv, S p = Gm p ( rpv r 3 pv + r ) sp rsp 3 According to Laplace the domain of influence of a planet is defined as the set of all points for which (65) S p A s S s A p. The domain of influence of a planet is approximately a ball of radius (m p /m s ) /5 r sp. According to this definition the moon is well inside the domain of influence of the earth. 6. Patched conics Within the domain of influence (sphere of influence) of a planet the two body problem vehicleplanet is considered. The exit velocity is approximatively equal to v. Outside of the domain of influence of the planet the two body problem vehiclesun is considered. The initial velocity is equal to v v = v planet + v 6.3 Flyby or gravity assist Entrance into the domain of influence with v, exit with v +. In the sun vehicle system this leads to v = v v. For the magnitude of v one has v (66) v = 1 + ( v ) r p v0 r 0 where r 0 is the radius of the planet, r p is the radius of the periapsis and v 0 is the velocity on a circular orbit of radius r 0 (note r 0 v0 = ). 6.4 The restricted three body problem The two primaries with masses m 1, m move on circular orbits around their centre of mass. In a rotating frame with scaled distances the primaries have fixed positions (, 0, 0), ( 1, 0, 0) where 1 = m /(m 1 + m ), = m 1 /(m 1 + m ). The equations of motion for a test particle (or a vehicle) (67) (68) (69) ẍ ẏ = U x ÿ + ẋ = U y z = U z 10
11 where U = 1 (x + y ) + 1 More explicitely r 1 + r with r 1 = (x + ) + y + z, r = (x 1 ) + y + z. (70) (71) (7) ẍ ẏ x = 1 (x + r1 3 ) (x r 3 1 ) ( 1 ÿ + ẋ y = + ) y r 3 z = r1 ( 3 1 r r 3 ) z The Jacoby integral (73) ( C := x + y ) ẋ ẏ ż r 1 r is a constant of motion. There are 5 equilibria: the three Euler points L 1 between the two primaries, L and L 3 on the positive and negative x axis and the two Lagrange points L 4,5 = (( 1 )/, ± 3/) 7 Perturbations The Keplerian motion of satellites is perturbed by the oblateness of the earth, the atmospheric drag, the influence of the sun and the moon, the radiation pressure, electomagnetic forces, etc. General assumption: The perturbation is much smaller than gravitation. 7.1 The perturbation equations The perturbation is given as an acceleration (force/mass). The perturbation (74) F = F ξ e ξ + F η e η + F ζ e ζ is given in a satellite oriented frame (e ξ, e η, e ζ ) with e ξ = (1/r)r, e η e ξ in the orbital plane and e ζ = e ξ e η. For given r(t), v(t) = ṙ(t) the osculating elements a(t), e(t), i(t), Ω(t), ω(t), M(t) are the elements of the unperturbed Kepler motion corresponding to r(t) and v(t) = ṙ(t). For the unperturbed two body problem the elements are constant, for the perturbed problem the osculating elements vary slowly as time evolves. The osculating elements 11
12 satisfy the following differential equations. (75) ȧ = a [ Fξ e sin ϑ + F (1 e η (1 + e cos ϑ) ] ) (76) ė = a(1 e )[ Fξ sin ϑ + F η (cos ϑ + cos E) ] (77) (78) (79) i = Ω = ω = 1 e a(1 e ) a(1 e ) a(1 e ) [ Ω cos i e cos ϑ F ζ cos(ϑ + ω) e cos ϑ F sin(ϑ + ω) ζ sin i ( + e cos ϑ + cos E) sin ϑ ] F ξ cos ϑ + F η 1 + e cos ϑ Introduce the two variables ν := ndt and χ := nt 0 where n is the angular velocity of the mean anomaly M and t 0 is the time of passing through periapsis. The equation for χ is a (1 e ) [ (80) χ = Fξ (e cos ϑ e cos ϑ) + F η ( + e cos ϑ) sin ϑ ] e(1 + e cos ϑ) To determine M = ν χ one has to integrate the equation (81) ν = n = a. 3 It is often advantageous to take the variable u := ω + ϑ a independent variable. A lengthy transformation leads to (8) (83) (84) (85) (86) where dp du = r3 Γ p F η de du = r Γ[ Fξ sin ϑ + F η (cos ϑ + cos E) ] e di du = r3 Γ p cos(u)f ζ dω du = r3 Γ sin u p sin i F ζ dω du = r Γ [ ( + e cos ϑ + cos E) sin ϑ F ξ cos ϑ + F η e 1 + e cos ϑ e sin u ] F ζ 1 + e cos ϑ tan i (87) Γ = 1 1 r3 sin u F p tan i ζ 1.
13 7. The method of averaging Consider the differential equation (88) ẋ = εf(t, x) where ε is a small parameter and where f is T periodic with respect to t. Then the solutions of the averaged equation (89) ẏ = εf(y) with f(y) := (1/T ) T 0 f(t, y)dt satisfy (90) x(t) y(t) Cε for t [0, L/ε]. 7.3 Oblateness of the earth The gravitational potential of the earth is approximated by (91) U = U 0 + U J = r r 0 r 3 J (3 sin ϕ 1)/ where U J describes the influence of the oblateness of the earth. The perturbation F = grad U J is (9) F = 3r 0J r 4 [1 (1 3 sin3 i sin u) e ξ + sin i sin u cos u e η + sin i cos i sin u e ζ ] Setting Γ = 1 in (8) (86) one gets after rescaling to the variable t (93) (94) (95) (96) (97) Ω = 3 ( r0 ) 3.5 a r0 3 ω = 3 4( r0 ȧ = 0 i = 0 ė = 0 a ) 3.5 r 3 0 J cos i ( (1 e ) = r0 ) 3.5 cos i /4h a (1 e ) ( r0 ) cos i 1 /4h a (1 e ) J 5 cos i 1 (1 e ) = 4.98 If i < 90 then Ω < 0, i.e., the node line drifts westward. If i > 90 then Ω > 0, i.e., the node line advances eastward. If i < 63.4 or i > then ω > 0, meaning that the perigee advances in the direction of the satellite. If 63.4 < i < then the perigee regresses, it moves oposite to the direction of motion. 7.4 Atmospheric drag For nearly circular LEO orbits one has approximately (98) F ξ = 0, F η = 1 ρc wav /m, F ζ = 0. 13
14 From (75) one gets with a = r and v = /r (99) ṙ = rρc w A/m < 0. From (84) (86) one gets i = 0, Ω = 0, ω = 0, meaning that the orbit plane and the direction of the perigee remain constant. For noncircular orbits one gets (100) (101) (10) (103) (104) ṗ < 0 ė < 0 Ω = 0 i = 0 ω = 0 The orbit becomes smaller and closer to a circular orbit. direction of the perigee remain constant. The orbit plane and the 8 Attitude dynamics An Example: the dumbbell satellite Two masses m are connected with a massless rod of length l. Positions of the masses: (105) (106) x 1, = r cos ϕ ± l cos(ϕ + ϑ) y 1, = r sin ϕ ± l sin(ϕ + ϑ) With the kinetic and the potential energy (107) (108) T = m[ṙ + r ϕ + l ( ϕ + ϑ) ] ( U = m + ) r 1 r and the Lagrange function L = T U one derives the equations of motion from the Lagrange equation (109) d L dt q L q = 0 for q = r, ϕ, ϑ. Taking the limit l 0 one gets (110) (111) (11) ϑ + 3 r r ϕ = r d dt (r ϕ) = 0 sin(ϑ) r 3 = ϕ Equations (110) and (111) are the equations for the Kepler problem in polar coordinates. They are decoupled from (11). Equation (11) describes the attitude dynamics of the dumbbell satellite. 14
15 For circular orbits (11) degenerates to the pendulum equation (113) ϑ + 3 r 3 0 sin(ϑ) = 0. The radial equilibrium solution ϑ = 0 is stable, the tangential equilibrium solution ϑ = π/ is unstable. To investigate (11) it is convenient to replace the time t by the excentric anomaly E. One gets ϑ ϑ e sin E 1 e cos ϑ + 3 sin(ϑ) 1 e cos ϑ = e 1 e sin E (114) (1 e cos ϑ) where denotes the derivative d/de with respect to E. Appendix Vector identities Astronomical constants The Sun a (b c) = a, c b a, b c a, b c = c, a b mass = kg radius = km sun = G m sun = km 3 /s The Earth mass = kg radius = km earth = G m earth = km 3 /s mean distance from sun = 1 au = km The Moon mass = kg radius = km moon = G m moon = km 3 /s mean distance from earth = km orbit eccentricity = orbit inclination (to ecliptic) =
16 Physical characteristics of the planets Planet Equatorial Mass Siderial Inclination of radius rotation equator to (units of R earth ) (units of M earth ) period orbit plane Mercury d 16h Venus d(retro) Earth 1, h 56m 04s 3 7 Mars h 37m 3s 5 11 Jupiter h 50m 3 07 Saturn h 14m 6 44 Uranus h 54m 97 5 Neptune h 1m 9 36 Pluto d 9h 18m 1 46 Elements of the planetary orbits Planet semimajor axis eccentricity siderial inclination to (in au) period ecliptic plane Mercury d 7 00 Venus d 3 4 Earth d 0 00 Mars y 31.73d 1 51 Jupiter y d 1 19 Saturn y 167d 30 Uranus y 7.4d 0 46 Neptune y 80.3d 1 47 Pluto y 49d
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