Previous Lecture. Orbital maneuvers: general framework. Single-impulse maneuver: compatibility conditions

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2 2 / 48 Previous Lecture Orbital maneuvers: general framework Single-impulse maneuver: compatibility conditions closed form expression for the impulsive velocity vector magnitude interpretation coplanar examples

3 3 / 48 Previous Lecture Two-impulse maneuvers: The Hohmann Transfer Three-impulse maneuvers: the Bi-Elliptic transfer Comparisons

4 4 / 48 Previous Lecture: Single-Impulse Maneuver Impulsive velocity expression The impulsive velocity to change a Keplerian orbit (h 0, E 0 ) to another Keplerian orbit (h 1, E 1 ), at a possible maneuver point (r 0, v 0 ) : v = h 1 r 0 r 2 0 ( ± 2E 1 + 2µ ) 1/2 h2 1 r0 r 0 r0 2 v 0 (1) r 0 The maneuver should be performed where the two orbits intersect The compatibility condition: 2E 1 + 2µ r 0 h2 1 r (2)

5 5 / 48 Phasing Maneuvers Usefullness Constellation (deployment or replacement of a failed satellite) GEO First phase of a rendezvous procedure

6 Phasing Maneuvers 6 / 48

7 Phasing Maneuvers 7 / 48

8 Phasing Maneuvers 8 / 48

9 9 / 48 Phasing Maneuvers Input data: a, e, f

10 10 / 48 Phasing Maneuvers Output data: v

11 Phasing Maneuvers f given; obtain E from: cos E = e + cos f 1 e ; sin E = 2 sin f ; (3) 1 + e cos f 1 + e cos f Obtain t from Kepler s equation: t = 1 n ( E e sin E) = a3/2 µ ( E e sin E) (4) Obtain the period of the phasing orbit: T phas = T t = a3/2 µ (2π E + e sin E) (5) 11 / 48

12 12 / 48 Phasing Maneuvers Obtain the semimajor axis of the phasing orbit: ( ) µt 2 1/3 ( ) phas E e sin E 2/3 a phas = 4π 2 = a 1 (6) 2π The eccentricity of the phasing orbit: e phas = 1 a (1 e) a phas = 1 ( 1 1 e E e sin E 2π The specific angular momentum of the phasing orbit: ( ) h phas = µa phas 1 e 2 phas The impulsive velocity: ) 2/3 ( ) (7) (8) v = h h phas a (1 e) (9)

13 13 / 48 Phasing Maneuvers Phasing maneuver was made at perigee (Remember Oberth effect). Ignore Oberth s effect: optimal maneuver is at apogee. Apogee maneuver: restricted by the perigee of the phasing orbit (greater the difference between target/interceptor lower the perigee of the phasing orbit). Phasing at apogee: in the tutorial.

14 14 / 48 Non-Coplanar Impulsive Maneuver The impulsive velocity magnitude: v 2 = v 2 rad + v2 h (10) v rad : change in radial velocity v h : change in angular momentum Change of orbital plane: v h > 0 (reciprocal is not true) Changes in the orbital plane are extremely costly: v 2 h = h2 0 + h2 1 2h 0 h 1 r 2 0 (11)

15 15 / 48 Non-Coplanar Impulsive Maneuver Given two non-coplanar orbits (h 0, E 0 ), (h 1, E 1 ), what remains to be determined is the possible maneuver point r 0 Given the known properties of the Keplerian orbits, look for r 0 in the following form: r 0 = ±r 0 h 0 h 1 h 0 h 1 (12) In the tutorial: example of RAAN change example de inclination change

16 16 / 48 Example: Space Shuttle and Hubble Space Shuttle Capabilities Space Shuttle: capable of a plane change in orbit of only 3, a maneuver which would exhaust its entire fuel capacity. Can Hubble be moved to the ISS? Hubble: 28.5 inclination, 569 km altitude ISS: 51.6 inclination, 370 km altitude Too costly! Instead, NASA has chosen to have another shuttle ready to lift off to retrieve the astronauts if needed.

17 17 / 48 Example: Inclination Change Inclination change (circular orbits assumption): should occur only when crossing the equatorial plane (Why?) The impulsive velocity maneuver is: µ i v h = 2 sin R 2 = 2v 0 sin i 2 Numerical example: 1 degree inclination change at 600 km altitude R = R e + H = = m µ = m 3 /s 2 Impulsive velocity magnitude: v = m/s (13)

18 18 / 48 Lambert s Problem Lambert s Problem Given two different times t 1 and t 2, and given two position vectors r 1 and r 2, determine the Keplerian arc characterized by the law of motion r = r (t) such that r (t 1 ) = r 1 and r (t 2 ) = r 2 Johann Heinrich Lambert, Swiss mathematician

19 19 / 48 Lambert s Problem Lambert s Conjecture The orbital transfer time depends only upon the semimajor axis, the sum of the distances of the initial and final points of the arc from the center of the force, and the length of the chord joining these points. Mathematical Formulation If t 1,2 are the two moments of time, then Lambert theorem states that: (t 2 t 1 ) = 1 µ F (a, r 1 + r 2, r 2 r 1 )

20 20 / 48 Lambert s Problem: Why? Design of interplanetary missions: initial approximation of travel from Earth to another planet (Mars, for example): t 1 = time of launch from Earth; t 2 = time of arrival; r 1 = position of Earth at the launch; r 2 = position of Mars at the time of arrival elliptic heliocentric orbit Orbit determination: by knowing two succesive positions (GPS measured) r 1 = r 1 (t 1 ) and r 2 = r 2 (t 1 ) Rendez-vous missions

21 Solving Lambert s Problem 21 / 48

22 Solving Lambert s Problem 22 / 48

23 23 / 48 Solving Lambert s Problem Two-point boundary value problem: Unperturbed: Find the vector valued function r = r (t) such that: r + µ r 3 r = 0; r (t 1) = r 1 ; r (t 2 ) = r 2 (14) The Lambert problem in the Keplerian case has a closed-form analytic solution Perturbations might be included as well: Find the vector valued function r = r (t) such that: r + µ r 3 r = f (r, ṙ, t) ; r (t 1) = r 1 ; r (t 2 ) = r 2 (15) General method: numerical

24 24 / 48 Solving Lambert s Problem: Keplerian Lagrange was first to supply the analytic proof of Lambert s theorem for elliptic orbits just one year before Lambert died. Euler (1743) developed the special case of Lambert s theorem for the parabola, but he did not extend it to the ellipse and hyperbola. Gauss, in his Theoria Motus, presented almost the same solution. The problem is solvable analytically for any conic: elliptic, parabolic, hyperbolic

25 25 / 48 Solving Lambert s Problem: Keplerian Further Reading: Chapter 6 Two-Body Orbital Boundary-Value Problem Section 6.6: Lambert s Theorem

26 26 / 48 Solving Lambert s Problem: General Input data: r 1,2 ; t = t 2 t 1 Output data: ṙ (t 1 ) = v (t 1 ) ; Shooting method: Start with an initial guess v 1 = ṙ (t 1 ) 1 Integrate the initial value problem: r + µ r 3 r = f (r, ṙ, t) ; r (t 1) = r 1 ; ṙ (t 1 ) = v 1 (16) obtain the solution r (t) and evaluate r (t 2 ) ; 2 Evaluate the targeting error δr (t 2 ) = r (t 2 ) r 2 3 Update the initial velocity estimate by using the local-linearization: ṙ (t 1 ) }{{} new = ṙ (t 1 ) }{{} old [ ] r (t2 ) 1 δr (t 2 ) (17) ṙ (t 1 ) 4 Return to step (1) until δr (t 2 ) has become sufficiently small.

27 27 / 48 Solving Lambert s Problem: General Further Reading: Chapter 13 Transfer Orbits

28 28 / 48 Maneuvers: Delta v budget A satellite / spacecraft has a limited amount of fuel a Delta v budget It is the sum of the velocity changes required throughout the space mission life. It is a good starting point for early design decisions. As important as power and mass budgets. In some cases, it may become a principal design driver and impose complex trajectories to deep space probes. Maneuvers: minimize the total expense of Delta v. Maneuvers: several parameters involved restrictions.

29 Delta v: Examples 29 / 48

30 30 / 48 Maneuvers: Rocket Engines Maneuvers are performed using firings of onboard rocket motors. Chemical rocket engines: Assumption of impulsive thrust in this lecture: because the burn times are small compared with the time intervals between burns, the thrust can be idealized as having infinitely small duration (no thrust included in the equation of motion). Electric propulsion: Continuous and low thrust.

31 Rocket Engines: Monopropellant 31 / 48

32 Rocket Engines: Bipropellant 32 / 48

33 Rocket Engines: Solid 33 / 48

34 Rocket Engines: Low-Thrust 34 / 48

35 Specific Impulse, I sp 35 / 48

36 Rocket Engines: I sp 36 / 48

37 Further Reading on the Web Site 37 / 48

38 Appendix: The Lagrangian Coefficients 38 / 48

39 39 / 48 Triple Vector Product Given three vectors a, b, c in V 3, their triple vector product is defined as: (a, b, c) = a (b c) Geometrical interpretation: numerically equal to the volume of the parallelepiped with the edges a, b, c.

40 40 / 48 Reciprocal (dual) basis Given three non-coplanar vectors e 1, e 2, e 3 V 3, they form a basis B. The reciprocal basis B = {e1, e 2, e 3 } is defined as: e k e p = δ kp = { 1, k = p 0, k = p, k, p {1, 2, 3} δ kp are called the Kronecker symbols. The vectors of the reciprocal basis B are determined as: e 1 = e 2 e 3 (e 1, e 2, e 3 ) ; e 2 = e 3 e 1 (e 1, e 2, e 3 ) ; e 3 = e 1 e 2 (e 1, e 2, e 3 )

41 41 / 48 The Lagrangian Coefficients The functions F, G, F t, G t are the components of the position and velocity respectively in the non-orthogonal basis {r 0, v 0, h 0 = r 0 v 0 } : { r (t) = Fr0 + Gv 0 v (t) = F t r 0 + G t v 0 (18) The Lagrangian coefficients are obtained by dot-multiplying each equation by r0, v 0 : { F = r (t) r 0 G = r (t) v0 F t = v (t) r0 G t = v (t) v0 (19) The expressions of r0, v 0 are computed as in the previous slide: r0 = v 0 h 0 h 2 ; v0 = h 0 r 0 0 h 2 0 (20)

42 42 / 48 The Lagrangian Coefficients The Lagrangian coefficients may formally be written as: [ ] [ ] F G r (t) T [ = r F t G t v (t) T 0 v0 ] (21) where: r (t) = r 0 = 1 r 0 [ [ a (cos E e) b sin E cos E0 a b sin E 0 ] ] ; v (t) = na [ a sin E r b cos E ; v 0 = 1 h [ ] b sin E 0 a (e cos E 0 ) ] (22)

43 The Lagrangian Coefficients The following expressions are obtained: F = a [cos (E E 0 ) e cos E 0 ] ; r 0 G = 1 n [sin (E E 0) e (sin E sin E 0 )] ; F t = na2 rr 0 sin (E E 0 ) ; (23) G t = b r [cos (E E 0) e cos E] ; where: r = a (cos E e) ; r 0 = a (cos E 0 e) (24) cos E 0, sin E 0 are given in Lecture 4A. The Analytic Methods. 43 / 48

44 44 / 48 The Lagrangian Coefficients The vectors r, v, r0, v 0 may also be expressed as: [ ] cos f r (t) = r ; v (t) = h [ sin f sin f p e + cos f r 0 = 1 p [ e + cos f0 sin f 0 ] ; v0 = r [ 0 sin f0 h cos f 0 ] ] (25)

45 The Lagrangian Coefficients The Lagrangian coefficients F, G, F t, G t are now expressed equivalently as: F = r p [cos ( f f 0) + e cos f ] ; G = rr 0 h sin( f f 0) F t = h p 2 [sin ( f f 0) + e (sin f sin f 0 )] ; (26) where: G t = r 0 p [cos ( f f 0) + e cos f 0 ] ; r = p 1 + e cos f ; r 0 = and f 0 is deduced from the initial conditions (r 0, v 0 ). p 1 + e cos f 0 (27) 45 / 48

46 46 / 48 This Lecture Phasing impulsive maneuvers. Non-coplanar impulsive maneuvers. Lambert s Problem. Delta v budget and propulsion systems. The Lagrangian coefficients.

47 47 / 48 Next Lecture Spacecraft Formation Flying

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