Astrodynamics (AERO0024)

Transcription

1 Astrodynamics (AERO0024) 5B. Orbital Maneuvers Gaëtan Kerschen Space Structures & Systems Lab (S3L)

2 Previous Lecture: Coplanar Maneuvers 5.1 INTRODUCTION Why? How? How much? When? 5.2 COPLANAR MANEUVERS One-impulse transfer Two-impulse transfer Three-impulse transfer Nontangential burns Phasing maneuvers 2

3 Motivation Coplanar transfers have wide applicability (e.g., phasing maneuvers and orbit raising). They alter 4 orbital elements: 1. Semi-major axis. 2. Eccentricity. 3. True anaomaly. 4. Argument of perigee. Noncoplanar transfers requires applying v out of the orbit plane and can therefore alter the last two elements: 1. Inclination. 2. RAAN. 3

4 5. Orbital Maneuvers 5.3 Noncoplanar transfers 5.4 Rendez-vous 4

5 5. Orbital Maneuvers 5.3 Noncoplanar transfers Inclination change RAAN change 5

6 Gaussian VOP (Lecture 4) θ F = eˆ + Teˆ + Neˆ R R T N θ 2 i& = a N cosθ2 µ 1+ ecosθ 2 (1 e ) ( ) Ω & = a N sinθ2 µ sin i 1+ ecosθ 2 (1 e ) ( ) Inclination change 6

7 Where? Cranking maneuver: with a single v maneuver, one wants to change only the inclination of the orbit plane. Where should we apply this maneuver? Inclination change 7

8 Where? At an Equator Crossing The two orbit planes intersect in the equatorial plane. We can therefore change the inclination at an equator crossing, by a simple rotation of the velocity vector. v r δ v v View down the line of intersection of the two orbital planes Inclination change 8

9 V Computation for a Circular Orbit ( ) v = v v = v v uˆ + v uˆ v uˆ 2 1 r 2 r1 r ( ) 2 v = v v + v + v 2v v cosδ r 2 r v = v, v = v r1 r v = δ 2vsin Inclination change 9

10 Inclination Changes Are Very Expensive δ v (% of v) How to minimize V? Is this cranking maneuver the minimum-fuel transfer? (60º, 100%) (24º, 41.6%) δ, degrees Inclination change 10

11 Real-Life Examples: Space Shuttle and Hubble The Space Shuttle is capable of a plane change in orbit of only about 3º, a maneuver which would exhaust its entire fuel capacity. What is Hubble space telescope s inclination? (HST) vs (KSC) Inclination change 11

12 What s Unique with this Picture? Too costly! Instead, NASA has chosen to have another shuttle ready to lift off to retrieve the astronauts if needed Inclination change 12

13 Minimizing V 1. Do not modify the radial velocity. 2. The maneuver should occur where the azimuth component is smallest, which is at apogee Inclination change 13

14 Minimum-Fuel Transfer: Bi-Elliptic Transfer One can always save fuel over a one-impulse maneuver by using three-impulses: noncoplanar bi-elliptic transfer. J.E. Prussing, B.A. Conway, Orbital Mechanics, Oxford University Press Inclination change 14

15 Case Study at DLR Inclination change 15

16 Direct Transfer from GTO (Hohmann) Direct transfer possible if the line of nodes of the GTO and the line of nodes of the Moon s orbit coincide, i.e., Ω=0. In that case, no inclination change is needed. For a GTO, this occurs twice a year because the nodal and apsidal lines coincide (power reasons, few eclipses) Inclination change 16

17 Direct Transfer from GTO (Hohmann-like) For small Ω, a small mid-course maneuver V2 is needed for inclination change Inclination change 17

18 Bi-elliptic Transfer from GTO For large Ω, the inclination change is too expensive. The apogee of the LTO is chosen to be 1e6 kms and the inclination change performed there is 17 degrees. Increased flight time of 50 days but savings of 115m/s Inclination change 18

19 Motivation for RAAN Change Constraints on the line of nodes may be imposed to a spacecraft (e.g., for rendez-vous): Launch window to match up the line of nodes. If we wait too long for such a window, the satellite can be launched sooner and thrust can be applied to change the line of nodes RAAN change 19

20 Why is RAAN Change Complicated? Ω & = i& = a a 2 (1 e ) N sinθ2 µ sin i 1 cosθ N cosθ2 µ θ 2 (1 e ) ( + e ) ( 1+ ecos ) 3 a a& = 2 Re sinθ T 1 e cos µ ( 1 e ) ( θ ) 2 a(1 e ) e& = Rsinθ + T cos + cos E µ ( θ ) ( + e θ ) 2 1 a(1 e ) T sinθ 2 cos & ω = Ω & cosi + R cosθ + e µ 1+ ecosθ M = nt χ, with & χ = A change in the node affects the argument of perigee (except for circular and polar orbits)! RAAN change ( θ θ ) θ ( θ ) a µ e θ 2 2 (1 e ) R 2e cos ecos + T sin 2 + ecos ( 1+ ecos ) 20

21 RAAN Change: Circular Orbits One-burn is sufficient. But where? Applying a change in velocity at any other points than the common points of intersection of the orbital planes will change both inclination and RAAN RAAN change 21

22 5. Orbital Maneuvers 5.4 Rendez-vous Hill Clohessy Wiltshire equations Gemini and Space Shuttle 22

23 Rendez-vous: Two Successive Phases 1. Long-distance navigation: basic maneuvers bring the spacecraft in a vicinity (<100 km) of the target in the same orbit plane. 2. Terminal rendez-vous: the interest is in the relative motion, i.e. the motion of the maneuvering spacecraft in relation to the target (spacecraft or neighboring orbit). Focus of the present discussion H-C-W equations 23

24 ISS & ATV Rescue of Intelsat-6: new perigee kick motor (STS-49) STS & HST

25 Basic Assumption The rendez-vous can be solved using the classical 2- body problem. But nonlinear equations of motion have to be solved and one should resort to numerical methods (e.g., Lambert s problem). Can we use approximate equations considering that distances are small relative to the dimensions of the orbit? Yes! Use linearization procedures H-C-W equations 25

26 Problem Statement chase r r 0 δr target (known orbit) δ r r = r0 + δr, <<< 1 r 0 What is the equation governing the relative motion δr? A rendez-vous occurs when δr = H-C-W equations 26

27 Mathematical Developments Relative motion as well! r&& = µ r r 3 r = r0 + δr 0 + δ δ && r = && r0 µ r r 3 r? r = ( r + δr).( r + δr) = r + 2 r. δr + δ r r r r δr = r0 3 2 Binomial theorem r r r δr = r H-C-W equations 27

28 Mathematical Developments 1 3 r0. δr δ && r = && r0 µ 3 5 r0 + δr r0 r0 ( ) Higher-order terms neglected δ && r = && r r µ µ 3 r. δr δr r r0 r0 r0 r&& = µ r r µ 3 0. δ δ r r && r = δ 3 r r 2 r0 r0 0 Linear ODE. Observer in an inertial frame H-C-W equations 28

29 Co-Moving Frame & Circular Orbit r r 0 δr ˆk ĵ î Local vertical coordinate frame Our focus is on a rotating observer in the local frame (e.g., an ISS astronaut watching the ATV) dω && r = && r0 + δr + Ω ( Ω δr) + 2Ω δ vrel + δarel = && r0 + δ&& r dt H-C-W equations ( ) 2 δ&& r = Ω Ω. δr Ω δr + 2Ω δ vrel + δarel 29

30 Co-Moving Frame & Circular Orbit δr = δ x ˆi + δ y ˆj + δ z kˆ δ v = δ x& ˆi + δ y& ˆj + δ z& kˆ rel δa = δ&& x ˆi + δ&& y ˆj + δ&& z kˆ rel Ω = nkˆ ( ) 2 δ&& r = Ω Ω. δr Ω δr + 2Ω δ vrel + δarel Interpretation of the terms? δ&& δ δ & δ&& ˆ δ δ & δ&& ˆ δ&& ˆ 2 2 r = ( n x 2 n y + x) i + ( n y + 2 n x + y) j+ z k Centrifugal Coriolis H-C-W equations 30

31 Co-Moving Frame & Circular Orbit δ&& δ δ & δ&& ˆ δ δ & δ&& ˆ δ&& ˆ 2 2 r = ( n x 2 n y + x) i + ( n y + 2 n x + y) j+ z k µ 3 r0. δr 2 ˆ ˆ ˆ 3 r0. δ x ˆ δ&& r = 3 δr r 2 0 = n δ x i + δ y j+ δ z k r 2 0i r0 r0 r0 δ&& x n δ x nδ y& = δ&& y + 2nδ x& = 0 2 δ&& z + n δ z = 0 Hill Clohessy Wiltshire equations: acceleration relative to a rotating observer fixed in the moving frame H-C-W equations 31

32 An Obvious Solution δ&& x n δ x nδ y& = δ&& y + 2nδ x& = 0 2 δ&& z + n δ z = 0 δ x =? δ y =? δ z =? δx=0, δy=constant. δz=harmonic motion (the out-of plane motion is decoupled from the in-plane motion). Physical interpretation? Two spacecraft on the same orbit H-C-W equations 32

33 In Matrix Form Linear ODEs with constant coefficient: we can solve them! 2 δ&& x 0 2n 0 δ x& 3n 0 0 δ x 0 δ&& y 2n 0 0 δ y& δ y = 0 2 δ z δ z 0 0 n δ z 0 && & Gyroscopic damping: imposes a rotation if there is a relative velocity H-C-W equations Centrifugal + gravitational terms: move away from the target if not on the same orbit 33

34 Interpretation Impulse to the left rotation due to Coriolis Forward impulse backward motion! Backward impulse forward motion! H-C-W equations 34

35 Solution δ&& y + 2nδ x& = 0 δ y& + 2nδ x = Cte = δ y& + 2nδ x 0 0 δ&& x 3n δ x 2nδ y& = 0 δ&& x + n δ x = 2nδ y& + 4n δ x δ x = Asin nt + B cos nt + δ y& + 4δ x n H-C-W equations 35

36 Solution: Overall δr( t) Φ ( t) δr ( t) Φ rv ( t 0 ) δ v ( t) 0 rr ( nt) sin nt 2 1 cos 0 n n δ x 4 3cos nt 0 0 δ x δ x& 0 2( cos nt 1) ( 4sin nt 3nt ) 0 δ y = 6( sin nt nt) 1 0 δ y + 0 δ y& 0 0 n n δ z 0 0 cos nt δ z 0 δ z & sin nt n δ x& 3nsin nt 0 0 δ x0 cos nt 2sin nt 0 δ x& 0 δ y& = 6n cos nt δ y + 2sin nt 4 cos nt 3 0 δ y& ( ) 0 0 δ z 0 0 nsin nt δ z cos nt δ z & & 0 δ v( t) Φ ( t) δr ( t) Φ vv ( t 0 ) δ v ( t) 0 vr H. Curtis, Orbital Mechanics for Engineering Students, Elsevier H-C-W equations 36

37 Key Feature? ( nt) sin nt 2 1 cos 0 n n δ x 4 3cos nt 0 0 δ x δ x& 0 2( cos nt 1) ( 4sin nt 3nt ) 0 δ y = 6( sin nt nt) 1 0 δ y + 0 δ y& 0 0 n n δ z 0 0 cos nt δ z 0 δ z & sin nt n Secular terms! What do they mean? H-C-W equations 37

38 Example: HST and STS The HST is released from the Space Shuttle, which is in a circular orbit of 590 km altitude. The relative velocity of ejection is 0.1 m/s down, 0.04 m/s backwards and 0.02 m/s to the right. Find the position of HST after 5,10,20 minutes. ( nt) sin nt 2 1 cos 0 n n δ x 4 3cos nt 0 0 δ x δ x& 0 2( cos nt 1) ( 4sin nt 3nt ) 0 δ y = 6( sin nt nt) 1 0 δ y + 0 δ y& 0 0 n n δ z 0 0 cos nt δ z 0 δ z & n = = rad/s sin (-0.1;-0.04;-0.02) T nt n H-C-W equations 38

39 Example: HST and STS 5 minutes: ( ; ; ) T 10 minutes: ( ; ; ) T 20 minutes: ( ; ; ) T H-C-W equations 39

40 Example: HST and STS δ y (m) δ x (m) H-C-W equations 40

41 Further Reading on the Web Site They use Hill Clohessy Wiltshire equations! H-C-W equations 41

42 Solution: Overall δr( t) = Φ ( t) δr + Φ ( t) δ v rr 0 rv 0 δ v( t) = Φ ( t) δr + Φ ( t) δ v vr 0 vv 0 Linear analog of Kepler s equation Assumptions: 1. The two satellites are within a few kms of each other. 2. The target is in a circular orbit. 3. There are no external forces on the interceptor. Starting from t=0, how to exploit these equations for rendez-vous after a time t f? H-C-W equations 42

43 Two-Impulse Rendez-vous Maneuvers t = 0, δr and δ v known 0 0 Compute δ v + 0 for rendez-vous in a time t f 0 = Φ ( t ) δr + Φ ( t ) δ v + rr f 0 rv f 0 δ v = Φ ( t ) Φ ( t ) δr rv f rr f 0 v = δ v δ v Compute δ v = Φ ( t ) δr + Φ ( t ) δ v + f vr f 0 vv f 0 = Φ ( t ) δr + Φ ( t ) Φ ( t ) Φ ( t ) δr 1 vr f 0 vv f rv f rr f 0 Impose δ + = 0 v = δ v v f f f H-C-W equations 43

44 Physical Interpretation The initial velocity change places the vehicle on a trajectory which will intercept the origin at time t f. The final velocity change cancels the arrival velocity of the vehicle at the origin and completes the rendez-vous H-C-W equations 44

45 Example: HST and STS After 10 minutes, the STS astronauts want to retrieve HST. Compute the δv 0+ to rendez-vous in 5 and 15 minutes. 5 minutes: ( ; ; ) T 15 minutes: ( ; ; ) T H-C-W equations 45

46 Real Data: Gemini 10 and Agena 8 A 4-orbit rendez-vous plan similar to that of Gemini 6-7: O1: no maneuvers to allow spacecraft checkout. N H N CI N SR TPI TPF O2: height adjustment at perigee (0.2 m/s for orbital decay) phase adjustment maneuver at apogee (17 m/s). O3: switch to a circular orbit in the same plane as Agena (16 m/s). O4: terminal phase initiation (10 m/s) velocity matching (13 m/s) Total: ± 56 m/s Gemini and Space Shuttle 46

47 5.4.2 Gemini and Space Shuttle Nc: nominal corrective burn Nh: nominal height adjust burn

48 Real Data: Gemini 10 and Agena Gemini and Space Shuttle 48

49 Real Data: Gemini 10 and Agena Gemini and Space Shuttle 49

50 Real Data: Gemini 10 and Agena 8 50

51 STS-120 (Rendez-vous with ISS, Oct. 07) Date Time UTC Name Impulsion Orbite (m/s) (km x km) 23 15:46:44 MECO x :16:38 OMS-2 70, x :32:45 NC1 26, x :50:56 NC2 6, x :27:09 NC3 0, x :30:35 NH 7, x :24:45 NC4 2, x :55:43 Ti 2, x :12:45 MC1 0, x :26:20 MC2 0, x 334 Pour STS123: Gemini and Space Shuttle 51

52 STS-120 (Rendez-vous with ISS, Oct. 07) Gemini and Space Shuttle 52

53 5. Orbital Maneuvers 5.1 INTRODUCTION Why? How? How much? When? 5.2 COPLANAR MANEUVERS One-impulse transfer Two-impulse transfer Three-impulse transfer Nontangential burns Phasing maneuvers 53

54 5. Orbital Maneuvers 5.3 NONCOPLANAR TRANSFERS Inclination change RAAN change 5.4 RENDEZ-VOUS Hill Clohessy Wiltshire equations Gemini and Space Shuttle 54

55 In Summary Maneuvers can find many practical applications and may drive spacecraft design (fuel consumption). Focus on impulsive maneuvers (chemical propulsion). Most important maneuvers in Earth s orbit have been described (plane change, circularization, phasing, rendezvous). Maneuvers are not always intuitive: to catch a spacecraft one the same orbit, one must first decelerate! 55

56 Other maneuvers Combined maneuvers (e.g., circularization and inclination change for a GEO satellite). Fixed V maneuvers. 56

57 Astrodynamics (AERO0024) 5. Orbital Maneuvers Gaëtan Kerschen Space Structures & Systems Lab (S3L)