Patch Conics. Basic Approach

Transcription

1 Patch Conics Basic Approach Inside the sphere of influence: Planet is the perturbing body Outside the sphere of influence: Sun is the perturbing body (no extra-solar system trajectories in this class...) SOI Mars SOI Earth 1

2 Orbital Maneuvers Necessary to achieve the spacecraft s final destination Assumption: Maneuvers are treated as instantaneous changes to velocity (ΔV) Good approximation for this class In reality, burns take place over a finite amount of time Numerically integrate to compute 2

3 Coplanar Transfers: Hohmann Two impulse minimum ΔV Transfers between two circular/elliptical* coplanar orbits Necessary Conditions: Tangential Burns! Excludes parabolic/hyperbolic transfers Final Orbit Initial Orbit ΔV2 ΔV1 Transfer Ellipse *Technically, Walter Hohmann only studied transfers between circular orbits. Others (Lawden, Thompson) extended his work to circular and determined that tangential burns minimized the cost 3

4 Coplanar Transfers: Bi-elliptic Variant of Hohmann Requires 3 ΔVs Consists of two half elliptical orbits Final Orbit, rf Initial Orbit, r0 4

5 Coplanar Transfers: Bi-elliptic ΔV1 at r 0 to raise apoapsis to r B r B > r f Final Orbit, rf rb ΔV1 Initial Orbit, r0 5

6 Coplanar Transfers: Bi-elliptic ΔV2 at r B (apoapsis of transfer orbit) Raise periapsis of transfer orbit to that of final orbit ΔV3 at r f to inject into final orbit Final Orbit, rf ΔV2 rb Initial Orbit, r0 ΔV1 ΔV3 6

7 Coplanar Transfers: Bi-elliptic Bi-elliptic is more economical than Hohmann if Drawbacks? r f r 0 > Final Orbit, rf ΔV2 rb ΔV1 ΔV3 Initial Orbit, r0 7

8 Coplanar Transfers: Bi-elliptic Bi-elliptic is more economical than Hohmann if Drawbacks? Increased time Extra burn (complexity) r f =6.4 r 0 for 185 km LEO to GEO. ΔV2 rb r f r 0 > Final Orbit, rf Initial Orbit, r0 ΔV1 ΔV3 8

9 Parabolic Transfer Similar to Bi-elliptic transfer r B = ΔV to inject onto minimum energy escape trajectory At, perform an infinitesimal ΔV to change inclination Basic idea: Move the spacecraft between two parabolas with different inclinations 9

10 Drawbacks of simple transfers Consider interplanetary transfer (e.g., Earth to Mars) What are major drawbacks to the simple transfers? Hohmann Transfer INCLINATION! Mars orbit is inclined to Earth Would need a large ΔV to get into the correct orbit How could we avoid high ΔV penalty and still use Hohmann? Need Earth at the node of Mars Strict Launch Requirements for Launch/Arrival Time of Flight: ~260 days. (A bit long...) 10

11 Drawbacks of simple transfers Consider interplanetary transfer (e.g., Earth to Mars) What is a major drawbacks to the simple transfers? Parabolic Transfer TIME OF FLIGHT! We probably don t have time to transfer to infinity with respect to the Sun to perform an infinitesimal ΔV to change inclinations 11

12 Modeling trajectory segments At Earth departure and Mars arrival, we can model the trajectory with respect to each planet Hyperbola with respect to the planet How do we solve for the orbit with respect to the Sun? SOI Mars SOI Earth 12

13 What do we know? Assume we want to travel from Earth to Mars and we are given a launch date and arrival date 13

14 What do we know? Assume we want to travel from Earth to Mars and we are given a launch date and arrival date States of Earth and Mars Gives transfer radii Time of Flight How do we solve for the transfer? r 2 Mars at arrival r 1 Earth at Departure 14

15 Lambert s Problem Lambert s Problem: Known: Two position vectors and the time of flight between them Unknown: Orbit between the endpoints Lambert first formed the solution ~1761 Lambert s is the boundary value problem for the differential equation r = µ ˆr r 2 A Keplerian orbit is the general solution 15

16 Lambert s Problem Given r 1 r 2 TOF at Launch Date at Arrival Date Solve Lambert s Problem to determine the orbital elements of the transfer Get the departure and arrival velocities Transfer path = Conic section r 1 r 2 16

17 Lambert s Theorem Given the Planets P1 and P2 The Time of Flight from P1 to P2 depends only on the sum of the magnitudes of the position vectors, the semimajor axis, and the length of the chord joining P1 and P2 Alternatively, if the TOF from P1 to P2 is specified, the conic connecting P1 and P2 is unique 17

18 One Conic: Two Travel Directions Even though the conic connecting P1 and P2 is unique... Short Way : Δν < 180 o : Type I Long Way : Δν > 180 o : Type II r 2 r 2 Δν Δν r 1 r 1 Assume planets both rotate CCW. Notice any impracticalities? 18

19 One Conic: Two Travel Directions For this class, it would not be wise to attempt a CW transfer if the planets are rotating CCW relative to the Sun r 2 r 2 Δν Δν r 1 r 1 Does this mean you ll never have Δν > 180 o? 19

20 Two solutions Takeaway: There are two solutions to Lamberts! Pick the good one. Good news: For this class, the algorithm for your Lambert Solver will assume Δν is in the planetary direction of motion r 2 r 1 Δν r 2 r 1 Δν 20

21 Lambert s Problem Geometry Considerations Ellipse is defined by 2 Foci P 1 P 2 P1 = Planet 1 P2 = Planet 2 O = Origin (Sun), Focus #1 F = Empty Focus r 1 r 2 O F 21

22 Lambert s Problem Geometry Considerations Ellipse is defined by 2 Foci If empty focus is located, then ellipse is defined Else: Infinite # of conics F P 1 P 2 P1 = Planet 1 P2 = Planet 2 O = Origin (Sun), Focus #1 F = Empty Focus r 1 r 2 O F 22

23 Lambert s Problem Lambert s techniques rely on the geometry of the problem Define the chord between the two position vectors Chord is the shortest distance between the endpoints c = r1 2 + r2 2 2r 1 r 2 cos( ) P 1 P 2 c r 1 r 2 O F 23

24 Lambert s Problem Sum of the distances from a point on the orbit to each foci is twice the semimajor axis: r PO + r PF =2a Substitute r PF =2a r PO into (1) above to get: 2a = r PO +(2a r PO ) P rpf rpo F O 24

25 Lambert s Problem This relationship holds for any point on the orbit 2a = r +(2a r) Use this knowledge to find possible locations for the empty focus P 2a-r1 r1 2a-r2 F r2 O 25

26 Locate the Empty Focus For each position (P1 and P2), select a value for a Draw circles about each position point with radii 2a r 1 and 2a r 2 Intersection of dashed circles = empty focus 2am - r 1 P1 c Fm 2am - r2 P2 r1 r2 O Note that for am (2a m r 1 )+(2a m r 2 )=c Minimum Energy Orbit 26

27 Locate the Empty Focus For any a>am, there will be two possibilities for the empty foci am represents the minimum energy solution Fn 2am - r 1 P1 c Fm 2am - r2 P2 2a n - r2 2an - r1 O Fn 27

28 Locate the Empty Focus Draw a line connecting the empty foci Fp Fn 2am - r 1 P1 c Fm 2am - r2 P2 2an - r 1 2ap - r 1 O Fn 2ap - r2 2a n - r2 Fp 28

29 Locate the Empty Focus Locus of secondary foci lies on a hyperbola Use the Time of Flight to find the empty focus Fp Fn 2am - r 1 P1 c Fm 2am - r2 P2 2an - r 1 2ap - r 1 O Fn 2ap - r2 2a n - r2 Fp 29

30 Locate the Empty Focus Conic with TOF that defines Fp Fn Fp P1 c Fm P2 O Fn Fp 30

31 Locate the Empty Focus Conic with TOF that defines Fp Conic with TOF that defines Fn Fn Fp P1 c Fm P2 O Fn Fp 31

32 Empty Focus: Hyperbolic transfer Recall hyperbola has semimajor axis <0 Draw circles with radii 2a i +r 2a i + r1 2ai + r2 P1 r 1 c O r 2 P2 32

33 Empty Focus: Hyperbolic transfer Again, empty foci form a hyperbola 2a i + r1 2ai + r2 P1 r 1 c O r 2 P2 33

34 Solving Lambert s Problem We use the chord, TOF and the direction of motion to determine Lambert s solution Direction of Motion +1 for Δν<180 o -1 for Δν>180 o Many techniques are available for solving Lambert s We will use Universal Variables 34