colorado center for astrodynamics research -...

Colorado Center for Astrodynamics Research

1/27/15

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...)

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SOI Earth

SOI Mars


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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

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Coplanar Transfers: Hohmann

✦ Two impulse minimum ΔV ✦ Transfers between two circular/elliptical* coplanar orbits ✦ Necessary Conditions: Tangential Burns! ✦ Excludes parabolic/hyperbolic transfers

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ΔV2

ΔV1

Initial Orbit

Final Orbit

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


1/27/15

Coplanar Transfers: Bi-elliptic

✦ Variant of Hohmann ✦ Requires 3 ΔVs ✦ Consists of two half elliptical orbits

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Initial Orbit, r0

Final Orbit, rf


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✦ ΔV1 at r0 to raise apoapsis to rB ✦ rB > rf

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ΔV1

Initial Orbit, r0

Final Orbit, rf

rB


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✦ ΔV2 at rB (apoapsis of transfer orbit) ✦ Raise periapsis of transfer orbit to that of final orbit

✦ ΔV3 at rf to inject into final orbit

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ΔV2

ΔV1ΔV3

Initial Orbit, r0

Final Orbit, rf

rB


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✦ Bi-elliptic is more economical than Hohmann if

✦ Drawbacks?

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ΔV2

ΔV1ΔV3

rfr0

> 15.58

Initial Orbit, r0

Final Orbit, rf

rB


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✦ Bi-elliptic is more economical than Hohmann if

✦ Drawbacks? ✦ Increased time ✦ Extra burn (complexity) ✦ for

185 km LEO to GEO.

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ΔV2

ΔV1ΔV3

rfr0

> 15.58

rfr0

= 6.4

Initial Orbit, r0

Final Orbit, rf

rB


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Parabolic Transfer✦ Similar to Bi-elliptic transfer ✦ rB = ∞

✦ 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

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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


1/27/15

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

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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?

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SOI Earth

SOI Mars


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What do we know?✦ Assume we want to travel from Earth to Mars and we

are given a launch date and arrival date

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1/27/15

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?

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r2

r1

Mars at arrival

Earth at Departure


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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

✦ A Keplerian orbit is the general solution

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r̈ = �µr̂

r2


1/27/15

Lambert’s Problem✦ Given ✦ at Launch Date ✦ at Arrival Date ✦ TOF

✦ Solve Lambert’s Problem to determine the orbital elements of the transfer ✦ Get the departure and

arrival velocities ✦ Transfer path = Conic section

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r2

r1

r1

r2


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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

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1/27/15

One Conic: Two Travel Directions

✦ Even though the conic connecting P1 and P2 is unique.... ✦ “Short Way”: Δν < 180o: Type I

✦ “Long Way”: Δν > 180o: Type II

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r2

r1

Δν

r2

r1

Δν

✦ Assume planets both rotate CCW. ✦ Notice any impracticalities?


1/27/15

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

✦ Does this mean you’ll never have Δν > 180o?19

r2

r1

Δν

r2

r1

Δν


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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

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r2

r1

Δν

r2

r1

Δν


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Lambert’s Problem✦ Geometry Considerations ✦ Ellipse is defined by 2 Foci

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P1 P2

F⇥r1 ⇥r2

O

P1 = Planet 1P2 = Planet 2O = Origin (Sun), Focus #1F = Empty Focus


1/27/15

Lambert’s Problem✦ Geometry Considerations ✦ Ellipse is defined by 2 Foci ✦ If empty focus is located, then ellipse is defined ✦ Else: Infinite # of conics

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P1 P2

F

F

⇥r1 ⇥r2

O

P1 = Planet 1P2 = Planet 2O = Origin (Sun), Focus #1F = Empty Focus


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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

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c =�

r21 + r22 � 2r1r2 cos(��)

P1 P2

F

⇥r1 ⇥r2

c

O

��


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Lambert’s Problem✦ Sum of the distances from a point on the orbit to each

foci is twice the semimajor axis:

✦ Substitute into (1) above to get:

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OF

rPOrPF

P

rPO + rPF = 2a

rPF = 2a� rPO

2a = rPO + (2a� rPO)


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Lambert’s Problem✦ This relationship holds for any point on the orbit

✦ Use this knowledge to find possible locations for the empty focus

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2a = r + (2a� r)

OF

r12a-r1

r22a-r2

P


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Locate the Empty Focus✦ For each position (P1 and P2), select a value for a ✦ Draw circles about each position point with radii and

✦ Intersection of dashed circles = empty focus

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P1 P2

O

cFm

2am - r12am -

r2

2a� r1 2a� r2

Note that for am

Minimum Energy Orbit(2am � r1) + (2am � r2) = c

r1r2


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Locate the Empty Focus✦ For any a>am, there will be two possibilities for the empty foci

✦ am represents the minimum energy solution

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P1 P2

O

cFm

2am - r12am -

r2

2an - r1

Fn

2an - r2

Fn


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Locate the Empty Focus✦ Draw a line connecting the empty foci

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P1 P2

O

cFm

2am - r12am -

r2

2an - r1

2an - r2

Fn

2ap -

r 1Fp

2ap -

r 2Fp

Fn


1/27/15

Locate the Empty Focus✦ Locus of secondary foci lies on a hyperbola

✦ Use the Time of Flight to find the empty focus

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P1 P2

O

cFm

2am - r12am -

r2

2an - r1

2an - r2

Fn

2ap -

r 1Fp

2ap -

r 2Fp

Fn


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Locate the Empty Focus✦ Conic with TOF that defines Fp

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P1 P2

O

cFm

Fn Fp

Fp

Fn


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Locate the Empty Focus✦ Conic with TOF that defines Fp

✦ Conic with TOF that defines Fn

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P1 P2

O

cFm

Fn Fp

Fp

Fn


1/27/15

Empty Focus: Hyperbolic transfer

✦ Recall hyperbola has semimajor axis <0 ✦ Draw circles with radii 2ai+r

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P1 P2

O

c

2ai + r12ai + r2

r2r1


1/27/15

Empty Focus: Hyperbolic transfer

✦ Again, empty foci form a hyperbola

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P1

O

c

r1

P1 P2

O

c

2ai + r12ai + r2

r2r1


1/27/15

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<180o ✦ -1 for Δν>180o

✦ Many techniques are available for solving Lambert’s ✦ We will use Universal Variables

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