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### OPTIMAL TWO IMPULSE EARTH-MOON Trajectories

John McGreevy

Advisor: Manoranjan Majji

Committee:

John Crassidis

TarunSingh

Outline

- Motivation
- Problem Formulation
- Results for Planar Trajectories
- Comparison with previous work
- Effect of varying initial conditions
- Effect of varying time of flight
- Addition of Terminal Inclination Constraint
- Results
- Conclusions

Motivation

- Lunar Missions
- Moon observing
- Moon landing
- Reduce necessary thrust
- Save fuel/money
- Increase payload
- Short duration
- Not low-energy trajectories

Problem Statement

- Find trajectory that minimizes cost function
- Mayer cost
- Quadratic cost allows closed form solutions
- Fixed initial conditions
- Circular, low Earth orbit, in Earth-Moon plane
- Terminal manifold
- Circular, low lunar orbit
- Fixed time
- Non-dimensional units
- Distance ~ Moon orbit radius
- Time ~ Inverse of angular frequency of lunar orbit

Solution Method

- Earth centered non-rotating frame
- Using necessary conditions of optimal control theory
- Hamiltonian:

Terminal Constraints

- Form augmented cost function:
- are terminal Lagrange multipliers

Terminal Constraints

- Solve internal optimization problem for
- Take first variation of with respect to
- Because states are free at final time, the co-states are fixed
- Transversality conditions:

Necessary Conditions

- Can also verify solution using direct method
- Take variation of augmented cost function with respect to and set = 0

Computations

- Utilized Shooting Method
- Guess values of unknowns
- 6 initial values of co-states
- 3 Lagrange multipliers
- Solve 12 differential equations
- Evaluate terminal boundary conditions
- 6 transverality conditions
- 3 terminal constraint equations
- Update guess using Newton’s Method and repeat until convergance

Computations

- Provided initial conditions based on Miele1
- Fixed initial conditions
- f0 = -116.88°
- Orbit altitude of 463 km
- Circular orbit
- In Earth-Moon plane
- Fixed terminal manifold
- Orbit altitude of 100km
- Circular orbit
- Did not require orbit to be in Earth-Moon plane

[1] Miele, A. and Mancuso, S., “Optimal Trajectories For Earth-Moon-Earth Flight,” ActaAstronautica, Vol. 49, No. 2, 2001, pp. 59 – 71.

Comparison Results

- Verify accuracy of current method
- Reasons for some expected differences:
- Different cost function
- Weighted quadratic with respect to state variables instead of 2-norm
- Additional term in dynamics

Comparison Results

Inertial Frame, 4.5 days, -116.88°, 463km

Comparison Results

Rotating Frame, 4.5 days, -116.88°, 463km

Comparison Results

Rotating Frame, 4.5 days, -116.88°, 463km

Comparison Results

Inertial Frame, 4.5 days, -116.88°, 463km

- Comparable results for both trajectories
- Small increase due to additional term in dynamics

Varying Initial Conditions

- Vary initial true anomaly
- Range of 50° from -116.88° to -166.88°
- Initial altitude of 160 km
- Match Apollo
- Same terminal manifold as before
- Orbit altitude of 100km
- Circular orbit
- Did not require orbit to be in Earth-Moon plane
- Use previous solution to provide initial guess of unknowns for similar initial conditions

Varying Time of Flight

- Want to find time of flight which minimizes cost function
- Use initial true anomaly of -135° based on previous section
- Initial altitude of 160 km
- Same terminal manifold as before
- Orbit altitude of 100km
- Circular orbit
- Did not require orbit to be in Earth-Moon plane

Summary So Far

- Take advantage of quadratic cost formulation
- Multiple solutions for each set of initial conditions
- Retrograde and Prograde
- Retrograde orbit is more sensitive to initial guess
- Lowest cost occurs at tf=3.15 days
- Next:
- Non-planar orbits
- Specify terminal orbit inclination

Additional Constraint

- Using equation for orbital inclination:
- Put into quadratic form:
- Form new augmented cost:

Terminal Constraints

- New equation for
- where
- Because states are free at final time, the co-states are fixed
- Transversality conditions:

Solution failure at 90° inclination

- Inverted matrix in becomes ill-conditioned at 90°
- M is singular (rank 1)
- Large value of
- M dominates
- Possible cause is constraint formulation

Summary

- Take advantage of quadratic cost formulation
- Multiple solutions for each set of initial conditions
- Retrograde and Prograde
- Obtained solutions for desired terminal inclinations
- Discovered convergence problem at 90° inclination
- Future Work
- Vary multiple initial conditions and time simultaneously
- Non-planar orbits
- Specify initial orbit inclination
- Multiple revolution trajectories

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