Using Direct Transcription to Compute Optimal Low Thrust Transfers Between Libration Point Orbits

Abstract

The direct transcription method has been used to solve many challenging optimal control problems. One such example involves the calculation of a low thrust orbit transfer between libration point orbits. The recent implementation of high order discretization techniques is first described and then illustrated by computing optimal low thrust trajectories between orbits about the L ₁ and L ₂ Earth-Moon libration points.

Appendix: Lobatto IIIA Method Coefficients

$$\displaystyle\begin{array}{rcl} S = 2& & {}\\ \begin{array}{c|cc} \rho _{1} & \alpha _{11} & \alpha _{12}\\ \rho _{ 2} & \alpha _{21} & \alpha _{22} \\\hline & \beta _{1} & \beta _{2}\end{array} \qquad = \qquad \begin{array}{c|cc} 0 & 0 & 0 \\ 1 &\frac{1} {2} & \frac{1} {2} \\\hline &\frac{1} {2} & \frac{1} {2} \end{array} & & {}\\ \end{array}$$

$$\displaystyle\begin{array}{rcl} S = 3& & {}\\ \begin{array}{c|ccc} \rho _{1} & \alpha _{11} & \alpha _{12} & \alpha _{13}\\ \rho _{ 2} & \alpha _{21} & \alpha _{22} & \alpha _{23}\\ \rho _{3}&\alpha _{31}&\alpha _{32}&\alpha _{33} \\\hline & \beta _{1} & \beta _{2} & \beta _{3}\end{array} \qquad = \qquad \begin{array}{c|ccc} 0 & 0 & 0 & 0 \\ \frac{1} {2} & \frac{5} {24} & \frac{1} {3} & - \frac{1} {24} \\ 1 & \frac{1} {6} & \frac{2} {3} & \frac{1} {6} \\\hline & \frac{1} {6} & \frac{2} {3} & \frac{1} {6} \end{array} & & {}\\ \end{array}$$

$$\displaystyle\begin{array}{rcl} S = 4& & {}\\ \begin{array}{c|cccc} \rho _{1} & \alpha _{11} & \alpha _{12} & \alpha _{13} & \alpha _{14}\\ \rho _{ 2} & \alpha _{21} & \alpha _{22} & \alpha _{23} & \alpha _{24}\\ \rho _{3}&\alpha _{31}&\alpha _{32}&\alpha _{33}&\alpha _{34} \\ \rho _{4} & \alpha _{41} & \alpha _{42} & \alpha _{43} & \alpha _{44} \\\hline & \beta _{1} & \beta _{2} & \beta _{3} & \beta _{4}\end{array} \qquad = \qquad \begin{array}{c|cccc} 0 & 0 & 0 & 0 & 0 \\ \frac{1} {2} -\frac{\sqrt{5}} {10} & \frac{11+\sqrt{5}} {120} & \frac{25-\sqrt{5}} {120} & \frac{25-13\sqrt{5}} {120} & \frac{-1+\sqrt{5}} {120} \\ \frac{1} {2} + \frac{\sqrt{5}} {10} & \frac{11-\sqrt{5}} {120} & \frac{25+13\sqrt{5}} {120} & \frac{25+\sqrt{5}} {120} & \frac{-1-\sqrt{5}} {120} \\ 1 & \frac{1} {12} & \frac{5} {12} & \frac{5} {12} & \frac{1} {12} \\\hline & \frac{1} {12} & \frac{5} {12} & \frac{5} {12} & \frac{1} {12} \end{array} & & {}\\ \end{array}$$

$$\displaystyle\begin{array}{rcl} S = 5& & {}\\ \begin{array}{c|ccccc} \rho _{1} & \alpha _{11} & \alpha _{12} & \alpha _{13} & \alpha _{14} & \alpha _{15}\\ \rho _{ 2} & \alpha _{21} & \alpha _{22} & \alpha _{23} & \alpha _{24} & \alpha _{25}\\ \rho _{3}&\alpha _{31}&\alpha _{32}&\alpha _{33}&\alpha _{34}&\alpha _{35} \\ \rho _{4} & \alpha _{41} & \alpha _{42} & \alpha _{43} & \alpha _{44} & \alpha _{45}\\ \rho _{ 5} & \alpha _{51} & \alpha _{52} & \alpha _{53} & \alpha _{54} & \alpha _{55} \\\hline & \beta _{1} & \beta _{2} & \beta _{3} & \beta _{4} & \beta _{5}\end{array} & & {}\\ \qquad = \qquad \begin{array}{c|ccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{1} {2} -\frac{\sqrt{21}} {14} & \frac{119+3\sqrt{21}} {1960} & \frac{343-9\sqrt{21}} {2520} & \frac{392-96\sqrt{21}} {2205} & \frac{343-69\sqrt{21}} {2520} & \frac{-21+3\sqrt{21}} {1960} \\ \frac{1} {2} & \frac{13} {320} & \frac{392+105\sqrt{21}} {2880} & \frac{8} {45} & \frac{392-105\sqrt{21}} {2880} & \frac{3} {320} \\ \frac{1} {2} + \frac{\sqrt{21}} {14} & \frac{119-3\sqrt{21}} {1960} & \frac{343+69\sqrt{21}} {2520} & \frac{392+96\sqrt{21}} {2205} & \frac{343+9\sqrt{21}} {2520} & \frac{-21-3\sqrt{21}} {1960} \\ 1 & \frac{1} {20} & \frac{49} {180} & \frac{16} {45} & \frac{49} {180} & \frac{1} {20} \\\hline& \frac{1} {20} & \frac{49} {180} & \frac{16} {45} & \frac{49} {180} & \frac{1} {20} \end{array} & & {}\\ \end{array}$$