Low-energy transfers to the Moon with long transfer time

Abstract

This paper globally explores two-impulse, low-energy Earth–Moon transfers in the planar bicircular restricted four-body problem with transfer time of up to 200 days. A grid search combined with a direct transcription and multiple shooting technique reveals numerous families of optimal low-energy solutions, including some that have not been reported yet. We investigate characteristics of solutions in terms of parameters in two- and three-body dynamics, and discuss a trade-off between cost and transfer time based on Pareto-optimal solutions, with and without lunar gravity assists. Analysis of orbital characteristics reveals the role of the Sun, the Earth, and the Moon in the transfer dynamics.

Appendices

Appendix 1

Since the majority of the studies in Table 1 use the initial and final circular orbits around the Earth and the Moon of “reference” altitudes 167 km and 100 km, respectively, this study transforms the values of $\varDelta v$ of the solutions using different altitudes (Yamakawa 1993; Morcos 2010; Peng et al. 2010; Moore et al. 2012) for a fair comparison. The following calculation shows the case of transforming Earth departure maneuvers, but the transformation of Moon arrival maneuvers is similar.

The Kepler energy at departure from an initial Earth circular orbit of a different altitude from the reference altitude is

$$\begin{aligned} H_{E1}=\frac{1}{2} {v_{E1}}^2-\frac{1-\mu }{r_{E1}}, \end{aligned}$$

(22)

where $v_{E1}$ and $r_{E1}$ are the magnitude of the spacecraft’s velocity in the Earth-centered inertial frame and the distance from the center of the Earth, respectively.

The required velocity $v_{E2}$ to achieve $H_{E1}$ from the reference altitude at the Earth is

$$\begin{aligned} v_{E2}=\sqrt{2\left( H_{E1}+\frac{1-\mu }{r_{E2}}\right) }, \end{aligned}$$

(23)

where $r_{E2}$ is the distance between the reference circular orbit and the center of the Earth.

The transformed Earth departure maneuver can be computed by assuming the tangential maneuver as

$$\begin{aligned} \varDelta v_E=v_{E2}-\sqrt{\frac{1-\mu }{r_{E2}}}. \end{aligned}$$

(24)

Appendix 2

This appendix presents analytic derivatives of the objective function and the constraints with respect to the NLP variables in the optimization problem in Sect. 4.2.1. For the sake of simple notations, we assume $N=4$ in the following expressions, but the generalization is straightforward.

The derivative of the objective function J with respect to the NLP variables ${\varvec{y}}$ can be expressed as

$$\begin{aligned} \frac{\partial J}{\partial {\varvec{y}}}=\begin{bmatrix} P_1&O&P_N&O \end{bmatrix}, \end{aligned}$$

(25)

where

$$\begin{aligned}&P_1 :=\frac{\partial J}{\partial {\varvec{x}}_1} = \frac{1}{\sqrt{(\dot{x}_1-y_1)^2+(\dot{y}_1+x_1+\mu )^2}} \begin{bmatrix} \dot{y}_1+x_1+\mu&y_1-\dot{x}_1&\dot{x}_1-y_1&\dot{y}_1+x_1+\mu \end{bmatrix}, \nonumber \\\end{aligned}$$

(26)

$$\begin{aligned}&P_N :=\frac{\partial J}{\partial {\varvec{x}}_N} = \frac{1}{\sqrt{(\dot{x}_N-y_N)^2+(\dot{y}_N+x_N+\mu -1)^2}}\nonumber \\&\begin{bmatrix} \dot{y}_N+x_N+\mu -1&y_N-\dot{x}_N&\dot{x}_N-y_N&\dot{y}_N+x_N+\mu -1 \end{bmatrix}. \end{aligned}$$

(27)

The derivative of the equality constraints ${\varvec{c}} :=\{{\varvec{\zeta }}_j, {\varvec{\psi }}_1, {\varvec{\psi }}_N\}={\varvec{0}}$ with respect to the NLP variables ${\varvec{y}}$ can be expressed as

$$\begin{aligned} \frac{\partial {\varvec{c}}}{\partial {\varvec{y}}}=\begin{bmatrix} {\varvec{\varPhi }}(t_1, t_2)&-I_4&O&O&Q^1_1&Q^1_N \\ O&{\varvec{\varPhi }}(t_2, t_3)&-I_4&O&Q^2_1&Q^2_N \\ O&O&{\varvec{\varPhi }}(t_{N-1}, t_N)&-I_4&Q^{N-1}_1&Q^{N-1}_N \\ R_1&O&O&O&O&O \\ O&O&O&R_N&O&O \\ \end{bmatrix}, \end{aligned}$$

(28)

where

$$\begin{aligned}&Q^j_1 :=\frac{\partial {\varvec{\zeta }}_j}{\partial {\varvec{t}}_1} = -\frac{N-j}{N-1}{\varvec{\varPhi }}(t_{j}, t_{j+1}){\varvec{f}}({\varvec{x}}_j, t_j) \nonumber \\&\quad +\frac{N-j-1}{N-1}{\varvec{f}}({\varvec{\varphi }}({\varvec{x}}_j, t_j, t_{j+1}), t_{j+1}), \qquad j = 1, \ldots , N-1, \end{aligned}$$

(29)

$$\begin{aligned}&Q^j_N :=\frac{\partial {\varvec{\zeta }}_j}{\partial {\varvec{t}}_N} = -\frac{j-1}{N-1}{\varvec{\varPhi }}(t_{j}, t_{j+1}){\varvec{f}}({\varvec{x}}_j, t_j) \nonumber \\&\quad +\frac{j}{N-1}{\varvec{f}}({\varvec{\varphi }}({\varvec{x}}_j, t_j, t_{j+1}), t_{j+1}), \qquad j = 1, \ldots , N-1, \end{aligned}$$

(30)

$$\begin{aligned}&R_1 :=\frac{\partial {\varvec{\psi }}_1}{\partial {\varvec{x}}_1} = \begin{bmatrix} 2(x_1+\mu )&2y_1&0&0 \\ \dot{x}_1&\dot{y}_1&x_1+\mu&y_1\\ \end{bmatrix}, \end{aligned}$$

(31)

$$\begin{aligned}&R_N :=\frac{\partial {\varvec{\psi }}_N}{\partial {\varvec{x}}_N} = \begin{bmatrix} 2(x_N+\mu -1)&2y_N&0&0 \\ \dot{x}_N&\dot{y}_N&x_N+\mu -1&y_N \\ \end{bmatrix}. \end{aligned}$$

(32)

The derivative of the inequality constraints ${\varvec{g(y)}} :=\{{\varvec{\eta }}_j, \tau \}<{\varvec{0}}$ with respect to the NLP variables ${\varvec{y}}$ can be expressed as

$$\begin{aligned} \frac{\partial {\varvec{g}}}{\partial {\varvec{y}}}=\begin{bmatrix} S_1&O&O&O&O \\ O&S_2&O&O&O \\ O&O&S_3&O&O \\ O&O&O&S_N&O \\ O&O&O&O&S_t \\ \end{bmatrix}, \end{aligned}$$

(33)

where

$$\begin{aligned}&S_j :=\frac{\partial {\varvec{\eta }}_j}{\partial {\varvec{x}}_j} = \begin{bmatrix} -2(x_j+\mu )&-2y_j&0&0 \\ -2(x_j+\mu -1)&-2y_j&0&0 \\ \end{bmatrix}, \qquad j = 1, \ldots , N, \end{aligned}$$

(34)

$$\begin{aligned}&S_t :=\begin{bmatrix} \dfrac{\partial \tau }{\partial t_1}&\dfrac{\partial \tau }{\partial t_N} \end{bmatrix} = \begin{bmatrix} 1&-1 \end{bmatrix}. \end{aligned}$$

(35)