A family of periodic orbits in the three-dimensional lunar problem

Abstract

A family of periodic orbits is proven to exist in the spatial lunar problem that are continuations of a family of consecutive collision orbits, perpendicular to the primary orbit plane. This family emanates from all but two energy values. The orbits are numerically explored. The global properties and geometry of the family are studied.

Appendices

Appendix: Regularization in coordinates

Moser regularization is based on $n-$dimensional stereographic projection. The position and momentum variables are given by $q = (q_1, q_2, \ldots , q_n) \in \mathbb {R}^n$ , $p = (p_1, p_2, \ldots , p_n) \in \mathbb {R}^n$. We denote by $(q,p) \in T^*\mathbb {R}^n$ a point in the (co)-tangent bundle of $\mathbb {R}^n$, where we think of $\mathbb {R}^n$ as a chart for $S^n = \{|\xi |^2 = \varSigma _{i=0}^{n} \xi _i^2 = 1\}$, $\xi = (\xi _0, \xi _1, \ldots , \xi _n)$. We set $x= - p$ and $y=q$, and define the (co)-tangent bundle of $S^n$ as

$$\begin{aligned} T^*S^n =\{ (\xi ,\eta ) \in T^*\mathbb {R}^{n+1} ~|~ |\xi |^2 =1 , \quad \langle \xi ,\eta \rangle \equiv \varSigma _{i=0}^{n} \xi _i \eta _i = 0 \}. \end{aligned}$$

To go from $T^*S^n$ to $T^*\mathbb {R}^n$ we use the map

$$\begin{aligned} \begin{aligned} x&= \frac{\tilde{\xi }}{1-\xi _0} \\ y&= \eta _0 \tilde{\xi } +(1-\xi _0) \tilde{\eta } , \end{aligned} \end{aligned}$$

(14)

where $\tilde{\xi } = (\xi _1, \xi _2, \ldots , \xi _n)$. Collision corresponds to $\xi _0 = 1$.

To go from $T^*\mathbb {R}^n$ to $T^*S^n$, we use the inverse given by

$$\begin{aligned} \begin{aligned} \xi _0&= \frac{|x|^2-1}{|x|^2+1} \\ \tilde{\xi }&= -\frac{2 x}{|x|^2+1} \\ \eta _0&= -\langle x,y \rangle \\ \tilde{\eta }&= \frac{|x|^2+1}{2}y - \langle x, y \rangle x . \end{aligned} \end{aligned}$$

(15)

The Belbruno transform employs a Möbius transformation which sends to the collision point $|p| = \infty$ to $P=(1,0,\ldots ,0)\in \mathbb {R}^n$. In coordinates for three dimensions (the index $j=2,3$), the forward Belbruno transformation is given by

$$\begin{aligned} \begin{aligned} Q_1&= \frac{1-|p|^2}{2}q_1 +\langle q, p\rangle (p_1+1) \\ Q_j&= \frac{|p|^2 + 1}{2}q_j + p_1 q_j - p_j q_1 - \langle q, p\rangle p_j\\ P_1&= \frac{|p|^2 -1 }{|p+1|^2} \\ P_j&= \frac{2p_j}{|p+1|^2} . \end{aligned} \end{aligned}$$

The inverse Belbruno transform is given by

$$\begin{aligned} \begin{aligned} q_1&= \frac{1-|P|^2}{2}Q_1 + \langle Q,P\rangle (P_1 - 1)\\ q_j&= \frac{|P|^2 +1}{2}Q_j - P_1 Q_j + P_j Q_1 - \langle Q,P\rangle P_j\\ p_1&= \frac{1-|P|^2}{ |P-1|^2 } \\ p_j&= \frac{2P_j}{ |P-1|^2 } . \end{aligned} \end{aligned}$$

Appendix: Hamiltonian vector field with constraints

The setup is the following. We are given a manifold M, which is a symplectic submanifold of the symplectic manifold $(N,\varOmega )$. We denote the inclusion by $\iota : M\rightarrow N$, and the induced symplectic form on M by $\omega :=\iota ^*\varOmega$. We assume that $M=f_1^{-1}(0) \cap f_2^{-1}(0)$. In addition, we are given a Hamiltonian function $H_N:N \rightarrow \mathbb {R}$, and we have the induced Hamiltonian $H_M=\iota ^*H_N$. In our case $N=T^*\mathbb {R}^{n+1}$ and $M:=T^*S^n$.

The functions that define M are

$$\begin{aligned} f_1=\frac{1}{2} |\xi |^2-\frac{1}{2},\quad f_2=\langle \xi ,\eta \rangle . \end{aligned}$$

In our case, the symplectic manifold $N=T^*\mathbb {R}^{n+1}$ has a global chart, but $T^*S^n$ has not. We will give a formula for the Hamiltonian vector field $X_H$ on M in terms of Hamiltonian vector field on N. In our example, this means that we can use the global coordinates on $N=T^*\mathbb {R}^{n+1}$. We have

$$\begin{aligned} X_H=X_{H_N}+c_1 X_{f_1}+c_2 X_{f_2}, \end{aligned}$$

(16)

where

$$\begin{aligned} \begin{aligned} X_{H_N}&=\sum _{j=0}^n \frac{\partial H_N}{\partial \eta _j} \frac{\partial }{\partial \xi _j}-\frac{\partial H_N}{\partial \xi _j} \frac{\partial }{\partial \eta _j} \\ X_{f_1}&=\sum _{j=0}^n \frac{\partial f_1}{\partial \eta _j} \frac{\partial }{\partial \xi _j}-\frac{\partial f_1}{\partial \xi _j} \frac{\partial }{\partial \eta _j} = -\sum _j \xi _j \frac{\partial }{\partial \eta _j}\\ X_{f_2}&=\sum _{j=0}^n \frac{\partial f_2}{\partial \eta _j} \frac{\partial }{\partial \xi _j}-\frac{\partial f_2}{\partial \xi _j} \frac{\partial }{\partial \eta _j} = \sum _j \xi _j \frac{\partial }{\partial \xi _j}-\eta _j \frac{\partial }{\partial \eta _j}\\ c_1&= \frac{df_2(X_{H_N})}{df_2(X_{f_1})} = -\frac{ \{ f_2,H_N\} }{\{ f_1,f_2\}} = -\{ f_2, H_N\} \\ c_2&= \frac{df_1(X_{H_N})}{df_1(X_{f_2})} = \frac{ \{ f_1,H_N\} }{\{ f_1,f_2\}} = \{ f_1, H_N\} . \end{aligned} \end{aligned}$$

The Poisson brackets defined by $\{ f, g\}:=\omega (X_f, X_g\}$ are of course not needed to do the computations, but they clarify the situation if M is a symplectic submanifold of higher codimension, where a matrix filled with $\{ f_i, f_j \}$ has to be inverted. A computation shows that the above vector field is tangent to the submanifold M and that it is the Hamiltonian vector field.