## 3 Optimization Problem Statement
\begin{array}{l}\mathbf{V}^{M}\Big{(}t_{1}+\Delta t_{f} \Big{)}=B\Big{(}t_{1}+\Delta t_{f}\Big{)}\left(\begin{array}{l}r\Big{(}t_{1} +\Delta t_{f}\Big{)}\\ 0\end{array}\right)-\left(\begin{array}{l}-\sin\Big{(}n_{S}\cdot\Big{(}t_{1} +\Delta t_{f}\Big{)}+\alpha_{M}\Big{(}t_{1}+\Delta t_{f}\Big{)}\Big{)}\\ \cos\Big{(}n_{S}\cdot\Big{(}t_{1}+\Delta t_{f}\Big{)}+\alpha_{M}\Big{(}t_{1}+ \Delta t_{f}\Big{)}\Big{)}\end{array}\right),\\ \mathbf{r}^{M}\Big{(}t_{1}+\Delta t_{f}\Big{)}=B\Big{(}t_{1}+\Delta t_{f}\Big{)} \left(\begin{array}{l}r\Big{(}t_{1}+\Delta t_{f}\Big{)}\\ 0\end{array}\right)-\left(\begin{array}{l}\cos\Big{(}n_{S}\cdot\Big{(}t_{1} +\Delta t_{f}\Big{)}+\alpha_{M}\Big{(}t_{1}+\Delta t_{f}\Big{)}\Big{)}\\ \sin\Big{(}n_{S}\cdot\Big{(}t_{1}+\Delta t_{f}\Big{)}+\alpha_{M}\Big{(}t_{1}+ \Delta t_{f}\Big{)}\Big{)}\end{array}\right),\\ \Delta V_{f}=\left|\mathbf{V}^{M}\Big{(}t_{1}+\Delta t_{f}\Big{)}-\sqrt{\frac{ \mu_{M}}{|\mathbf{r}^{M}\big{(}t_{1}+\Delta t_{f}\big{)}|^{\frac{3}{2}}}} \left(\begin{array}{l}x^{M}\Big{(}t_{1}+\Delta t_{f}\Big{)}\\ y^{M}\Big{(}t_{1}+\Delta t_{f}\Big{)}\end{array}\right)\right|\text{, } \mathbf{r}^{M}=\left(x^{M},y^{M}\right)^{T}\end{array} \right\} \tag{24}\]  
Now we proceed to the formulation of the problem of the third (and final) stage: it is required to design a low-energy trajectory of the spacecraft's center of mass, starting from a low circular orbit of the Earth and reaching a given selenocentric orbit. As an optimality criterion, the total characteristic velocity of the flight is considered, which is determined as the sum of the departure impulse near the Earth and the braking impulse near the Moon. It is assumed that the vehicle's trajectory still passes through the vicinity of the libration point \(L_{2}\) (considered at the previous stages of solving the main problem). The above formulation generally corresponds to a problem of the form (17). The functional of the third stage problem can be represented as:  
\[J_{2}(\mathbf{y})=\Delta V_{0}(\mathbf{y})+\Delta V_{f}(\mathbf{y}),\]  
where the magnitude of the initial (departure) velocity impulse \(\Delta V_{0}\) can be determined as \[\Delta V_{0}(\mathbf{y})=\sqrt{\frac{2\mu_{E}}{r_{0}}-\frac{2\mu_{E}}{r_{0}+r_{ \alpha_{0}}}}-\sqrt{\frac{\mu_{E}}{r_{0}}},\]  
and the value of \(\Delta V_{f}\) is given by expression (24). Then finally, the optimization problem of the third stage can be formalized as follows:  
\[\left. \begin{array}{l}\underset{\mathbf{y}\in Q}{\mathbf{\min}}J_{2}( \mathbf{y})\\ Q=\left\{\begin{array}{l}\mathbf{y}\in\mathbf{R}^{4}:\sqrt{r^{2}+{r_{L}}_{2} ^{2}-2\;r\;r_{L}\cos(\beta-\alpha_{M})}\Big{|}_{t_{1}}-\Delta r_{L_{2}}\leq 0,\\ \sqrt{1+r^{2}-2\;r\cos(\beta-\alpha_{M})}\Big{|}_{t_{1}+\Delta t_{f}}-(R_{ Moon}+h_{Moon})=0\end{array}\right\}\\ \mathbf{y}=\left(\beta_{0},r_{\alpha_{0}},\alpha_{M_{0}},t_{1},\Delta t_{f} \right)^{\mathsf{T}},\;\mathbf{y}\in\mathbf{R}^{5}\end{array} \right\} \tag{25}\]  
Finally, let us summarized the main aspects of the given 3-stage algorithm for solving a flat auxiliary problem in a form of a simple scheme (with some additional comments) (see Box 1):  
**Box 1. ** 3-stage algorithm for solving a flat auxiliary problem.