# Numerical Solution for the Single-Impulse Flyby Co-Orbital Spacecraft Problem
## 3 Mathematical Formulation of the Problem
### Nonlinear Equations for Coplanar Lambert Problem  
For a given Lambert problem, the eccentricity vector \(\mathbf{e}\) can be divided into a component \(\mathbf{e}_{T}\) perpendicular to the transfer chord and a component \(\mathbf{e}_{F}\) parallel to the transfer chord. Then, the eccentricity of the orbit becomes  
\[\mathbf{e}=\sqrt{\mathbf{e}_{T}^{2}+\mathbf{e}_{F}{}^{2}} \tag{16}\]As shown in Figure 3, the length of the chord can be calculated using  
\[c=\left\|\mathbf{r}_{1,t_{1}}-\mathbf{r}_{0,t_{0}}\right\| \tag{17}\]  
Additionally, the component parallel to the transfer chord can be obtained using  
\[\mathbf{e}_{F}=\frac{\left|r_{1,t_{1}}-r_{0,t_{0}}\right|}{c} \tag{18}\]  
It is clearly shown that \(\mathbf{e}_{F}\) is a constant for a given Lambert problem.  
For the transfer orbit with the eccentricity vector \(\mathbf{e}_{2}\):  
\[\mathbf{e}_{2}=e_{2F}\hat{i}_{F}+e_{2T}\hat{i}_{T} \tag{19}\]  
where  
\[\left\{\begin{array}{l}\hat{i}_{F}=\frac{\mathbf{r}_{1,t_{1}}-\mathbf{r}_{0,t_{0}}}{\left|\mathbf{r}_{1,t_{1}}-\mathbf{r}_{0,t_{0}}\right|}\\ \hat{i}_{T}=i_{h}\times i_{F}\end{array}\right. \tag{20}\]  
Furthermore, \(\hat{i}_{h}\) is the unitized angular momentum vector. Then, the true anomaly of the impulse position and interception position 1 in the transfer orbit can be obtained using  
\[\theta_{0,t_{1}}=\begin{cases}\arccos\left(\frac{\mathbf{e}_{2}\cdot\mathbf{r }_{1,t_{1}}}{\left|\mathbf{e}_{2}\right|\cdot\left|\mathbf{r}_{1,t_{1}}\right| }\right)&(\mathbf{e}_{2}\cdot\mathbf{r}_{1,t_{1}}\
eq 0)\\ \frac{\alpha}{2}+k\pi(k=0,1)&(\mathbf{e}_{2}\cdot\mathbf{r}_{1,t_{1}}=0)\end{cases} \tag{21}\]  
Meanwhile, the true anomaly of the transfer orbit at the time \(t_{0}\) can be obtained using a transformation relation analogous to Equation (3).  
\[\left\{\begin{array}{l}\theta_{0,t_{0},\rho_{2}}=\theta_{0,t_{0},\rho_{0}}+ \Delta\omega_{\rho_{2},\rho_{0}}\\ \Delta\omega_{\rho_{2},\rho_{0}}=\arccos(\frac{\mathbf{e}_{2}\cdot\mathbf{e}_ {0}}{\left|\mathbf{e}_{2}\right|\cdot\left|\mathbf{e}_{0}\right|})(\mathbf{e}_ {2}\cdot\mathbf{r}_{1,t_{1}}\
eq 0)\\ \Delta\omega_{\rho_{2},\rho_{0}}=\frac{\pi}{2}+k\pi(k=0,1)(\mathbf{e}_{2} \cdot\mathbf{r}_{1,t_{1}}=0)\end{array}\right. \tag{22}\]  
Then, the mean anomaly of the transfer orbit at the time \(t_{0}\) and the time \(t_{1}\) can be obtained using Equation (14). The semi-latus rectum of the transfer orbit can also be calculated using  
\[p_{2}=r_{1,t_{1}}+e_{2}\cos\theta_{1,t_{1}} \tag{23}\]  
Figure 3: Geometric interpretation of coplanar Lambert problem.  
Then, defining the semimajor axis with \(a=p/1-e^{2}\), the orbital angular velocity can be calculated using \(n=\sqrt{\frac{H}{a^{3}}}\). Finally, the orbital transfer time yields  
\[\Delta t=\frac{M_{1,t_{1}}-M_{0,t_{0},\rho_{2}}}{n_{2}}=f\big{(}\mathbf{r}_{0,t_ {0}},\mathbf{r}_{1,t_{1}},\mathbf{e}_{0};e_{T}\big{)} \tag{24}\]  
where the transverse eccentricity component \(e_{T}\) is the only unknown, considering that only when \(0\leq e_{2}<1\) is meaningful can the value \(e_{2T}\) be a constraint in \(\left(-\sqrt{1-e_{2F}^{2}}\sqrt{1-e_{2F}^{2}}\right)\).