# Orbital mechanics terminal velocity vectors

For a given space triangle, derive expressions for the terminal velocity vectors ##\mathbf{v}_1## and ##\mathbf{v}_2## between points ##P_1## and ##P_2## in terms of the unit vectors ##\mathbf{u}_1##, ##\mathbf{u}_2## and ##\mathbf{u}_c##.

I know that ##\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{s}{2a}}##, ##\sin\left(\frac{\beta}{2}\right) = \sqrt{\frac{s-c}{2a}}##, ##s = \frac{r_1+r_2+c}{2}##, ##a_m = \frac{s}{2}## (where ##a_m## is the minimum semi-major axis), ##A = \sqrt{\frac{\mu}{4a}}\cot\left(\frac{\alpha}{2}\right)##, and ##B = \sqrt{\frac{\mu}{4a}}\cot\left(\frac{\beta}{2}\right)##.

\begin{alignat}{2}
0\leq \Delta\nu\leq \pi,\quad & \beta = \beta_0\\
\pi < \Delta\nu < 2\pi,\quad & \beta = -\beta_0\\
t_2 - t_1\leq t_m,\quad & \alpha = \alpha_0\\
t_2 - t_1 > t_m,\quad & \alpha = 2\pi - \alpha_0
\end{alignat}
where $t_m$ greater than or equal to $t_2 - t_1$ is the shortest time of flight and the other situation is the longest.

The solution is
$$\mathbf{v}_1 = (A+B)\mathbf{u}_c + (B-A)\mathbf{u}_1$$
and
$$\mathbf{v}_2 = (A+B)\mathbf{u}_c - (B-A)\mathbf{u}_2.$$

I don't see how to obtain this.

http://img199.imageshack.us/img199/4710/spacetriangle.png [Broken]

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