# Orbital mechanics terminal velocity vectors

1. Mar 23, 2013

### Dustinsfl

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]

Last edited by a moderator: May 6, 2017