Orbital mechanics terminal velocity vectors

[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 $B = \sqrt{\frac{\mu}{4a}}\cot\left(\frac{\beta}{2}\right)$.

Additionally,
\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

 

[Valkarie]

The solution can be obtained by using the law of cosines, which states that for a triangle with sides of length $a,b,c$, $$c^2 = a^2 + b^2 - 2ab\cos\gamma.$$This equation can be applied to the space triangle to give$$c^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\Delta\nu.$$Solving for $\cos\Delta\nu$ yields$$\cos\Delta\nu = \frac{r_1^2+r_2^2-c^2}{2r_1r_2}.$$Using the fact that $$\mathbf{v}_1 = v_{1,\parallel}\mathbf{u}_c + v_{1,\perp}\mathbf{u}_1$$and$$\mathbf{v}_2 = v_{2,\parallel}\mathbf{u}_c + v_{2,\perp}\mathbf{u}_2$$we can then solve for the components $v_{1,\parallel}$, $v_{1,\perp}$, $v_{2,\parallel}$, and $v_{2,\perp}$ in terms of $\mathbf{u}_1$, $\mathbf{u}_2$, and $\mathbf{u}_c$.Using the law of sines, we have $$\frac{\sin\alpha}{r_1} = \frac{\sin\beta}{r_2},$$which can be rearranged to get$$\frac{r_2\sin\alpha}{r_1} = \sin\beta.$$Using the fact that$$\sin\beta = \sin(\pi-\alpha) = \sin\alpha\cos\Delta\nu + \cos\alpha\sin\Delta\nu,$$we can solve for $\cos\Delta\nu$:$$\cos\Delta\nu = \frac{\sin\beta -