Question 13.194: A shuttle is to rendezvous with a space station which is in ...

Conservation of energy:

\begin{aligned}T_{B} & =\frac{1}{2} m \nu_{0}^{2} & V_{B} & =-\frac{G M m}{r_{B}} \\T_{A} & =\frac{1}{2} m \nu_{A}^{2} & V_{A} & =-\frac{G M m}{r_{A}}\end{aligned}

G M=g R^{2} \quad \text { (Eq.12.30) }

T_{A}+V_{A}=T_{B}+V_{B}

\frac{1}{2} \cancel m \nu_{0}^{2}-\frac{g R^{2}}{r_{B}} \cancel m=\frac{1}{2} \cancel m \nu_{A}^{2} \frac{g R^{2}}{r_{A}} \cancel m

Conservation of angular momomentum: \quad r_{A} \nu_{A}=r_{B} \nu_{B} \sin \phi_{0}:

\nu_{A}=(4000 / 4210) \nu_{0} \sin 55^{\circ}=0.77829 \nu_{0}

Eqs. (2) and (1) \quad\left[1-(0.77829)^{2}\right] \nu_{0}^{2}=66.495 \times 10^{6}

\nu_{0}=12,990 \mathrm{\ ft} / \mathrm{s\blacktriangleleft}