Question

An astronaut in a circular orbit around earth observes a celestial body moving in a lower circular orbit around earth in same plane as his orbit and in the same sense. He observes that the body moves at a speed of $5m/s$ relative to himself when it is closest to him. The minimum distance between him and the body if he is moving at a speed of $5000m/s$ is $\alpha km$ . Find the value of $\frac{\alpha }{4}$ . (Mass of earth $6 \times 10^{24}kg$ round off the value of $\alpha $ to the nearest integer).

Answer 1

For astronaut $\sqrt{\frac{G M}{R_{2}}}=v_{a s t}$
For body, $\sqrt{\frac{G M}{R_{1}}}=v_{b o d y }$
given, $v_{b o d y}-v_{a s t}=5m/s$
$v_{b o d y}=5000m/s$
$\therefore R_{2}=\frac{G M}{\left(5000\right)^{2}},R_{1}=\frac{G M}{\left(5005\right)^{2}}$
$\therefore R_{2}-R_{1}=GM\left[\frac{1}{\left(5000\right)^{2}} - \frac{1}{\left(5005\right)^{2}}\right]\approx32km$