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

Answer: 8

For astronaut $\sqrt{\frac{GM}{R_2}}=v_{ast}$
For body, $\sqrt{\frac{GM}{R_1}}=v_{body}$
given, $v_{body}-v_{ast}=5m/s$
$v_{body}=5000m/s$
$\therefore R_2=\frac{GM}{{\left(5000\right)}^2},R_1=\frac{GM}{{\left(5005\right)}^2}$
$\therefore R_2-R_1=GM\left[\frac{1}{{\left(5000\right)}^2}-\frac{1}{{\left(5005\right)}^2}\right]\approx 32km$