Question

Two satellites $S_1$ and $S_2$ revolve round a planet in coplanar circular orbits in same sense. Their periods of revolution are 1 hr and 8 hrs respectively. Radius of orbit of $S_1$ is ${10}^4$ km. When $S_1$ is closest to $S_2$, relative speed of $S_1$ with respect to $S_2$ is $n\times {10}^4$ km/h, here n is (Answer upto two digits after the decimal point.)

Answer 1

Since $T^2\propto r^3\Rightarrow {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{r_1}{r_2}\right)}^3$

${\left(\frac{1}{8}\right)}^2={\left(\frac{{10}^4}{r_2}\right)}^3\Rightarrow r_2=4\times {10}^4$ km

$V_1=r_1{w}_1={10}^4\times \frac{2\pi }{1}=2\pi \times {10}^4$ km/h
$V_2=r_2{w}_2=4\times {10}^4\times \frac{2\pi }{8}=\pi \times {10}^4$ km/h

$V_1-V_2=\pi \times {10}^4$ km/h