Explanation

(a) $\frac{mv^{2}}{R} = \frac{RMm}{R^{2}} \Rightarrow v^{2} = \frac{GM}{R}$
$v = \frac{2\pi R}{T} \Rightarrow v^{2} = \frac{4\pi^{2}R^{2}}{T^{2}} = \frac{GM}{R}$
$\therefore T^{2} = \frac{4\pi^{2}R^{3}}{GM}$
If $T_{1}$ and $T_{2}$ are the time periods for satellite $S_{1}$ and $S_{2}$ respectively
$\left( \frac{T_{1}}{T_{2}} \right)^{2} = \left( \frac{R_{1}}{R_{2}} \right)^{3} \Rightarrow R_{2} = \left( \frac{T_{2}}{T_{1}} \right)^{2/3}R_{1}$
$T_{1} = 1\ h,\ T_{2} = 8\ h = 10^{4}$km
$R_{2} = \left( \frac{8}{1} \right)^{3/2} \times 10^{4}\ km = 4 \times 10^{4}\ km$
$v_{1} = \frac{2\pi R_{1}}{T_{1}} = \frac{2\pi \times 10^{4}}{1} = 2\pi \times 10^{4}\ kmh^{- 1}$
$v_{2} = \frac{2\pi R_{2}}{T_{2}} = \frac{2\pi \times 4 \times 10^{4}}{8} = \pi \times 10^{4}\ kmh^{- 1}$
Relative velocity of $S_{2}$ with respect to $S_{1}$ is
$v = v_{2} - v_{1}\left( \pi \times 10^{4} - 2\pi \times 10^{4} \right)\ kmh^{- 1}$
$|v| = \pi \times 10^{4}\ kmh^{- 1}$