Question

Two satellites $S_1$ and $S_2$ are revolving round a planet in coplanar and concentric circular orbit of radii $R_1$ and $R_2$ in the same direction respectively. Their respective periods of revolution are $1 \,hr$ and $8 \,hr$. The radius of the orbit of satellite $S_1$ is equal to $10^4 \,km$. Find the relative speed in kmph when they are closest.

• A. $2 \pi \times 10^4 \,kmph$
• B. $\pi \times 10^4 \,kmph$
• C. $\frac{\pi}{2} \times 10^4 \,kmph$
• D. $\frac{\pi}{3} \times 10^4 \,kmph$

Answer 1

Answer: (B)

By Kepler’s $3^{rd}$ law, $\frac{T^2}{R^3} =$ constant
$\therefore \frac{T^{2}_{1}}{R^{3}_{1}} = \frac{T^{2}_{2}}{R^{3}_{2} }$ or $\frac{1}{\left(10^{4}\right)^{3}} = \frac{64}{R^{3}_{2}}$ or
$R_{2} = 4 \times 10^{4} km$
Distance travelled in one revolution,
$S_1 = 2\pi R_1 = 2\pi \times 10^4$
and $S_2 = 2\pi R_2 = 2\pi \times 4 \times 10^4$
$v_1 = \frac{S_1}{t_1} = \frac{2\pi \times 10^4}{1} = 2\pi \times 10^4$ kmph
and $v_2 = \frac{S_2}{t_2} = \frac{2\pi \times 4 \times 10^4}{8}$
$ = \pi \times 10^4\,kmph$
$\therefore $ Relative velocity
$v_1 = v_2 = 2\pi \times 10^4 - \pi \times 10^4$
$ = \pi \times 10^4$ kmph