Question

Two satellites $S_1$ and $S_2$ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours respectively. The radius of the orbit of $S_1$ is ${10}^4\text{km}$.

When $S_2$ is closest to $S_1$, find the speed of $S_2$ relative to $S_1$(in $kmph$)

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

Answer 1

The correct option is A $\pi \times {10}^4$


Given, $R_1={10}^{4}km,T_1=1hr,T_2=8hr$
let the velocity of satellite 1 is $v_1$ and satellite 2 is $v_2$.

For satellite 1:
$T_1=\frac{2\pi R_1}{v_1}{v}_1=\frac{2\pi R_1}{T_1}$ ... (1)

For satellite 2:
$v_2=\frac{2\pi R_2}{T_2}$ ... (2)

using Kepler's 3rd law $\to T^2\propto R^3$
$\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}{R}_2^3=\frac{T_2^2}{T_1^2}\times R_1^3$

Here, $T_2=8hr,T_1=1hr$
$R_2^3=(\frac{8}{1}{)}^2\times R_1^3$
so, $R_2=4R_1$
Given, $R_1={10}^{4}km$
$R_2=4\times {10}^{4}km$

then, relative velocity =
$v_2-v_1=\frac{2\pi \times 4\times {10}^4}{8}-\frac{2\pi \times {10}^4}{1}{v}_2-v_1=\pi \times {10}^{4}kmph$