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 $S_1$, angular speed (in $rad/hr$) of $S_2$ as observed by an astronaut in $S_1$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is B $\frac{\pi }{3}$

Angular velocity ${\omega }_{s_2,s_1}=\frac{v_{s_2,s_1}}{R_{s_2,s_1}}$


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

Here, $v_{s2,s1}=v_2-v_1=\pi \times {10}^{4}km/h$
and $R_{s2,s1}=R_2-R_1=3\times {10}^{4}km$

Hence, ${\omega }_{s_2,s_1}=\frac{\pi \times {10}^4}{3\times {10}^4}\Rightarrow \frac{\pi }{3}rad/h$