Question

Two satellites $S_1$ and $S_2$ revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are $1\text{hour}$ and $8\text{hours}$ respectively. The radius of the orbit of $S_1$ is ${10}^4\text{km}$. When $S_1$ is closest to $S_2$, the angular speed of $S_2$ as observed by an astronaut in $S_1$ is :

• A. $\pi \text{rad/hr}$
• B. $\pi /3\text{rad/hr}$
• C. $2\pi \text{rad/hr}$
• D. $\pi /2\text{rad/hr}$

Answer 1

The correct option is B $\pi /3\text{rad/hr}$


Angular frequency of $S_1$
${\omega }_1=\frac{2\pi }{1}\text{rad/hr}$
Angular frequency of $S_2$
${\omega }_2=\frac{2\pi }{8}\text{rad/hr}$
From kepler's law.
${\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3\Rightarrow \frac{R_2}{R_1}=4$
$\Rightarrow R_2=4\times {10}^4\text{km}$
Velocity of $S_1$
$V_1=\frac{2\pi R_1}{1h}=2\pi \times {10}^4\text{km/hr}$
Velocity of $S_2$
$V_2=\frac{2\pi R_2}{8h}=\pi \times {10}^4\text{km/hr}$
At closest seperation,
$\omega =\frac{V_{\text{rel}}\perp \text{line joining}}{\text{length of line journing}}=\frac{V_1-V_2}{R_2-R_1}$
$=\frac{\pi \times {10}^4\text{km/hr}}{3\times {10}^4\text{km}}=\frac{\pi }{3}\text{rad/hr}$