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 $1\text{hr}$ and $8\text{hr}$ respectively. The orbital radius of $S_1$ is ${10}^4\text{km}$ when $S_2$ is closest to $S_1$. Then,

• A. the speed of $S_2$ relative to $S_1$ is $2\pi \times {10}^4\text{km/hr}$.
• B. The angular speed of $S_1$ as actually observed by an astronaut in $S_2$ is $\frac{\pi }{6}\text{rad/hr}$
  
• C. The angular speed of $S_2$ as actually observed by an astronaut in $S_1$ is $\frac{7\pi }{4}\text{rad/hr}$
• D. The speed of $S_2$ as actually observed by an astronaut in $S_1$, when $S_1$ and $S_2$ are farthest from each other is $3\pi \times {10}^4\text{km/hr}$
  

Answer 1

The correct option is D The speed of $S_2$ as actually observed by an astronaut in $S_1$, when $S_1$ and $S_2$ are farthest from each other is $3\pi \times {10}^4\text{km/hr}$


By Kepler's third law of planetary motion, $T^2\propto r^3$

Therefore${\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{r_1}{r_2}\right)}^3$
$\Rightarrow {\left(\frac{1}{8}\right)}^2={\left(\frac{{10}^4}{r_2}\right)}^3$

Or, $r_2=4\times {10}^4\text{km}$

Time period of a body in circular motion is
$T=\frac{2\pi r}{V}$ $\Rightarrow V=\frac{2\pi V}{T}$

$\therefore |V_2-V_1|=2\pi |\frac{r_1}{T_1}-\frac{r_1}{T_2}|$

$=2\pi |\frac{{10}^4}{1}-\frac{4\times {10}^4}{8}|$
$=\pi \times {10}^4\text{km/hr}$

When the satellites are farthest from each other they move in opposite direction. Then
$|V_{rel}|=|V_2+V_1|$
$=2\pi |\frac{{10}^4}{1}+\frac{4\times {10}^4}{8}|=2\pi \times \frac{3}{2}\times {10}^4$
$=3\pi \times {10}^4\text{km/hr}$

$\therefore {\omega }_{rel}=\left|\frac{V_{rel}}{r_{rel}}\right|=\frac{3\pi \times {10}^4}{5\times {10}^4}$
$=\frac{3\pi }{5}\text{rad/hr}$