Question

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

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

Answer 1

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

Time period of revolution is $T^2=\frac{4{\pi }^2}{GM}{r}^3$
$\Rightarrow \frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}$
Given $T_1=$1 hr, $T_2=2$ hr and $r_1={10}^4$ km.
$\Rightarrow r_2=r_1\times \frac{T_2^{2/3}}{T_1^{2/3}}={10}^4\times \frac{8^{2/3}}{1^{2/3}}=4\times {10}^4$ km
Now,
$v_1=r_1{\omega }_1=r_1\frac{2\pi }{T_1}={10}^4\times \frac{2\pi }{1}=2\pi \times {10}^4$ km/hr $($ since we want the answer in kmph, we didn't convert km to m and hr to s while substituting $)$
$v_2=r_2{\omega }_2=r_2\frac{2\pi }{T_2}=4\times {10}^4\times \frac{2\pi }{8}=\pi \times {10}^4$ km/hr
Since we want the relative speed when the satellites are closer, direction of linear speed is in the same direction.
Hence, relative speed$=(v_1-v_2)=(2\pi -\pi )\times {10}^4=\pi \times {10}^4$ km/hr