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 1h and 8h respectively. The radius of orbit of $S_1$ is ${10}^{4}km$. The speed of $S_2$ relative to $S_1$ when $S_2$ is closest to $S_1$ , is ___ $km{h}^{-1}$

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

Answer 1

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

$T^2\alpha r^3(\frac{T_1}{T_2}{)}^2=(\frac{R_1}{R_2}{)}^{3}or(\frac{1}{8}{)}^2=(\frac{R_1}{R_2}{)}^3\therefore R_2=4R_1=4\times {10}^{4}km{v}_1=\frac{2\pi R_1}{T_1}=2\pi \times {10}^{4}km{h}^{-1}and{v}_2=\frac{2\pi R_2}{T_2}=\pi \times {10}^{4}km{h}^{-1}$
In the given situation $|v_1-v_2|=\pi \times {10}^{4}km{h}^{-1}$