AIIMS 2017 Physics Satellite Motion MCQ Question

To find the relative speed of the satellites when they are closest, we first calculate their speeds using the formula $v = \frac{2\pi r}{T}$. For satellite $S_1$ with a radius $r_1 = 10^4$ km and period $T_1 = 1$ h, the speed is $v_1 = \frac{2\pi \times 10^4}{1} = 2\pi \times 10^4$ km/h. For satellite $S_2$ with radius $r_2$ (which can be found using Kepler's third law) and period $T_2 = 8$ h, the speed is $v_2 = \frac{2\pi r_2}{8}$. When they are closest, the relative speed is $v_1 - v_2$, which simplifies to $\frac{2\pi \times 10^4}{1} - \frac{2\pi \times 10^4}{8} = \frac{2\pi \times 10^4 \times 7}{8} = \frac{14\pi \times 10^4}{8} = \frac{7\pi \times 10^4}{4}$. However, since we need the relative speed at the closest point, we find that it simplifies