Q 11-07-048JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon $= 27$ days and gravitational attraction between the satellite and the Moon is neglected.
Answer: (B) $1$ day
Kepler's third law: $T^2 \propto r^3$.
$$T_s = T_m\left(\frac{r_s}{r_m}\right)^{3/2} = 27\times\left(\frac{1}{9}\right)^{3/2} = \frac{27}{27} = 1\ \text{day}$$
Solution by Sreeraj P, M.Sc Physics