Q 11-07-097JEE MainJEE Main 2023 (10 Apr, Shift 2)Medium
The time period of a satellite, revolving above earth's surface at a height equal to $R$ will be (Given $g=\pi^2\ \text{m s}^{-2}$, $R$ = radius of earth)
Answer: (C) $\sqrt{32R}$
Orbit radius $r=2R$, and $GM=gR^2$:
$$T=2\pi\sqrt{\frac{(2R)^3}{gR^2}}=2\pi\sqrt{\frac{8R}{\pi^2}}=2\sqrt{8R}=\sqrt{32R}$$
Solution by Sreeraj P, M.Sc Physics