JEE Main 2024 — Gravitation Question with Solution

$\Rightarrow \frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h}{)}^2}=\frac{{\mathrm{mv}}^2}{(\mathrm{R}+\mathrm{h})}$ $\Rightarrow \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}={\mathrm{v}}^2$ ....(1) $\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h})\omega$ $\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h})\frac{2\pi }{\mathrm{T}}$ ....(2) $\Rightarrow \frac{\mathrm{GM}}{{\mathrm{R}}^2}=\mathrm{g}$ $\Rightarrow \mathrm{GM}={\mathrm{gR}}^2$ ....(3) Put value from (2) \& (3) in eq. (1) $\begin{aligned} & \Rightarrow \frac{\mathrm{gR}^2}{(\mathrm{R}+\mathrm{h})}=(\mathrm{R}+\mathrm{h})^2\left(\frac{2 \pi}{\mathrm{T}}\right)^2 \\ & \Rightarrow \frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}=(\mathrm{R}+\mathrm{h})^3 \\ & \Rightarrow\left[\frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}\right]^{1 / 3}-\mathrm{R}=\mathrm{h} \end{aligned}$