Q 11-07-081JEE MainJEE Main 2023 (30 Jan, Shift 2)Easy
An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be
Answer: (B) $\sqrt{gR}$
Energy conservation from $r=2R$ to $r=R$:
$$\frac12v^2=GM\left(\frac1R-\frac1{2R}\right)=\frac{GM}{2R}\ \Rightarrow\ v=\sqrt{\frac{GM}{R}}=\sqrt{gR}$$
Solution by Sreeraj P, M.Sc Physics