JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Gravitation

Take the fall from a height $R$ above the surface, a distance $2R$ from the earth's centre, down to the surface at distance $R$. Conserving mechanical energy with gravitational potential energy $U=-\frac{GMm}{r}$, $-\frac{GMm}{2R}=-\frac{GMm}{R}+\frac{1}{2}m{v}^2$, which rearranges to $\frac{1}{2}m{v}^2=\frac{GMm}{2R}$. Writing $GM=g{R}^2$, this gives $v^2=\frac{GM}{R}=gR$, so the speed on striking the surface is $v=\sqrt{gR}$.