Q 11-07-094JEE MainJEE Main 2023 (15 Apr, Shift 1)Easy
A body is released from a height equal to the radius ($R$) of the earth. The velocity of the body when it strikes the surface of the earth will be (given $g$ = acceleration due to gravity on the earth)
Answer: (B) $\sqrt{gR}$
$\tfrac12v^2=GM\left(\dfrac1R-\dfrac1{2R}\right)=\dfrac{GM}{2R}\Rightarrow v=\sqrt{\dfrac{GM}{R}}=\sqrt{gR}$.
Solution by Sreeraj P, M.Sc Physics