Q 11-07-090JEE MainJEE Main 2023 (25 Jan, Shift 2)Easy
A body of mass $m$ is taken from earth surface to the height $h$ equal to twice the radius of earth ($R_e$), the increase in potential energy will be ($g$ = acceleration due to gravity on the surface of earth)
Answer: (C) $\dfrac23mgR_e$
$$\Delta U=GMm\left(\frac1{R}-\frac1{3R}\right)=\frac23\frac{GMm}{R}=\frac23mgR_e$$
Solution by Sreeraj P, M.Sc Physics