JEE Main 2026 — Gravitation Question with Solution

Initial potential energy of the body at the surface of the earth is given by:

$U_i=-\frac{GMm}{R_e}$

Final potential energy of the body at a height $h=2R_e$ from the surface of the earth is:

$U_f=-\frac{GMm}{R_e+h}=-\frac{GMm}{R_e+2R_e}=-\frac{GMm}{3R_e}$

The increase in potential energy is:

$\Delta U=U_f-U_i=-\frac{GMm}{3R_e}-\left(-\frac{GMm}{R_e}\right)$

$\Delta U=\frac{GMm}{R_e}\left(1-\frac{1}{3}\right)=\frac{2GMm}{3R_e}$

Acceleration due to gravity at the surface of the earth is $g=\frac{GM}{R_e^2}$, which gives $GM=g{R}_e^2$.

Substituting this value, we get:

$\Delta U=\frac{2(g{R}_e^2)m}{3R_e}=\frac{2}{3}mg{R}_e$

Answer: $\frac{2}{3}mg{R}_e$