JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Gravitation

A body of mass

m

is taken from the surface of earth to a height equal to twice the radius of earth (

R_{e}

). The increase in potential energy will be ______.
(

g

is acceleration due to gravity at the surface of earth)

Answer: (D)

Solution

The gravitational potential energy of a body at distance

r

from the centre of the earth is

U = - \frac{GM m}{r}

. At the surface

U_{1} = - \frac{GM m}{R _{e}}

, and at a height

2 R_{e}

(a distance

3 R_{e}

from the centre)

U_{2} = - \frac{GM m}{3 R _{e}}

. The increase is

Δ U = U_{2} - U_{1} = \frac{GM m}{R _{e}} - \frac{GM m}{3 R _{e}} = \frac{2}{3} \frac{GM m}{R _{e}}

. Using

g = \frac{GM}{R _{e}^{2}}

, so

GM = g R_{e}^{2}

, the increase in potential energy is

\frac{2}{3} m g R_{e}

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option