JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Gravitation

A body of mass 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)

(A) $3 mgR_{e}$
(B) $\frac{1}{3} mgR_{e}$
(C) $\frac{2}{3} mgR_{e}$ ✓ Correct
(D) $\frac{1}{2} mgR_{e}$

Answer: (C)

Solution

Measured from the Earth's centre, the potential energy of the mass is

U = - \frac{G M _{e} m}{r}

. At the surface

r = R_{e}

, while at height

h = 2 R_{e}

the centre distance is

r = 3 R_{e}

. The gain in potential energy is

Δ U = - \frac{G M _{e} m}{3 R _{e}} + \frac{G M _{e} m}{R _{e}} = \frac{2}{3} \frac{G M _{e} m}{R _{e}}

. Writing

g = \frac{G M _{e}}{R _{e}^{2}}

turns this into an increase of

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

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option

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