JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Gravitation

An object is kept at rest at a distance of 3 R above the earth's surface where

R

is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is :
(Assume

M =

mass of earth,

G =

Universal gravitational constant)

(A) $\frac{GM}{2 R}$ ✓ Correct
(B) $\frac{GM}{R}$
(C) $\frac{3 GM}{R}$
(D) $\frac{2 GM}{R}$

Answer: (A)

Solution

Resting a height

3 R

above the surface places the object at a distance

r = R + 3 R = 4 R

from the centre. Escape means the total mechanical energy is just zero at infinity, so

\frac{1}{2} m v^{2} - \frac{GM m}{4 R} = 0

. Solving for the launch speed gives

v = \frac{2 GM}{4 R} = \frac{GM}{2 R}

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option

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