The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $\frac{\mathrm{g}}{9}$, where $g$ is acceleration due to gravity on earth's surface, is $\_\_\_\_$ .

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Gravitation

The height in terms of radius of the earth

(R)

, at which the acceleration due to gravity becomes

\frac{g}{9}

, where

g

is acceleration due to gravity on earth's surface, is

____

(A) $3 R$
(B) $22 R$
(C) $2 R$ ✓ Correct
(D) $\frac{4}{9} R$

Answer: (C)

Solution

Above the surface the gravitational acceleration is

g_{h} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}

. Setting

g_{h} = \frac{g}{9}

gives

(1 + \frac{h}{R})^{2} = 9

, so

1 + \frac{h}{R} = 3

and

\frac{h}{R} = 2

. The required height is

h = 2 R

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option

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