JEE Main 2026 — Apr 4, Shift 2 — Physics solution

JEE Main2026Apr 4, Shift 2Physics

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
$2 R$
D
$\frac{4}{9} R$

Solution

Gravitational acceleration above the surface: $g_{h} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}$ .

Setting $g_{h} = \frac{g}{9}$ gives $(1 + \frac{h}{R})^{2} = 9$ , so $1 + \frac{h}{R} = 3$ .

$∴ h = 2 R$ .

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