The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

d = 2 k m

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d = \frac{1}{2} k m

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d = 1 k m

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d = \frac{3}{2} k m

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Solution

The correct option is A $d = 2 k m$
Acceteration due to gravity at height h,
$g_{n} = g_{0} (1 - \frac{2 h}{R})$ $h = 1 k m$
Acceleration due to gravity at depth d,
$d_{d} = g_{0} (1 - \frac{d}{R})$
$g_{h} = g_{d}$
$g_{0} (1 - \frac{2 h}{R}) = g_{0} (1 - \frac{d}{R})$
$\to d = 2 h$
$= 2 \times 1 k m$
$d = 2 k m$

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The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

The correct option is A d=2kmAcceteration due to gravity at height h, gn=g0(1−2hR) h=1kmAcceleration due to gravity at depth d,dd=g0(1−dR)gh=gdg0(1−2hR)=g0(1−dR)→d=2h=2×1kmd=2km