The acceleration due to gravity at a height $1 k m$ 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 $d = 2 k m$
Acceleration due to gravity at height h,
$g_{h} = g_{0} (1 - \frac{2 h}{R})$
Acceleration due to gravity at depth d,
$g_{d} = g_{0} (1 - \frac{d}{R})$
$g_{h} = g_{d}$
$g_{0} (1 - \frac{2 h}{R}) = g_{0} (1 - \frac{d}{R})$
$d = 2 h$
$d = 2 \times 1 k m$
$d = 2 k m$

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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=2 kmAcceleration due to gravity at height h, gh=g0(1−2hR) Acceleration due to gravity at depth d, gd=g0(1−dR) gh=gd g0(1−2hR)=g0(1−dR) d=2h d=2×1 km d=2 km

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

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