At a certain depth “d” below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3R above earth surface. Where R is Radius of earth (Take R=6400 km). The depth d is equal to

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5260km

2560 km

4800km

640km

answer is D.

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Detailed Solution

We know,

Acceleration due to gravity at a depth d is

$g_{d} = g (1 - \frac{d}{R})$

And acceleration due to gravity at a height h is

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

Given in the question

$g_{d} = 4 \times g_{h = 3 R}$

$g [1 - \frac{d}{R}] = 4 g {[1 + \frac{3 R}{R}]}^{- 2}$

$\Rightarrow [1 - \frac{d}{R}] = \frac{1}{4} \Rightarrow d = \frac{3 R}{4}$

d = 4800km

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