The depth at which acceleration due to gravit | vedclass

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Question

(A) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula: $g^{\prime} = g \left(1 - \frac{d}{R}\right

Vedclass · Accepted Answer

$\frac{R(n-1)}{n}$

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(A) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula: $g^{\prime} = g \left(1 - \frac{d}{R}\right)$ Given that the acceleration due to gravity at depth $d$ is $\frac{g}{n}$,we substitute this into the equation: $\frac{g}{n} = g \left(1 - \frac{d}{R}\right)$ Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{d}{R}$ Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{n}$ $\frac{d}{R} = \frac{n-1}{n}$ $d = \frac{R(n-1)}{n}$

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is [ $R$ = radius of earth,$g$ = acceleration due to gravity,$n=$ integer].

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