At what depth does the acceleration due to gr | vedclass

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

Vedclass · Accepted Answer

$\frac{3R}{4}$

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(B) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula: $g' = g\left(1 - \frac{d}{R}\right)$ Given that the acceleration due to gravity at depth $d$ is $\frac{g}{4}$,we substitute $g' = \frac{g}{4}$ into the equation: $\frac{g}{4} = g\left(1 - \frac{d}{R}\right)$ Dividing both sides by $g$: $\frac{1}{4} = 1 - \frac{d}{R}$ Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{4}$ $\frac{d}{R} = \frac{3}{4}$ $d = \frac{3R}{4}$ Thus,at a depth of $\frac{3R}{4}$,the acceleration due to gravity becomes $\frac{g}{4}$.

At what depth does the acceleration due to gravity become $\frac{g}{4}$? ($R$ = radius of the Earth)

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