At what depth d below the surface of the Eart | vedclass

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

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

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(B) The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula: $g' = g \left( 1 - \frac{d}{R} \right)$.Given that the acceleration due to gravity at depth $d$ is $\frac{1}{n}$ times the value at the surface,we have $g' = \frac{g}{n}$.Substituting this into the formula: $\frac{g}{n} = g \left( 1 - \frac{d}{R} \right)$.Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{d}{R}$.Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{n}$.$\frac{d}{R} = \frac{n-1}{n}$.Therefore,the depth $d$ is $d = R \left( \frac{n-1}{n} \right)$.

At what depth $d$ below the surface of the Earth does the acceleration due to gravity become $\frac{1}{n}$ times its value at the surface? ($R$ = radius of the Earth)

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