The depth 'd' at which the value of acceleration due to gravity becomes n − 1 1 times the value at the earth's surface is ( R = radius of the earth)

Question

The depth 'd' at which the value of acceleration due to gravity becomes n − 1 1 ​ times the value at the earth's surface is ( R = radius of the earth)

TestHub · Accepted Answer

The acceleration due to gravity at a depth $d$ below the earth's surface is given by:

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

Given that the acceleration due to gravity at depth $d$ is $\frac{1}{n - 1}$ times the value at the surface:

$g_{d} = \frac{g}{n - 1}$

Substituting this into the formula, we get:

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

$\frac{1}{n - 1} = 1 - \frac{d}{R}$

$\frac{d}{R} = 1 - \frac{1}{n - 1}$

$\frac{d}{R} = \frac{n - 1 - 1}{n - 1}$

$\frac{d}{R} = \frac{n - 2}{n - 1}$

$d = R (\frac{n - 2}{n - 1})$

Answer: $R (\frac{n - 2}{n - 1})$

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