The change in the value of g at a height h ab | vedclass

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(C) The acceleration due to gravity at a height $h$ above the surface is given by $g_h = g(1 - \frac{2h}{R})$. The change in $g$ at height $h$ is $\De

Vedclass · Accepted Answer

$d = 2h$

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(C) The acceleration due to gravity at a height $h$ above the surface is given by $g_h = g(1 - \frac{2h}{R})$. The change in $g$ at height $h$ is $\Delta g_h = g - g_h = \frac{2hg}{R}$.The acceleration due to gravity at a depth $d$ below the surface is given by $g_d = g(1 - \frac{d}{R})$. The change in $g$ at depth $d$ is $\Delta g_d = g - g_d = \frac{dg}{R}$.Given that the change in the value of $g$ is the same at height $h$ and depth $d$,we equate the two expressions:$\frac{dg}{R} = \frac{2hg}{R}$.Simplifying this,we get $d = 2h$.

The change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $d$ below the surface of the earth. When both $d$ and $h$ are much smaller than the radius of the earth,then which one of the following is correct?

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