Variation of g

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Question

Imagine earth to be a solid sphere of mass M and radius R. If the value of acceleration due to gravity at a depth 'd' below earth's surface is same as its value at a height 'h' above its surface and equal to $\frac{g}{4}$ (where g is the value of acceleration due to gravity on the surface of earth), the ratio of $\frac{h}{d}$ will be:

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Solution

$\begin{array}{l} g_{d} = g_{h} = \frac{g}{4} \\ g (1 - \frac{d}{R}) = \frac{g}{{(1 + \frac{h}{R})}^{2}} = \frac{g}{4} \end{array}$
Solving we get $d = \frac{3 R}{4}, h = R$
$\frac{h}{d} = \frac{R}{3 R / 4} = \frac{4}{3}$

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Variation of g

Remember concepts with our Masterclasses.

gd = gh=g4g1−dR = g1+hR2=g4Solving we get d=3R4, h=Rhd=R3R/4=43

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$\begin{array}{l} g_{d} = g_{h} = \frac{g}{4} \\ g (1 - \frac{d}{R}) = \frac{g}{{(1 + \frac{h}{R})}^{2}} = \frac{g}{4} \end{array}$
Solving we get $d = \frac{3 R}{4}, h = R$
$\frac{h}{d} = \frac{R}{3 R / 4} = \frac{4}{3}$