Question

please answer this question​

Answer 1

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Let $\rm W_{h}$ be the weight of the stone at the height half of the radius of the earth. Let $\rm W_{d}$ be the weight of the stone at the depth half of the radius of the earth. Required ratio is given by

$\rm \frac{W_{h}}{W_{d}}=\frac{m g_{h}}{m g_{d}}=\frac{g_{h}}{g_{d}} \ldots(1) ,$

where m is the mass of the stone, h is the height of the stone and d is the depth of the stone.

Acceleration due to gravity at a height h, is given by

$\rm g_{h}=\frac{g R^{2}}{(R+h)^{2}}=\frac{g R^{2}}{(R+R / 2)^{2}}=\frac{4}{9} g$

Acceleration due to gravity at a depth d ,is given by

$\rm g_{d}=g\left(1-\frac{d}{R}\right)=g\left(1-\frac{R / 2}{R}\right)=\frac{g}{2}$

From equation (1),we have

$\rm\[ \frac{W_{h}}{W_{d}}=\frac{g_{h}}{g_{d}}=\frac{\frac{4}{9} g}{\frac{g}{2}}=\frac{8}{9}=0.9 \]$