For a body of mass m,the acceleration due to | vedclass

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(C) The acceleration due to gravity at a distance $r$ from the center of the earth is given by $g' = \frac{GM}{r^2}$.In the first case,the distance fr

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$\frac{4g}{9}$

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(C) The acceleration due to gravity at a distance $r$ from the center of the earth is given by $g' = \frac{GM}{r^2}$.In the first case,the distance from the surface is $R$,so the distance from the center is $r_1 = R + R = 2R$. Given $g_1 = \frac{g}{4}$.In the second case,the distance from the surface is $\frac{R}{2}$,so the distance from the center is $r_2 = R + \frac{R}{2} = \frac{3R}{2}$.Using the ratio: $\frac{g_2}{g_1} = \left(\frac{r_1}{r_2}ight)^2 = \left(\frac{2R}{3R/2}ight)^2 = \left(\frac{4}{3}ight)^2 = \frac{16}{9}$.Therefore,$g_2 = \frac{16}{9} 	imes g_1 = \frac{16}{9} 	imes \frac{g}{4} = \frac{4g}{9}$.

For a body of mass $m$,the acceleration due to gravity at a distance $R$ from the surface of the earth is $\frac{g}{4}$. Its value at a distance $\frac{R}{2}$ from the surface of the earth is ($R = \text{radius of the earth}$,$g = \text{acceleration due to gravity at the surface}$)

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