The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ( R= radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth d below the surface of the earth. The ratio $(\frac{d}{R})$ equals

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$\frac{4}{9}$

$\frac{5}{9}$

$\frac{7}{9}$

$\frac{1}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} A t h e i g h t : \\ g = \frac{GM}{{(R+h)}^{2}} = \frac{GM}{{(R+ \frac{R}{2})}^{2}} = \frac{4}{9} \frac{GM}{{(R)}^{2}} \\ A t d e p t h : \\ g = g (1- \frac{d}{R}) = \frac{GM}{R^{3}} (R-d) \\ (∵ g = \frac{G M}{R^{2}}) \\ ∴ \frac{4}{9} \frac{GM}{{(R)}^{2}} = \frac{GM}{R^{3}} (R-d) \\ \Rightarrow \frac{d}{R} = \frac{5}{9} \end{array}$