# 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 suface of the earth. The ratio $\left ( \frac{d}{R} \right )$ equals : Option: 1 Option: 2 Option: 3 Option: 4

$g_{1}=\frac{G M}{\left(R+\frac{R}{2}\right)^{2}} \ldots (1)$

\begin{aligned} &g_{2}=\frac{G M(R-d)}{R^{3}} \ldots (2) \\ &g_{1}=g_{2}\\ &\frac{\mathrm{GM}}{\left(\frac{3 \mathrm{R}}{2}\right)^{2}}=\frac{\operatorname{GM}(\mathrm{R}-\mathrm{d})}{\mathrm{R}^{3}}\\ &\Rightarrow \frac{4}{9}=\frac{(\mathrm{R}-\mathrm{d})}{\mathrm{R}}\\ &4 R=9 R-9 d\\ &5 \mathrm{R}=9 \mathrm{~d} \Rightarrow \frac{\mathrm{d}}{\mathrm{R}}=\frac{5}{9} \end{aligned}

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