The value of the acceleration due to gravity is $\mathrm{g}_{1}$ at a height $\mathrm{h}=\frac{\mathrm{R}}{2}(\mathrm{R}=$ radius of the earth $)$ from the surface of the earth. It is again equal to $\mathrm{g}_{1}$ at a depth $\mathrm{d}$ below the surface of the earth.
The ratio $\left(\frac{\mathrm{d}}{\mathrm{R}}\right)$ equals :
Correct Option: 4
$g_{2}=\frac{\operatorname{GM}(\mathrm{R}-\mathrm{d})}{\mathrm{R}^{3}}$..(2)
$\mathrm{g}_{1}=\mathrm{g}_{2}$
$\frac{\mathrm{GM}}{\left(\frac{3 \mathrm{R}}{2}\right)^{2}}=\frac{\mathrm{GM}(\mathrm{R}-\mathrm{d})}{\mathrm{R}^{3}}$
$\Rightarrow \frac{4}{9}=\frac{(\mathrm{R}-\mathrm{d})}{\mathrm{R}}$
$4 R=9 R-9 d$
$5 \mathrm{R}=9 \mathrm{~d} \Rightarrow \frac{\mathrm{d}}{\mathrm{R}}=\frac{5}{9}$
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