Question:
In the reported figure of earth, the value of acceleration due to gravity is same at point A and $\mathrm{C}$ but it is smaller than that of its value at point $B$ (surface of the earth). The value of OA : AB will be $x: y$. The value of $x$ is
Solution:
$\mathrm{g}_{\mathrm{A}}=\frac{\mathrm{GM}(\mathrm{r})}{\mathrm{R}^{3}}$
$\mathrm{g}_{\mathrm{C}}=\frac{\mathrm{GM}}{\left(\mathrm{R}+\frac{\mathrm{R}}{2}\right)^{2}}$
$\mathrm{g}_{\mathrm{A}}=\mathrm{g}_{\mathrm{C}}$
$\frac{r}{R^{3}}=\frac{1}{\frac{9}{4} R^{2}} \Rightarrow r=\frac{4 R}{9}$
so $\mathrm{OA}=\frac{4 \mathrm{R}}{9} ; \mathrm{AB}=\mathrm{R}-\mathrm{r}=\frac{5 \mathrm{R}}{9}$
$\mathrm{OA}: \mathrm{AB}=4: 5$
