Question

In the reported figure of earth, the value of acceleration due to gravity is same at point $A$ and $C$ 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 $\dots \dots \dots$

Answer 1

Answer: 4

$g_A=\frac{GM(r)}{R^3}$
$g_C=\frac{GM}{{\left(R+\frac{R}{2}\right)}^2}$
$g_A=g_C$
$\frac{r}{R^3}=\frac{1}{\frac{9}{4}{R}^2}\Rightarrow r=\frac{4R}{9}$
so $OA=\frac{4R}{9};AB=R-r=\frac{5R}{9}$
$OA:AB=4:5$