Question

In the reported figure of earth, the value of acceleration due to gravity is the same at points $A$ 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 $X:5$. The value of $X$ is ______.

Answer 1

Let,
$E_g=$ Gravitational field of earth
$R$ is the radius of the earth.

For inside point $A$,
$E_g=\frac{GMr}{R^3}$, where, $r=OA$

For outside point $C$,
$E_g=\frac{GM}{r^2}$,
where, $r=OC=R+\frac{R}{2}=\frac{3R}{2}$

Since, acceleration due to gravity is same at both the points, so will be the gravitational field.

$\Rightarrow \frac{GMr}{R^3}=\frac{GM}{{\left(\frac{3R}{2}\right)}^2}$

$\Rightarrow r=OA=\frac{4R}{9}$

So, $AB=OB-OA=R-\frac{4R}{9}=\frac{5R}{9}$

Now,
$OA:AB=\frac{4R}{9}:\frac{5R}{9}=4:5=X:5$

$\Rightarrow X=4$