Question

In the reported figure of the 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

Step 1: Given data

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).

And, $OA:AB=x:5$

Step 2: Let, $E_g=$ The gravitational field of the earth and $R$the radius of the earth.

For inside point $A$,

$Eg=\frac{GMr,}{R^3}\text{where,}r=OA$ Here G is gravitational constant

For outside point $C$,

$Eg=\frac{GM,}{r^2}\text{where,}r=OC=R+\frac{R}{2}=\frac{3R}{2}$

Step 3: Acceleration due to gravity is the same at both points, and so will the gravitational field.

$\begin{array}{l}\Rightarrow \frac{GMr}{R^3}=\frac{GM}{{\frac{3R}{2}}^2}\\ \Rightarrow r=OA=\frac{4R}{9}\\ \therefore AB=OB-OA\\ AB=R-\frac{4R}{9}\\ AB=\frac{5R}{9}\end{array}$

Now,

$\begin{array}{l}OA:AB=\frac{4R}{9}:\frac{5R}{9}\\ OA:AB=4:5=X:5\\ \therefore X=4\end{array}$

Hence, the distance between the Earth's surface and its center is greater at the equator than at the poles, and the acceleration due to gravity is lowest there.