The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth).

By approximately how many meters is the radius of the Earth at the equator greater than the radius of the Earth at the poles (in $,000 \ m$)?

Considering Earth to be a sphere of radius $R$ having uniform density $\rho$,the value of acceleration due to gravity $g$ in terms of $R$,$\rho$,and $G$ is:

In the provided figure of the Earth,the value of acceleration due to gravity is the same at points $A$ and $C$,but it is smaller than its value at point $B$ (surface of the Earth). The value of $OA : AB$ is $x : y$. The value of $x$ is $\ldots \ldots \ldots$

$A$ tunnel is dug through the centre of the earth. Show that a body of mass $m$ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

If the radius of the earth were to shrink by $1\%$ while its mass remains the same,the acceleration due to gravity on the earth's surface would