Assuming the Earth to be a sphere of uniform mass density,how much would a body weigh (in $N$) halfway down to the centre of the Earth if it weighed $250 \; N$ on the surface?

Give the value of acceleration due to gravity at height $12\, km$ from the surface of earth. (in $, m/s^2$)

Assume that a tunnel is dug through the Earth from the North Pole to the South Pole and that the Earth is a non-rotating,uniform sphere of density $\rho$. The gravitational force on a particle of mass $m$ dropped into the tunnel when it reaches a distance $r$ from the center of the Earth is:

$A$ body has weight $W$ newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight at the equator becomes $\left(\frac{3}{5}\right)$ of its initial value? (Radius of Earth at the equator $R = 6400 \ km$,$g = 10 \ m/s^2$,$\cos 0^{\circ} = 1$)

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is [$R =$ radius of the earth].