Assuming the earth to be a sphere of uniform density,the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to

Derive the equation for the variation of $g$ due to height from the surface of the Earth.

At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).

The height at which the weight of a body becomes $\frac{1}{16}^{th}$ of its weight on the surface of the Earth (radius $R$) is: (in $R$)

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 weight of a body at the centre of the earth is