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

What should be the angular velocity of earth due to rotation about its own axis so that the weight at the equator becomes $(3/5)^{th}$ of its initial value? ($g$ = acceleration due to gravity, $R$ = radius of earth)

The diameters of two planets are in the ratio $4:1$. Their mean densities have the ratio $1:2$. The ratio of gravitational acceleration on the surface of the planets will be:

The acceleration due to gravity at the pole $(g_p)$ and at the equator $(g_e)$ are related as:

The angular speed of rotation of the Earth about its axis,at which the weight of a man standing on the equator becomes half of his weight at the poles,is given by:

The value of acceleration due to gravity at a depth $d$ from the surface of the Earth and at an altitude $h$ from the surface of the Earth are in the ratio: