At what distance from the centre of the earth,the value of acceleration due to gravity $g$ will be half that on the surface? ($R =$ radius of earth)

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{4}$ below the surface of the Earth,what is the new frequency of oscillation? ($R =$ radius of the 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 a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $0.5\, m$. If the mean density of the moon is two-thirds that of the earth and the radius is one-quarter that of the earth,what is the maximum vertical distance through which he can jump on the moon and the ratio of the duration of the jump on the moon to that on the earth?

The masses of two planets are in the ratio $1 : 2$. Their radii are in the ratio $1 : 2$. The ratio of the acceleration due to gravity on the planets is: