If two planets have their radii in the ratio $x: y$ and densities in the ratio $m: n$,then the ratio of the acceleration due to gravity on them is

At what depth below the surface of the earth will the acceleration due to gravity be half of its value at $1600 \ km$ above the surface of the earth?

If $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth's surface,the mean density of the earth is

Two spherical planets $A$ and $B$ have the same mass,but their densities are in a ratio $8:1$. For these planets,the ratio of acceleration due to gravity at the surface of $A$ to its value at the surface of $B$ is:

The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth,is (radius of earth is $6400 \ km$) (in $km$)

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 $m s^{-2}$)