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 ratio of the radii of a planet and the earth is $1: 2$, and the ratio of their mean densities is $4: 1$. If the acceleration due to gravity on the surface of the earth is $9.8 \,ms^{-2}$, then the acceleration due to gravity on the surface of the planet is: (in $\,ms^{-2}$)

If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of that planet is (in $\,g$)

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 mass of the moon is $(1/8)$ of the earth,but the gravitational pull is $(1/6)$ of the earth. This is due to the fact that:

As we move from the poles of the Earth towards the equator,what happens to the value of the acceleration due to gravity $(g)$?