$A$ body weighs $72 \, N$ on the surface of the earth. When it is taken to a height of $h = 2R$,where $R$ is the radius of the earth,it would weigh ........ $N$.

The value of gravitational acceleration $g'$ at a height $h$ above the earth's surface is $\frac{g}{4}$. Then,what is the value of $h$ in terms of the earth's radius $R$?

The variation of acceleration due to gravity $g$ with distance $d$ from the centre of the earth is best represented by ($R =$ Earth's radius)

The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth).

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 $km$)

At what height from the ground will the value of $g$ be the same as that in a $10 \, km$ deep mine below the surface of the Earth?