The value of acceleration due to gravity at a height of $4 R_E$ from the surface of the Earth is (where $R_E$ is the radius of the Earth and acceleration due to gravity on the surface of the Earth $g = 10 \,ms^{-2}$): (in $\,ms^{-2}$)

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 $R$)

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

$A$ narrow but tall cabin is falling freely near the earth's surface. Inside the cabin,two small stones $A$ and $B$ are released from rest (relative to the cabin). Initially $A$ is much above the centre of mass and $B$ is much below the centre of mass of the cabin. $A$ close observation of the motion of $A$ and $B$ will reveal that

Two equal masses $m$ and $m$ are hung from a balance whose scale pans differ in vertical height by $h$. The error in weighing in terms of density of the earth $\rho$ is:

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?