At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).

The depth at which acceleration due to gravity becomes $\frac{g}{2n}$ is ($R=$ radius of earth,$g=$ acceleration due to gravity on earth's surface,$n$ is an integer).

Assuming the Earth to be a sphere of uniform density, the acceleration due to gravity:

At a certain depth $d$ below the surface of the earth,the value of acceleration due to gravity becomes four times its value at a height $3R$ above the earth's surface. Where $R$ is the radius of the earth (Take $R = 6400 \ km$). The depth $d$ is equal to $............ \ km$.

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:

Write the equation of acceleration due to gravity at a height $h$ from the surface of the Earth,where $h << R_e$.