Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size (radius). If the acceleration due to gravity on the surface of the new planet is $g'$ and that on the surface of the earth is $g$, then:

Imagine Earth to be a solid sphere of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth $d$ below Earth's surface is the same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of Earth),the ratio of $\frac{h}{d}$ will be

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:

Two planets have radii $r_1$ and $r_2$ and their densities are $p_1$ and $p_2$ respectively. The ratio of the acceleration due to gravity on their surfaces will be:

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).

Since the Earth is not a perfect sphere,what is the effect on the acceleration due to gravity $(g)$?