The value of acceleration due to gravity at a depth $d$ from the surface of the Earth and at an altitude $h$ from the surface of the Earth are in the ratio:

What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface of the Earth? (radius of Earth $= R$)

The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth,assumed to be a sphere of radius $R$ of uniform density,is as shown in the figure below. The correct figure is:

If the earth suddenly shrinks (without changing mass) to half of its present radius,the acceleration due to gravity will be

If the value of $g$ (acceleration due to gravity) at the Earth's surface is $10 \ m/s^2$,what is its value in $m/s^2$ at the center of the Earth,assuming the Earth is a sphere of radius $R$ meters with uniform mass density?

$A$ newly discovered planet has a density eight times the density of the earth and a radius twice the radius of the earth. The time taken by a $2\, kg$ mass to fall freely through a distance $S$ near the surface of the earth is $1$ second. Then the time taken for a $4\, kg$ mass to fall freely through the same distance $S$ near the surface of the new planet is ....... $\sec$.