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

If earth is supposed to be a sphere of radius R, if $g_{30}$ is value of acceleration due to gravity at latitude of $ 30^\circ$ and g at the equator, the value of $g - {g_{{{30}^o}}}$ is

Assuming the earth to be a sphere of uniform density, the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to

At what distance from the centre of the earth, the value of acceleration due to gravity $g$ will be half that on the surface ($R =$ radius of earth)

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

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as $R$.)