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 $2R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth).

The rotation of the Earth,having radius $R$,about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightlessness. The duration of the day in such a case is:

The radius of Earth is about $6400 \,km$ and that of Mars is $3200 \,km$,and the mass of the Earth is about $10$ times the mass of Mars. An object weighs $200 \,N$ on the surface of Earth. Then,its weight on the surface of Mars will be (in $\,N$)

By approximately how many meters is the radius of the Earth at the equator greater than the radius of the Earth at the poles (in $,000 \ m$)?