If the radius of the earth shrinks by $1 \%$,its mass remaining the same,then the acceleration due to gravity on the earth's surface would

At a height of $h$ $km$ from the surface of the Earth,the gravitational potential and the value of $g$ are $-5.4 \times 10^7\, J kg^{-1}$ and $6.0\, m s^{-2}$ respectively. Take the radius of the Earth as $6400\, km$.

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is [$R =$ radius of the earth].

$A$ uniform solid sphere of radius $R$ produces a gravitational acceleration of $a_o$ on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes $\frac{a_o}{4}$ is,

The acceleration due to gravity at a height of $6400 \,km$ from the surface of the earth is $2.5 \,ms^{-2}$. The acceleration due to gravity at a height of $12800 \,km$ from the surface of the earth is (Radius of the earth $= 6400 \,km$) (in $\,ms^{-2}$)

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 $,000 \ m$)?