If the radius of the Earth shrinks by $1.5 \%$ (mass remaining same),then the value of gravitational acceleration changes by ......... $\%$

The value of acceleration due to gravity at a height of $4 R_E$ from the surface of the Earth is (where $R_E$ is the radius of the Earth and acceleration due to gravity on the surface of the Earth $g = 10 \,ms^{-2}$): (in $\,ms^{-2}$)

The masses of two planets are in the ratio $1 : 2$. Their radii are in the ratio $1 : 2$. The ratio of the acceleration due to gravity on the planets is:

$A$ 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,

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?

At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).