How does the force of gravity $F$ vary inside the Earth at a distance $r$ from the center,where $r < R_e$?

What should be the angular velocity of earth due to rotation about its own axis so that the weight at the equator becomes $(3/5)^{th}$ of its initial value? ($g$ = acceleration due to gravity, $R$ = radius of earth)

If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of that planet is (in $\,g$)

Acceleration due to gravity is $g$ on the surface of the earth. The value of acceleration due to gravity at a height of $32 \, km$ above the earth's surface is ........ $g$. (Radius of the earth $= 6400 \, km$)

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $.......... \, km$ : [Radius of earth $R = 6400 \, km$ and $\sqrt{3} = 1.732$]