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

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

The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

The percentage decrease in the weight of a rocket,when taken to a height of $32 \ km$ above the surface of the Earth,will be $.....\%$
(Radius of Earth $= 6400 \ km$)

Acceleration due to gravity on the moon is $\frac{1}{6}$ of the acceleration due to gravity on the earth. If the ratio of densities of earth $(\rho_e)$ and moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,then the radius of the moon $R_m$ in terms of $R_e$ will be:

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight at the equator becomes $\left(\frac{3}{5}\right)$ of its initial value? (Radius of Earth at the equator $R = 6400 \ km$,$g = 10 \ m/s^2$,$\cos 0^{\circ} = 1$)