Assuming Earth to be a sphere of uniform density,what is the value of gravitational acceleration in a mine $100 \, km$ below the Earth's surface? (Given $R = 6400 \, km$,$g = 9.8 \, m/s^2$)

Consider Earth to be a homogeneous sphere. Scientist $A$ goes deep down in a mine and scientist $B$ goes high up in a balloon. The value of $g$ measured by

If the radius of the Earth is $R$,then the height $h$ at which the value of $g$ becomes one-fourth is:

If it is assumed that the spinning motion of the Earth increases,then the weight of a body on the equator

The height $h$ from the surface of the earth at which the value of $g$ will be reduced by $64 \%$ from the value at the surface of the earth is ($R=$ radius of the earth).

The acceleration due to gravity on the moon is $\frac{1}{6}$ times the acceleration due to gravity on the earth. If the ratio of the density of the earth $\rho_e$ to the density of the moon $\rho_m$ is $\frac{5}{3}$,then the radius of the moon $R_m$ in terms of the radius of the earth $R_e$ is: