What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface of the Earth? (radius of Earth $= R$)

The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth).

$A$ research satellite of mass $200 \,kg$ circles the earth in an orbit of average radius $3R/2$,where $R$ is the radius of the earth. Assuming the gravitational pull on a mass of $1 \,kg$ on the earth's surface to be $10 \,N$,the pull on the satellite will be ........ $N$.

If a hole is bored along the diameter of the earth and a stone is dropped into the hole,what happens to the stone?

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400 \, km$. At the poles $g = 10 \, m/s^2$)

As we go from the equator to the poles,the value of $g$