The depth from the surface of the earth of radius $R$,at which the acceleration due to gravity will be $60 \%$ of its value on the earth's surface,is:

The radius of the earth is $6400 \, km$ and $g = 10 \, m/s^2$. In order that a body of $5 \, kg$ weighs zero at the equator,the angular speed of the earth is

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).

At which location,the equator or the poles of the Earth,is the value of acceleration due to gravity $g$ higher? Why?

Assuming the Earth to be a sphere of uniform mass density,the weight of a body at a depth $d = R/2$ from the surface of the Earth,if its weight on the surface of the Earth is $200 \, N$,will be $........... \, N$ (Given $R =$ Radius of Earth).

$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth (in $N$)?