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).

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 both the mass and the radius of the earth decrease by $1\%$,the value of the acceleration due to gravity will

The acceleration due to gravity on the earth's surface at the poles is $g$ and the angular velocity of the earth about the axis passing through the poles is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be the same,then $h$ is: ($h << R$,where $R$ is the radius of the earth)

If the value of $g$ (acceleration due to gravity) at the Earth's surface is $10 \ m/s^2$,what is its value in $m/s^2$ at the center of the Earth,assuming the Earth is a sphere of radius $R$ meters with uniform mass density?

If a man at the equator would weigh $(3/5)^{th}$ of his weight,the angular speed of the earth is