The weight of a body on the surface of the earth is $63 \ N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth? (in $N$)

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?

The angular speed of the Earth,such that an object on the equator may appear weightless,is ($g = 10\,m/s^2$,radius of the Earth $R = 6400\,km$).

If acceleration due to gravity at distance $d < R$ from the centre of the Earth is $\beta$,then its value at distance $d$ above the surface of the Earth will be [where $R$ is the radius of the Earth].

At a height of $10 \,km$ above the surface of the earth,the value of acceleration due to gravity is the same as that at a particular depth below the surface of the earth. Assuming uniform mass density for the earth,the depth is ............. $km$.

Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size (radius). If the acceleration due to gravity on the surface of the new planet is $g'$ and that on the surface of the earth is $g$, then: