As we move above the surface of the Earth,the value of acceleration due to gravity $g$ . . . . . . .

Calculate the difference in the value of $g$ at the equator and at the poles due to the rotation of the Earth.

The density of a newly discovered planet is twice that of Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth is $R$,the radius of the planet would be

The height of any point $P$ above the surface of the Earth is equal to the diameter of the Earth. The value of acceleration due to gravity at point $P$ will be: (Given $g = $ acceleration due to gravity at the surface of the Earth)

The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is $9 : 4$. The mass of the planet is $\frac{1}{9}^{th}$ of that of the Earth. If $R$ is the radius of the Earth,what is the radius of the planet? (Assume the planets have the same mass density)

Considering Earth to be a sphere of radius $R$ having uniform density $\rho$,the value of acceleration due to gravity $g$ in terms of $R$,$\rho$,and $G$ is: