$A$ planet has the same density as that of the Earth,and the universal gravitational constant $G$ is twice that of the Earth. The ratio of the acceleration due to gravity on the planet to that on the Earth is:

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

The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

The time period of a simple pendulum is $T_1$ when on the Earth's surface and $T_2$ when taken to a height $h = 2R$ above the Earth's surface, where $R$ is the radius of the Earth. The ratio $T_1 : T_2$ is:

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

The weight of an object in a coal mine,at sea level,and at the top of a mountain are $W_1$,$W_2$,and $W_3$ respectively. Then: