$A$ simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface,where $R$ is the radius of the earth. The value of $T_2/T_1$ is

What is the acceleration due to gravity at a distance of $2R$ from the surface of the Earth,where $R$ is the radius of the Earth?

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $0.5\, m$. If the mean density of the moon is two-thirds that of the earth and the radius is one-quarter that of the earth,what is the maximum vertical distance through which he can jump on the moon and the ratio of the duration of the jump on the moon to that on the earth?

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

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:

Acceleration due to gravity $g$ for a body of mass $m$ on the Earth's surface is proportional to (Radius of Earth $= R$,mass of Earth $= M$)