$A$ 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

$A$ body of mass $m$ is taken to the bottom of a deep mine. Then

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 acceleration due to gravity becomes $\left(\frac{g}{2}\right)$ ($g =$ acceleration due to gravity on the surface of the earth) at a height equal to

$A$ body weighs $500 \,N$ on the surface of the earth. At what distance below the surface of the earth will it weigh $250 \,N$ (in $\,km$)? (Radius of earth, $R = 6400 \,km$)

If the value of acceleration due to gravity on the surface of the Earth is $g$,what will be the value of $g$ at a height equal to the radius of the Earth from the surface?