The potential energy of a satellite of mass ' $m$ ' revolving around the Earth at a height of $R_e$ from the surface of the Earth is

( $R_e=$ Radius of Earth, $\mathrm{g}=$ acceleration due to gravity)

The time period of a simple pendulum on the surface of the Earth is $T$. If the pendulum is taken to a height equal to half of the radius of the Earth, then its time period is

If the escape velocity of a body from the surface of the Earth is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$, then the orbital velocity of a satellite in an orbit which is at a height equal to the radius of the Earth is

An artificial satellite is revolving around a planet of radius $R$ in a circular orbit of radius ' $a$ '. If the time period of revolution of the satellite. $T \propto a^{3 / 2} g^x R^y$, then the values of $x$ and $y$ are respectively

[ $g=$ acceleration due to gravity]