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]

A mass of $6 \times 10^{24} \mathrm{~kg}$ is to be compressed in the form of a solid sphere such that the escape velocity from its surface is $3 \times 10^4 \mathrm{~ms}^{-1}$. The radius of the sphere is

(Universal gravitational constant $=6.66 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2}$ )