If the time period of revolution of a satellite is $T$,then its kinetic energy is proportional to

$A$ satellite orbits the Earth at a height of $400 \; km$ above the surface. How much energy must be expended to rocket the satellite out of the Earth's gravitational influence? (Mass of the satellite $= 200 \; kg$; mass of the Earth $= 6.0 \times 10^{24} \; kg$; radius of the Earth $= 6.4 \times 10^{6} \; m$; $G = 6.67 \times 10^{-11} \; N m^{2} kg^{-2}$)

Match the following columns.

$A$. Potential energy of satellite	$I$. Positive
$B$. Total energy of satellite	$II$. Negative
$C$. Kinetic energy of satellite	$III$. Zero
$D$. Gravitational potential energy of satellite at infinity	$IV$. Infinity

The radius of the orbit of an Earth satellite is $R$. Its kinetic energy is proportional to:

The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$,to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2$ $(R_2 > R_1)$ is:

$A$ satellite of mass $m$,revolving around the Earth of radius $r$,has kinetic energy $E$. Its angular momentum is: