For a satellite,if the time of revolution is $T$,then its $K.E.$ is proportional to

The energy required to take a satellite to a height $h$ above the Earth's surface (radius of Earth $R = 6.4 \times 10^3 \, km$) is $E_1$,and the kinetic energy required for the satellite to be in a circular orbit at this height is $E_2$. The value of $h$ for which $E_1 = E_2$ is:

$A$ skylab of mass $m \ kg$ is first launched from the surface of the earth into a circular orbit of radius $2R$ (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius $3R$. The minimum energy required to shift the lab from the first orbit to the second orbit is:

$A$ satellite moves around the earth in a circular orbit of radius $r$ with speed $v$. If the mass of the satellite is $M$,its total energy is

$A$ planet is moving in an elliptical orbit around the sun. If $T, V, E$,and $L$ stand respectively for its kinetic energy,gravitational potential energy,total energy,and magnitude of angular momentum about the centre of force,which of the following is correct?

$A$ satellite of mass $m$ is in a circular orbit of radius $2R_E$ around the Earth. The energy required to transfer it to a circular orbit of radius $4R_E$ is (where $M_E$ and $R_E$ are the mass and radius of the Earth,respectively).