What is the ratio of the kinetic energies of two satellites at heights $R$ and $3R$ from the surface of the Earth? ($R$ = radius of the Earth)

Obtain an expression for the total energy of a satellite revolving around the Earth.

Initially,a satellite of $100 \ kg$ is in a circular orbit of radius $1.5 R_E$. This satellite can be moved to a circular orbit of radius $3 R_E$ by supplying $\alpha \times 10^6 \ J$ of energy. The value of $\alpha$ is . . . . . . .
(Take Radius of Earth $R_E = 6 \times 10^6 \ m$ and $g = 10 \ m/s^2$)

If the radius of an orbiting satellite is decreased,then its kinetic energy

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:

The ratio of energy required to raise a satellite to a height $h$ above the earth's surface to that required to put it into the orbit at the same height is ($R=$ radius of earth).