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

(N/A) Suppose a satellite of mass $m$ revolves around the Earth at a distance $r$ from the center of the Earth. Its orbital velocity is $v_{0}$.
The potential energy of the satellite in the gravitational field is given by:
$V = -\frac{GM_{E}m}{r}$
The kinetic energy of the satellite is:
$K = \frac{1}{2}mv_{0}^{2}$
Since the orbital velocity $v_{0} = \sqrt{\frac{GM_{E}}{r}}$,substituting this into the kinetic energy expression:
$K = \frac{1}{2}m\left(\sqrt{\frac{GM_{E}}{r}}\right)^{2} = \frac{GM_{E}m}{2r}$
The total energy $E$ of the satellite is the sum of its potential and kinetic energies:
$E = V + K = -\frac{GM_{E}m}{r} + \frac{GM_{E}m}{2r}$
Therefore,the total energy is:
$E = -\frac{GM_{E}m}{2r}$,where $r = R_{E} + h$.
The negative sign indicates that the satellite is in a bound state within the Earth's gravitational field. To escape this orbit,an energy equal to the magnitude of the total energy must be supplied to the satellite.