A body that revolves around another object under the action of gravitational force is called satellite. For example, the Moon revolves around the Earth and hence it is a satellite of the Earth. Similarly the Earth also revolves around the Sun and it is a satellite of the Sun. There are two types of satellites; natural and artificial. Satellites made by humans are artificial satellites while the Moon is a natural satellite of the Earth.
Consider a satellite of mass \(m\) which goes around the Earth in a circular orbit of radius \(r = R_E + h\) from the centre of the Earth where \( R_E\) is the radius of the Earth and \(h\) is the height of the satellite from the Earth's surface. We consider that the satellite-Earth system is isolated from other planetary bodies so that their gravitational effects are negligible or zero in our system. If the satellite moves with linear speed \(v\), the magnitude of centripetal force on the satellite due to the Earth's gravitational force is
\[{F_{{\rm{cen}}}} = m\frac{{{v^2}}}{r} \tag{1} \label{1}\]
The magnitude of the gravitational force of the interaction of the Earth and the satellite as given by Newton's law of gravitation is
\[{F_{{\rm{grav}}}} = G\frac{{mM_E}}{{{r^2}}} \tag{2} \label{2}\]
where \(M_E\) is the mass of the Earth. The satellite orbits the Earth means that it always falls towards the centre of the Earth with acceleration \(v^2/r\). The magnitude of the gravitational force the Earth exerts on the satellite in Eq. \eqref{2} is equal to the magnitude of the centripetal force given by Eq. \eqref{1}:
\[\begin{align*} m\frac{{{v^2}}}{r} = G\frac{{mM_E}}{{{r^2}}}\\ {\rm{or,}}\quad v = \sqrt {\frac{{GM_E}}{R_E + h}} \tag{3} \label{3} \end{align*}\]
The Eq. \eqref{3} shows that the linear speed \(v\) does not depend on the mass of the satellite but only depends on the distance \(r\) from the centre of the Earth. It means for a given distance \(r\) between the centre of the satellite and the centre of the Earth, the linear speed of the satellite is already fixed for the circular orbit.
In Figure 1 a satellite is launched at a distance \(r = R_E +h\) from the centre of the Earth. If the launching speed of the satellite is lower than enough, the satellite will go along orbit \(a\) or \(b\) and finally hit the Earth's surface. But if the Earth's mass is concentrated at the centre \(O\), the orbits \(a\) and \(b\) will be completed and the satellite will go along one of those elliptical orbits. For the orbit \(c\) the launching speed is enough and the satellite will always go along the orbit and never collide with the Earth. The orbits \(a\), \(b\) and \(c\) are called closed orbits. If the launching speed is increased further, the satellite will go along the orbit \(d\) or \(e\) away from the Earth into deep space. Orbits \(d\) and \(e\) are called open orbits.
What is the total mechanical energy of a satellite in orbit?
We consider the same satellite as in the above discussion. The satellite of mass \(m\) is at a height \(h\) from the Earth's surface at a distance \(r = R_E + h\) from the centre of the Earth. The gravitational potential energy of the satellite-Earth system is
\[U=-\frac{Gm{{M}_{E}}}{({{R}_{E}}+h)} \tag{4} \label{4}\]
The kinetic energy of the satellite moving with linear speed \(v\) in the orbit is
\[K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\left( \frac{Gm{{M}_{E}}}{{{R}_{E}}+h} \right) \tag{5} \label{5}\]
The total mechanical energy \(E\) of the satellite in its orbit is the sum of the kinetic and potential energies that is, \(E = K + U\):
\[\begin{align*} E&=\frac{1}{2}\left( \frac{Gm{{M}_{E}}}{{{R}_{E}}+h} \right)+\left( -\frac{Gm{{M}_{E}}}{({{R}_{E}}+h)} \right) \\ \text{or,}\quad E&=-\frac{Gm{{M}_{E}}}{2({{R}_{E}}+h)} \tag{6} \label{6} \end{align*}\]
Less negative value is greater than more negative value. The total mechanical energy is one half the potential energy which is greater than the potential energy.
