# Satellites

Something that revolves around another object under the action of gravitational force is 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.

## Energy Consideration of a Satellite

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.

## Escape Speed

Here we find the minimum initial speed required to give a body on the Earth's surface to escape from the gravitational attraction of the Earth. Since the only conservative force acting is gravity, the total mechanical energy is conserved. Note that we'll ignore the affects of air resistance. The conservation of mechanical energy tells us that the total initial mechanical energy is equal to the total final mechanical energy. If $v$ is the initial speed required to escape the gravitational field of the Earth called escape speed, initial kinetic energy is

$K_1 = \frac{mv^2}{2}$

The body is initially on the Earth's surface so the gravitational potential energy from Eq. \eqref{4} putting $h = 0$ is,

$U=-\frac{Gm{{M}_{E}}}{({{R}_{E}}+h)}=-\frac{Gm{{M}_{E}}}{{{R}_{E}}}$

The final potential energy $U_2$ at an infinite distance from the centre of the Earth is zero. The final kinetic energy $K_2$ is also zero at the infinite distance from the centre of the Earth. Note that the potential energy is maximum at the infinite distance from the centre of the Earth and gradually decreases (becomes more negative) when the body falls towards the centre of the Earth. But the kinetic energy is minimum which is zero at the infinite distance. The key to find the escape speed is that the final potential energy is maximum (zero) and the final kinetic energy is minimum (zero) at the infinite distance from the centre of the Earth. Therefore, $U_2 = 0$ and $K_2 = 0$. The law of conservation of mechanical energy gives

\begin{align*} {{K}_{1}}+{{U}_{1}}&={{K}_{2}}+{{U}_{2}} \\ \text{or,}\quad \frac{1}{2}m{{v}^{2}}-\frac{Gm{{M}_{E}}}{{{R}_{E}}+0}&=0+0 \\ \text{or,}\quad v&=\sqrt{\frac{2G{{M}_{E}}}{{{R}_{E}}}} \tag{7} \label{7} \end{align*}

The above equation for escape speed is the minimum speed required to give a body on the Earth's surface to reach the maximum distance or it means the infinite distance from the centre of the Earth. The escape speed does not depend on the mass of the body to be launched.