Satellite Motion

Satellite motion is physics surrounding the orbit of a small masses around larger masses. It is assumed that satellites have circular orbits.

Satellites can be either natural (planets, moons, etc) or artificial (objects sent up into orbit by humans).

Some artificial satellites are in a geostationary orbit; that is that they stay above the same fixed location on Earth’s surface throughout their orbit.
In order for this to occur these satellites must:

  • orbit directly above the equator and in the same direction as Earth’s rotation
  • have an orbital period equal to Earth’s rotational period – 24 hours.

Use this page to revise the following concepts within satellite motion:

Orbital motion

Orbital motion is motion of a body around another body, moving in a stable path, due to its gravitational attraction.

Consider firing a cannon at an angle from Earth at ever-increasing speeds yields one of three possibilities:

  • The speed of the projectile is too slow, and the projectile hits the ground
  • The speed of the projectile it too fast and the projectile is sent into space
  • The speed of the projectile is just right and the projectile enters a stable orbit around the Earth. In this case the projectile falls towards the Earth without ever getting closer to it, due to the curvature of the Earth.

An image of the Earth with a cannon sitting at polar north. 4 lines appear to be fired from the cannon and circle to, varying lengths, around the Earth. One line travels a short distance around the Earth and is labelled slow. Another line travels a longer distance around the Earth and is labelled faster. A third line curves away from the Earth and is labelled escape velocity. The final line curves all around the Earth and points back to the cannon, and this line is labelled orbital velocity

Any satellite orbiting a central mass obeys equations of circular motion, as we assume its orbit is circular:

\[v=\dfrac{2\pi r}{T}\]\[v=\sqrt\frac{GM}{r}\]

Where

\(v\) is the orbital velocity \((m/s)\)

\(r\) is the orbital radius \((m)\)

\(T\) is the orbital period \((s)\)

\(G\) is the gravitational constant \((6.67\times10^{-11}Nm^{2}/kg^{2})\)

\(M\) is the mass of the central mass being orbited \((kg)\)

A graphic of a circular orbital path. At the centre of the graphic is an image of the Earth. To the right of the Earth is an image of a satellite. A dotted line circles the Earth and travels through the satellite. This depicts the satellite moving about the Earth in a circular orbital path. An arrow pointing radially from the satellite to the Earth is labelled Gravitational force of attraction. An arrow at a tangent from the first arrow’s tail is labelled velocity

Notice that the mass of the satellite does not appear in these equations, hence orbital properties (velocity, radius and period) of a satellite are independent of the satellite’s mass.

By equating centripetal acceleration \((a=\frac{4\pi^{2}r}{T^{2}})\) and gravitational acceleration, or field strength, \((g=\frac{GM}{r^{2}})\) we arrive at the relationship:

\[\frac{r^{3}}{T^{2}}=\frac{GM}{4\pi^{2}}\]

Hence, if given the orbital radius of any satellite, we can determine the period of its orbit, if the central mass is known, and vice versa.

Importantly, for one central mass, the right-hand side (RHS) of the equation is a constant for all satellites around that mass. So a relationship between two satellites around the same mass can be found:

\[\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{r_{1}}{r_{2}}\right)^{3}\]

This can be useful to find characteristics of one satellite if the other is known, and due to both sides being ratios, so long as \(T_{1}\) and \(T_{2}\) have the same units and \(r_{1}\) and \(r_{2}\) have the same units, they don’t necessarily need to be SI units.

Note: SI units are International System of Units, and are the base unit to used in formulae unless otherwise specified, such as this situation.

Worked Example

Consider that Mercury has an orbital period of 88 days and an orbital radius of \(5.79\times10^{10}m\) around the Sun. Knowing that Venus has an orbital radius of \(1.08\times10^{11}\text{ m}\) the orbital period of Venus can be found:\[\begin{align}\left(\frac{T_{1}}{T_{2}}\right)^{2}&=\left(\frac{r_{1}}{r_{2}}\right)^{3}\\ \left(\frac{T_{1}}{88}\right)^{2}&=\left(\frac{1.08\times10^{11}}{5.79\times10^{10}}\right)^{3}\\ T_{1}&=224.182 \end{align}\]So the orbital period of Venus is 224 days.

Worked Example

Taking the mass of the Sun to be \(1.99\times10^{30}\text{ kg}\) and the orbital period of the Earth around the Sun to be 365.24 days. The orbital radius of the Earth can be found:\[\begin{align} r&=\left(\frac{GMT^{2}}{4\pi^{2}}\right)^\frac{1}{3} \\ r&=\left(\frac{(6.67\times10^{-11})(1.99\times10^{30})(365.24\times24\times60\times60^{2}}{4\pi^{2}}\right)^\frac{1}{3} \\ r&=1.50\times10^{11}\text{ m}\end{align}\]

Worked Example

Consider that Earth’s moon orbits at a distance of \(3.84\times10^{8}m\) with a period of \(27\) days. Using this information determine the orbital distance of a geostationary satellite. Let the moon be \(2:T_{2}=27\) day \(r_{2}=3.84\times10^{8}m\).To be geostationary, a satellite must have a period of \(1\) day. So \(T_{1}=1\) day.\[\begin{align} \left(\frac{T_{1}}{T_{2}}\right)^{2}&=\left(\frac{r_{1}}{r_{2}}\right)^{3} \\ \left(\frac{1}{27}\right)^{2}&=\left(\frac{r_{1}}{3.84\times10^{8}}\right)^{3} \\ r_{1}&=4.27\times10^{7}\text{ m} \end{align}\]