Fields Tutorial 3  Satellite Motion
A satellite is anything that orbits around another object in space. The Moon is a satellite of the Earth. A planet is a satellite of a star. The movement of planets and stars has been studied carefully for many thousands of years. Up to now we have always treated orbits of satellites as being circular. In fact they have orbits that are ellipses (slightly squashed circles). The shape of orbits obey Kepler's Laws.
Kepler's Laws were written in the early Seventeenth Century by a German mathematician, Johannes Kepler (1571  1630), based on the work of a Danish astronomer Tycho Brahe (1546  1601). He discovered that orbits were not perfect circles, but ellipses.
The three laws are:
Kepler I  the paths of planets are elliptical, with the star at one focus of the ellipse.
Maths Note An ellipse is a curved figure in which the sum of the distances from any point along the curve to the two foci are constant.
The general equation for an ellipse is:
Where:
A circle is a special ellipse where the
two foci are in exactly the same place. 
Kepler II  an imaginary line drawn from the star to a planet will sweep an equal area in an equal time period. It is called the Law of Equal Periods. The period is the time taken for the planet to do one complete orbit around the star.
As the area swept per unit time is the same, the orbital speed must be greater for the short wide area, and be lower for the long thin area. This is shown below:
Kepler III  the ratio of the squares of the periods of two planets is the ratio of the cubes of the average distances from the star to the planets. This is called the Law of Harmonies. Kepler III is of particular interest to us in considering the motion of satellites. Consider two planets. Planet 1 has orbital period of T_{1} and orbital radius of r_{1}. Planet 2 has orbital period of T_{2} and average orbital radius of r_{2}. Keppler III gives us this equation:
Motion of Masses in Gravitational Fields
Newton’s Laws of Gravitation can be used to explain the motion of planets and stars. Much of modern space exploration uses the three hundredyearold model. Orbiting satellites are NOT doing gravity defiance acts; instead they are actually falling in a curved path towards the Earth all the time. However they have a sufficient forwards velocity to miss the Earth all the time. Since the gravity field is radial, the force acts at 90^{o} to the direction of travel all the time. Therefore the path is circular. If we stopped gravity, the satellite would fly off tangentially into space in a straight line. If we stopped the satellite, it would fall straight back to Earth.
For a satellite to be in a particular orbit, a particular velocity is required or a given distance. Some satellites are placed so that they go in an easterly direction, completing one orbit each day. They remain above one given point on the Earth’s surface, so are called geostationary. This kind of satellite orbit is used in telecommunications and GPS.
Other satellites move in a polar orbit so that they can perform sweeps of the surface. Spy satellites use a polar orbit. Polar orbits are lower than geostationary orbits with a lower orbital period.
We can derive an expression for the geostationary satellite. We know that:
And
So we can write:
The m terms cancel out:
The term v^{2}/r = acceleration:
We also know that:
We can therefore write:
The minus signs cancel. We can simplify the expression to:
However, frequency, f, is not a very useful term; the period T is more helpful:
Since T = 1/f, we can write:
Our final rearrangement gives:
This relationship is called Kepler's Third Law.
Use the mass of the larger object in your calculation. So if planets are orbiting a star, use the mass of the star.
Strictly speaking we should use both the mass of the star and the planet, but planets tend to have much lower masses than stars, so we ignore the mass of the star. However if two stars are being considered, then we do need to change the equation to:

Worked Example A communications satellite is to be placed in a circular geostationary orbit. What must its height and speed be? 
Answer
w = 2pf.
We need to work out f. f = 1 ______ ___ = 1.16 ´ 10^{5} Hz 24 h ´ 60 min h^{1} ´ 60 s h^{1}
Þ w = 2 ´ p ´ 1.16 ´ 10^{5} Hz = 7.27 ´ 10^{5} rad s^{1}.
Now we need to consider the centripetal acceleration:
a = w^{2}r.
We also know that the acceleration is also given by g = GM/r^{2}
Þ w^{2}r = GM/r^{2} [The minus signs cancel.]
Þ r^{3} = GM = 6.67 ´ 10^{11} N m^{2} kg^{2} ´ 5.98 ´ 10^{24} kg = 7.55 ´ 10^{22} m^{3} (7.27 ´ 10^{5} rad s^{1})^{2}
Þ r = ^{3}Ö 7.55 ´ 10^{22} m^{3} = 4.24 ´ 10^{7} m
But we need to take away the radius of the Earth to get the height:
Height = 4.24 × 10^{7} m  6.37 ´ 10^{6} m = 3.60 × 10^{7} m = 36000 km
Now we can work out the forward velocity:
v = wr = 7.27 ´ 10^{5} rad s^{1} ´ 4.24 ´ 10^{7} m = 3100 m s^{1}. 
(a) State Newton's Law of Gravitation.
(b) Use Kepler III to show that the Earth orbits the Sun every 365 days. The mass of the Sun is 1.989 × 10^{30} kg and the orbital radius is 1.496 × 10^{11} m. 
If we plot a graph of T^{2} against r^{3}, we get a straight line.
If we observe the orbital period of four different satellites and know the radius of their orbits, we see that the graph is a straight line. The data fit the straight line neatly. Orbits are very slightly elliptical, but to a first approximation, they are circular.
We can also plot the graph of T^{2/3} against r. T^{2/3} is the cube root of the square of T.
Note that in some treatments on this topic, the distances, r, are measured in astronomical units (AU). 1 AU = 1.496 × 10^{11} m. The time periods, T, are measured in years (y). The masses, M, are in solar masses where 1 solar mass = 1.989 × 10^{30} kg. In this case we can write:
This makes:
The gravitational constant in this case is NOT 6.67 × 10^{11} N m^{2} kg^{2}.
Escape Velocity of a Rocket
This refers to the minimum velocity that is needed to move a rocket, mass m, from the Earth’s surface to infinity. We need to consider the work that needs to be done:
This work has to be done against gravity, and the energy for that work is provided by the fuel.
The object gains kinetic energy which is equal to the potential energy:
Therefore we can write:
Notice that the minus sign has gone. This is because we are doing a job of work against the gravity field. The masses cancel out, so we can say:
This then gives:
We know that gravitational field strength, g, is given by:
Which we can rearrange to:
Therefore:
The r^{2 }term cancels out to r. We take the square root to give our final expression:
What is the escape velocity of a rocket from the Earth? (g = 9.8 N kg^{1}; r = 6.37 × 10^{6} m) 

A black hole is an object that has such a high mass that light, travelling at 3 × 10^{8} m s^{1}, cannot escape. (a) What is the escape velocity of such a black hole? (b) Calculate the mass of such a black hole, assuming that it has the same radius of the Earth. (c) Compare the mass of the black hole with the mass of the Sun. Mass of the sun = 1.99 × 10^{30} kg 