Fields Tutorial 3 - Satellite Motion



Kepler's Laws

Motion of Masses

Graphs from Kepler III

Escape Velocity of a Rocket


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

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:


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:


  • x and y are the coordinates of any point on the ellipse,

  • a and b are the radius on the x and y axes.


A circle is a special ellipse where the two foci are in exactly the same place.




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:





Motion of Masses in Gravitational Fields

Newtons Laws of Gravitation can be used to explain the motion of planets and stars.  Much of modern space exploration uses the three hundred-year-old 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 90o 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 Earths 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:




So we can write:


The m terms cancel out:


The term v2/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?


  • This question seems to be remarkably lacking in information, but there is an answer.

  • Use Newtons Laws of Gravitation to solve this.  There is a single force acting on the satellite, gravitational attraction, so the satellite is acceleration all the time towards the centre of the Earth.

  • We need the satellite to be travelling at a sufficient forwards velocity so that it completes ONE orbit every 24 hours.  We need to work out the angular velocity before we can work out the linear speed.


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


w = 2 p 1.16 10-5 Hz = 7.27 10-5 rad s-1.


Now we need to consider the centripetal acceleration:


a = -w2r.


We also know that the acceleration is also given by g = -GM/r2


w2r = GM/r2 [The minus signs cancel.]


r3 = -GM = 6.67 10-11 N m2 kg-2 5.98 1024 kg = 7.55 1022 m3

(7.27 10-5 rad s-1)2


r = 3 7.55 1022 m3 = 4.24 107 m


But we need to take away the radius of the Earth to get the height:


Height = 4.24 107 m - 6.37 106 m = 3.60 107 m = 36000 km


Now we can work out the forward velocity:


v = wr = 7.27 10-5 rad s-1 4.24 107 m = 3100 m s-1.



Question 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 1030 kg and the orbital radius is 1.496 1011 m.




Graphs from Kepler III

If we plot a graph of T2 against r3, 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 T2/3 against rT2/3 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 1011 m.  The time periods, T, are measured in years (y).  The masses, M, are in solar masses where 1 solar mass = 1.989 1030 kg.  In this case we can write:


This makes:

The gravitational constant in this case is NOT 6.67 10-11 N m2 kg-2.



Escape Velocity of a Rocket

This refers to the minimum velocity that is needed to move a rocket, mass m, from the Earths 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:





The r2 term cancels out to r.  We take the square root to give our final expression:



Question 2

What is the escape velocity of a rocket from the Earth? (g = 9.8 N kg-1; r = 6.37 106 m)


Question 3

A black hole is an object that has such a high mass that light, travelling at 3 108 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 1030 kg