Examples

To completely determine the motion of an earth orbiting satellite we need six numbers. First we need to determine the plane of the ellipse, then the orientation of the ellipse on the plane, finally we need to determine the shape of the ellipse.

To start we use a frame of reference. This will be a Cartesian coordinate system (xyz) whose $z$ axis agrees with the mean direction of the axis of rotation of the earth. The axis of rotation is not constant, but has a precession once in about 23,000 years. However, in the time frames in which we are interested--several hours to several years--we assume it constant. As $x$-axis, we use a use an axes perpendicular to the $z$-axis which points to the sun at the vernal equinox. The $y$ is a third perpendicular line to complete a right handed coordinate system. The Earths' equatorial plane is then the $xy$-plane of this coordinate system. The plane of motion will be either equal to this plane or intersect it in a line, called the line of nodes. To uniquely determine the plane of the motion, we use two angles:

1. $\Omega$ is the angle of right ascension of the line of nodes from the $x$ axis, i.e. the angle between the $x$-axis and the line of nodes measures in mathematical positive direction.
2. $i$ is the angle between the $z$-axis and the normal $n$ of the plane of motion.

This uniquely establishes the plane of the motion. The next thing is to establish the direction of the ellipse. This is done by the angle between the line of modes and the perigee of the ellipse. This angle is denoted by $\omega$.

The shape of the orbit is now fixed by its semi-major axis $a$ and its eccentricity $e$. Finally, we need to know either the time or the mean or true anomaly to establish the position of the satellite. These orbital elements are illustrated in Figure 3, below¹.

Figure 3: Graphical representation of orbital elements

To measure time, we need a reference time, this is called the Epoch. The original Cartesian coordinate system depends on this reference time as well, since the direction of the $z$ and $x$-axes are time dependent. There are two common frames of reference. Most commonly we use the M50 frame of reference, in which the epoch starts with the beginning of 1950. Also in use is the J2K frame of reference, where the epoch is given by the beginning of year 2000.

The data for many satellites are available in so called two line element sets at

http://celestrak.com/NORAD/elements/.

For example one of the GPS satellites (GPS BII-05 (PRN 17)) is given as

1 20361U 89097A   01154.90156813 -.0000008400000-0  00000-0 0  7462
2 20361 56.2556 342.0793 0127851 179.5306 322.3780 2.00562298 74668

We now explain these data care fully. We start with the meaning of the first line:

Column	Description
01	Line Number
03-07	Satellite Number
08	Classification
10-11	Launch year
12-14	Launch number
15-17	Launch piece
19-20	Epoch year
21-32	Epoch day and fraction of a day
34-43	First time derivative of mean motion
45-52	Second time derivative of the mean motion
54-61	$B^*$ drag term
63	Ephemeris type
65-68	Element number
69	Check sum

We see that our satellite was launched in launch number 97 of 1989 as piece A of this launch. The time of this data is 01154.90156813 i.e. it is the 154th day of 2001 (i.e. June 3), 21h 38min 15.486432 sec.

The next data are of little relevance for us. It is element # 746. The check sum is the sum of all numbers modulo 10 and is used a a control datum whether or not the data was transmitted correctly.

More important data are in the second row:

Column	Description
01	Line number
03-07	Satellite number
09-16	Inclination $i$ in degrees
18-25	Right ascension of the ascending node $\Omega$
27-33	Eccentricity $e$
35-42	Argument of the perigee $\omega$
44-51	Mean anomaly $M$ (or $\nu$)
53-63	Mean motion (revolutions per day)
64-68	Revolution number at epoch
69	Check sum

Therefore in our example, we have:

The orbit is at an inclination of 56.2556 degrees, $\Omega=342.0793^{\circ}$. The eccentricity of the orbit is 0.0127851. The argument of the perigee is $\omega=179.5306^{\circ}$. The mean anomaly $\nu=322.3780^{\circ}$. The satellite completes 2.00562298 revolutions (orbits) per day, and was in its 7466th revolution at the time given.

We observe, that $a$ is not given in this set of data. However, from the mean motion and Kepler's third law we can compute $a$. Clearly $T=1/2.00562298 =.4985981962$ days. Kepler's third law is

$$a^3=\frac{GM}{4\pi^2}T^2.$$

and

$$GM=3986004.48\times10^8 \frac{{\rm m}^3}{{\rm s}^2}=2.9755364\times10^{15} \frac{{\rm km}^3}{{\rm d}^2}.$$

This gives a semi-major axis

$$a=26560.46326 {\rm km}$$