An explicit Example

In this section we will compute the cartesian coordinates of the satellite given as an example in a previous section. The two-line element was given by:

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 already calculated $a=26,560.46$ km. Furthermore, we have $e=0.0127851$. From this we get $b=26,558.29$km. The next step is to compute the ecc3entric anomaly from Kepler's equation. The mean anomaly was given by $\nu=322.378^{\circ}$. The numerical solution of Kepler's equation requires that we measure all angles in radians (You cannot do Calculus type operations with degrees ever!!). Transforming it into radians gives us $\nu=5.62639$. We next perform the fixed point iteration with $E_0=\nu=5.62639$. This gives us $E_1=5.63734$, $E_2=5.62767$, $E_3=5.62764$, and $E_4=5.62264$. Since $E_3=E_4$ both are approximations to the eccentric anomaly with 5 valid digits. We will use $E=5.62764$. Using this we can directly compute the $(x''',y''',z''')$ coordinates of the satellite.

$$\begin{eqnarray*} x''' &=& a\cos E-ae =20,566.52\\ y''' &=& b\sin E= -16,381.46\\ z''' &=& 0 \end{eqnarray*}$$

Next we compute the necessary transforms. First a rotation about $z'''$ by $-\omega$. From the two-line elements we get that $\omega=179.53^{\circ}$. Therefore,we have

$$\begin{eqnarray*} x''&=& \cos 179.53^{\circ} x'''-\sin 179.53^{\circ} y'''=-20,... ...3^{\circ}x'''+\cos 179.53^{\circ} y'''=16,549.62\\ z''&=&z''' \end{eqnarray*}$$

This is clearly a rotation of about $180^{\circ}$. In the nex step we need to rotate by $-i$ about the $x''$ axis. From the two line elements we get $i=56.26^{\circ}$. This gives us:

$$\begin{eqnarray*} x'&=&x''=-20,431.45\\ y'&=& \cos 56.26^{\circ}y''-\sin 56.2... ...\ z'&=& \sin 56.26^{\circ}y''+\cos 56.26^{\circ}z''=13,762.11 \end{eqnarray*}$$

We complete the procedure by rotating about the $z'$ axis by $-\Omega$. The two line elements give $\Omega=342.08^{\circ}$. This yields:

$$\begin{eqnarray*} x&=&\cos 342.08^{\circ}x'-\sin 342.08^{\circ}y'=-16,611.96\\ ... ...^{\circ}x'+\cos 342.08^{\circ}y'=15,032.66\\ z&=&z'=13,762.11 \end{eqnarray*}$$

Hence the cartesian coordinates of this satellite at the given time are $(-16,611,15,032.66,13,762.11)$, where these distances are measured in km. In this example we used an accuracy of 10m for the distances. One may of course need higher accuracy in some applications.