Numerical Solution of Kepler's equation

Since the mean anomaly is proportional to time it is an easily measured quantity for an orbiting body. However, in order to compute the location of this body we need to find either the eccentric or the true anomaly. For this one needs to solve Kepler's equation for $E$. Since this is a transcendental equation, the solution can normally not be found in closed form. However, for small $e$ we can devise a simple iteration scheme. To start we rewrite Kepler's equation as

$$E=M+e\sin E =\Phi(E).$$ (28)

In this form, the equation has the form of a fixed point equation. In Calculus one encounters the following Fixed Point Theorem

For a proof of this theorem we refer to a Calculus textbook. Kepler's equation perfectly fits mold of this theorem. First of all, the function $\Phi$ in (28) satisfies

$$\Phi: [0,2\pi]\to [0,2\pi].$$

Moreover, we have that

$$\left\vert\Phi'(E)\right\vert=\left\vert e\cos E\right\vert\le e<1.$$

Therefore we can use this iteration to find an approximate solution of the equation.

Example: A satellite moves on an elliptic orbit with ellipticity $e=0.00001$. The mean anomaly is $30^{\circ}$, Find an approximate value of the eccentric anomaly. In this case $M=\pi/6=.5235988$. We set $E_1=M$. Using the iteration we get

$$\begin{eqnarray*} E_2&=&M+e\sin E_1=.5235988+0.00001\sin(.5235988)=.5236038\\ E_3&=&M+e\sin E_2=.5235988+0.00001\sin(.5236038)=.5236038. \end{eqnarray*}$$

We see that after only two iterations, we get $E_3=E_2$ and the approximate value of the eccentric anomaly is $.5236038$ or $30.00028^{\circ}$. The method converges not always that fast, depending on the values of $e$ and $M$. For larger values of $e$, we can also use Newton's method. For this we rewrite the equation as

$$0=E-e\sin E-M=F(E).$$

Newton's method yields the iteration (starting with $E_1=M$):

$$E_{n+1} =E_n-\frac{F(E_n)}{F'(E_n)}= E_n-\frac{E_n-e\sin E_n-M}{1-e\cos E_n}.$$

which converges faster that the simple fixed point iteration above.