Eccentric anomaly

Formulas

Radius and eccentric anomaly

The eccentricity e is defined as:

${\displaystyle e=\sqrt{1-{\left(\frac{b}{a}\right)}^2}.}$

From Pythagoras's theorem applied to the triangle with r (a distance FP) as hypotenuse:

${\displaystyle \begin{array}{cc}{r}^2& =b^2{\sin }^{2}E+(ae-a\cos E{)}^2\\ & =a^2\left(1-e^2\right)\left(1-{\cos }^{2}E\right)+a^2\left(e^2-2e\cos E+{\cos }^{2}E\right)\\ & =a^2-2a^{2}e\cos E+a^2{e}^2{\cos }^{2}E\\ & =a^2{\left(1-e\cos E\right)}^2\end{array}}$

Thus, the radius (distance from the focus to point P) is related to the eccentric anomaly by the formula

${\displaystyle r=a\left(1-e\cos E\right).}$

With this result the eccentric anomaly can be determined from the true anomaly as shown next.

From the true anomaly

The true anomaly is the angle labeled ${\displaystyle \theta }$ in the figure, located at the focus of the ellipse. It is sometimes represented by f or v. The true anomaly and the eccentric anomaly are related as follows.

Using the formula for r above, the sine and cosine of E are found in terms of f :

${\displaystyle \begin{array}{cc}\cos E& =\frac{x}{a}=\frac{ae+r\cos f}{a}=e+(1-e\cos E)\cos f\\ \Rightarrow \cos E& =\frac{e+\cos f}{1+e\cos f}\\ \sin E& =\sqrt{1-{\cos }^{2}E}=\frac{\sqrt{1-e^2}\sin f}{1+e\cos f}.\end{array}}$

Hence,

${\displaystyle \tan E=\frac{\sin E}{\cos E}=\frac{\sqrt{1-e^2}\sin f}{e+\cos f}.}$

where the correct quadrant for E is given by the signs of numerator and denominator, so that E can be most easily found using an atan2 function.

Angle E is therefore the adjacent angle of a right triangle with hypotenuse ${\displaystyle 1+e\cos f,}$ adjacent side ${\displaystyle e+\cos f,}$ and opposite side ${\displaystyle \sqrt{1-e^2}\sin f.}$

Also,

${\displaystyle \tan \frac{f}{2}=\sqrt{\frac{1+e}{1-e}}\tan \frac{E}{2}}$

Substituting cos E as found above into the expression for r, the radial distance from the focal point to the point P, can be found in terms of the true anomaly as well:

${\displaystyle r=\frac{a\left(1-e^2\right)}{1+e\cos f}=\frac{p}{1+e\cos f}}$

where

${\displaystyle p\equiv a\left(1-e^2\right)}$

is called "the semi-latus rectum" in classical geometry.

From the mean anomaly

The eccentric anomaly E is related to the mean anomaly M by Kepler's equation:

${\displaystyle M=E-e\sin E}$

This equation does not have a closed-form solution for E given M. It is usually solved by numerical methods, e.g. the Newton–Raphson method. It may be expressed in a Fourier series as

${\displaystyle E=M+2\sum _{n=1}^{\mathrm{\infty }}\frac{J_n(ne)}{n}\sin (nM)}$

where ${\displaystyle J_n(x)}$ is the Bessel function of the first kind.