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Eccentric Anomaly

\begin{figure}\begin{center}\BoxedEPSF{Eccentric_Anomaly.epsf scaled 1000}\end{center}\end{figure}

The Angle obtained by drawing the Auxiliary Circle of an Ellipse with center $O$ and Focus $F$, and drawing a Line Perpendicular to the Semimajor Axis and intersecting it at $A$. The Angle $E$ is then defined as illustrated above. Then for an Ellipse with Eccentricity $e$,

\begin{displaymath}
AF = OF-AO = ae-a\cos E.
\end{displaymath} (1)

But the distance $AF$ is also given in terms of the distance from the Focus $r=FP$ and the Supplement of the Angle from the Semimajor Axis $v$ by
\begin{displaymath}
AF = r\cos(\pi-v) = -r\cos v.
\end{displaymath} (2)

Equating these two expressions gives
\begin{displaymath}
r = {a(\cos E-e)\over\cos v},
\end{displaymath} (3)

which can be solved for $\cos v$ to obtain
\begin{displaymath}
\cos v = {a(\cos E-e)\over r}.
\end{displaymath} (4)

To get $E$ in terms of $r$, plug (4) into the equation of the Ellipse
\begin{displaymath}
r = {a(1-e^2)\over 1+e\cos v}
\end{displaymath} (5)


\begin{displaymath}
r(1+e\cos v) = a(1-e^2)
\end{displaymath} (6)


\begin{displaymath}
r\left({1 + {ae\cos E\over r} - {e^2\over r}}\right)=r+ae\cos E-e^2= a(1-e^2)
\end{displaymath} (7)


\begin{displaymath}
r = a(1-e^2)-ea\cos E+e^2 a = a(1-e\cos E).
\end{displaymath} (8)

Differentiating gives
\begin{displaymath}
\dot r =ae\dot E\sin E.
\end{displaymath} (9)

The eccentric anomaly is a very useful concept in orbital mechanics, where it is related to the so-called mean anomaly $M$ by Kepler's Equation
\begin{displaymath}
M=E-e\sin E.
\end{displaymath} (10)

$M$ can also be interpreted as the Area of the shaded region in the above figure (Finch).

See also Eccentricity, Ellipse, Kepler's Equation


References

Danby, J. M. Fundamentals of Celestial Mechanics, 2nd ed., rev. ed. Richmond, VA: Willmann-Bell, 1988.

Finch, S. ``Favorite Mathematical Constants.'' http://www.mathsoft.com/asolve/constant/lpc/lpc.html



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© 1996-9 Eric W. Weisstein
1999-05-25