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Keplerian Orbit Theory

According to Kepler's first law of planetary motion, all planetary orbits are ellipses lying in a fixed plane which are confocal with the sun. Moreover, according to Kepler's second law, the radius vector which connects the sun to a given planet sweeps out equal areas in equal time intervals.

Figure 2: A Keplerian orbit.
\begin{figure}
\epsfysize =2.5in
\centerline{\epsffile{fig1.eps}}
\end{figure}

Consider Figure 2. $\Pi PBA$ is the arc of an ellipse which corresponds to a planetary orbit. Furthermore, $C$ is the geometric center of the orbit, $S$ the focus at which the sun is located, $P$ the instantaneous position of the planet, $\Pi$ the perihelion point (i.e., the point of closest approach to the sun), and $A$ the aphelion point (i.e., the point of furthest distance from the sun). The ellipse is symmetric about $\Pi A$, which is termed the major axis, and about $CB$, which is termed the minor axis. The length $CA\equiv a$ is called the orbital major radius. The length $CS$ represents the displacement of the sun from the geometric center of the orbit, and is generally written $e\,a$, where $e$ is termed the orbital eccentricity. The length $CB\equiv b = a\,(1-e^2)^{1/2}$ is called the orbital minor radius. The length $SP\equiv r$ represents the radial distance of the planet from the sun. Finally, the angle $RSP\equiv T$ is the angular bearing of the planet from the sun, relative to the major axis of the orbit, and is termed the true anomaly.

$\Pi QDA$ is the arc of a circle whose geometric center is $C$, and whose radius is $a$. Hence, the circle passes through the perihelion and aphelion points. $R$ is the point at which the perpendicular from $P$ meets the major axis $\Pi A$. The point where $RP$ produced meets circle $\Pi QDA$ is denoted $Q$. Finally, the angle $SCQ\equiv E$ is called the elliptic anomaly.

Now, the equation of the ellipse $\Pi PBA$ is

\begin{displaymath}
\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1,
\end{displaymath} (1)

where $x$ and $y$ are the perpendicular distances from the major and minor axes, respectively. Likewise, the equation of the circle $\Pi QDA$ is
\begin{displaymath}
\frac{x'^{\,2}}{a^2} + \frac{y'^{\,2}}{a^2} = 1.
\end{displaymath} (2)

Hence, if $x=x'$ then
\begin{displaymath}
\frac{y}{y'} = \frac{b}{a},
\end{displaymath} (3)

and it follows that
\begin{displaymath}
\frac{RP}{RQ} = \frac{b}{a}.
\end{displaymath} (4)

It is easily demonstrated that $SR=r\,\cos T$, $RP=r\,\sin T$, $CR= a\,\cos E$, $RQ= a\,\sin E$, and $CS= e\,a$. Consequently, Eq. (4) yields

\begin{displaymath}
r\,\sin T = b\,\sin E.
\end{displaymath} (5)

Also, since $SR = CR-CS$, we have
\begin{displaymath}
r\,\cos T = a\,(\cos E - e).
\end{displaymath} (6)

Taking the square root of the sum of the squares of the previous two equations, we obtain
\begin{displaymath}
r = a\,(1-e\,\cos E),
\end{displaymath} (7)

which can be combined with Eq. (6) to give
\begin{displaymath}
\cos T = \frac{\cos E - e}{1-e\,\cos E}.
\end{displaymath} (8)

Now, according to Kepler's second law,

\begin{displaymath}
\frac{{\rm Area}\,\Pi PS}{\pi\,a\,b} = \frac{t-t_\ast}{\tau},
\end{displaymath} (9)

where $t$ is the time at which the planet passes point $P$, $t_\ast$ the time at which it passes the perihelion point, and $\tau$ the orbital period. However,
$\displaystyle {\rm Area} \,\Pi PS$ $\textstyle =$ $\displaystyle {\rm Area}\, SRP + {\rm Area} \,\Pi RP$  
  $\textstyle =$ $\displaystyle \frac{1}{2}\,r^2\,\cos T\,\sin T + {\rm Area}\,\Pi RP.$ (10)

But,
\begin{displaymath}
{\rm Area}\,\Pi RP = \frac{b}{a}\,{\rm Area}\,\Pi RQ,
\end{displaymath} (11)

since $RP/RQ = b/a$ for all values of $T$. In addition,
$\displaystyle {\rm Area}\,\Pi RQ$ $\textstyle =$ $\displaystyle {\rm Area}\,\Pi QC - {\rm Area}\,RQC$  
  $\textstyle =$ $\displaystyle \frac{1}{2}\,E\,a^2 - \frac{1}{2}\,a^2\,\cos E\,\sin E.$ (12)

Hence, we can write
\begin{displaymath}
\left(\frac{t-t_\ast}{\tau}\right) \pi\,a\,b= \frac{1}{2}\,r...
... T\,\sin T + \frac{b}{a}\,\frac{a^2}{2}\,(E - \cos E\,\sin E).
\end{displaymath} (13)

It is easily demonstrated that $r\,\sin T = b\,\sin E$, and $r\,\cos T = a\,(\cos E - e)$, so the above equation reduces to
\begin{displaymath}
M = E - e\,\sin E,
\end{displaymath} (14)

where
\begin{displaymath}
M = \left(\frac{2\pi}{\tau}\right)\,(t-t_\ast)
\end{displaymath} (15)

is an angle which is zero at the perihelion point, increases uniformly in time, and has a repetition period which matches the period of the planetary orbit. This angle is termed the mean anomaly.

In summary, the radial and angular polar coordinates, $r$ and $T$, respectively, of a planet in a Keplerian orbit about the sun are specified as implicit functions of the mean anomaly, which is a linear function of time, by the following three equations:

$\displaystyle M$ $\textstyle =$ $\displaystyle E - e\,\sin E,$ (16)
$\displaystyle r$ $\textstyle =$ $\displaystyle a\,(1- e\,\cos E),$ (17)
$\displaystyle \cos T$ $\textstyle =$ $\displaystyle \frac{\cos E - e}{1-e\,\cos E}.$ (18)

It turns out that the earth and the five visible planets all possess low eccentricity orbits characterized by $e\ll 1$. Hence, it is a good approximation to expand the above three equations using $e$ as a small parameter. To second-order, we get
$\displaystyle E$ $\textstyle =$ $\displaystyle M + e\,\sin M + (1/2)\,e^2\,\sin 2M,$ (19)
$\displaystyle r$ $\textstyle =$ $\displaystyle a\,(1-e\,\cos T - e^2\,\sin^2 T),$ (20)
$\displaystyle T$ $\textstyle =$ $\displaystyle E + e\,\sin E + (1/4)\,e^2\,\sin 2 E.$ (21)

These equations can be combined to give $r$ and $T$ as explicit functions of the mean anomaly:
$\displaystyle r$ $\textstyle =$ $\displaystyle a\,(1 -e\,\cos M + e^2\,\sin^2 M),$ (22)
$\displaystyle T$ $\textstyle =$ $\displaystyle M + 2\,e\,\sin M + (5/4)\,e^2\,\sin 2M.$ (23)

It is interesting to note that the above formulae are almost identical to those obtained by expanding Ptolemy's equant theory to second-order in the eccentricity: i.e.,
$\displaystyle r$ $\textstyle =$ $\displaystyle a\,[1 -e\,\cos M + (3/2)\,e^2\,\sin^2 M],$ (24)
$\displaystyle T$ $\textstyle =$ $\displaystyle M + 2\,e\,\sin M + e^2\,\sin 2M.\ $ (25)


next up previous
Next: The Celestial Sphere Up: Almagest Previous: Dates
Richard Fitzpatrick 2008-02-25