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In a nutshell, the so-called Kepler problem consists of determining
the radial and angular coordinates,
and
, respectively, of
an object in a Keplerian orbit about the Sun as a function of time.
Consider an object in a general Keplerian orbit about the Sun which
passes through its perihelion point,
and
, at
. It
follows from the previous analysis that
![\begin{displaymath}
r = \frac{r_p\,(1+e)}{1+e\,\cos\theta},
\end{displaymath}](img790.png) |
(276) |
and
![\begin{displaymath}
{\cal E} = \frac{\dot{r}^{\,2}}{2} + \frac{h^2}{2\,r^2} - \frac{G\,M}{r},
\end{displaymath}](img791.png) |
(277) |
where
,
, and
are the orbital eccentricity, angular momentum per unit mass, and
energy per unit mass, respectively. The above equation can be rearranged to
give
![\begin{displaymath}
\dot{r}^{\,2} = (e-1)\,\frac{G\,M}{r_p} - (e+1)\,\frac{r_p\,G\,M}{r^2}
+ \frac{2\,G\,M}{r}.
\end{displaymath}](img794.png) |
(278) |
Taking the square-root, and integrating, we obtain
![\begin{displaymath}
\int_{r_p}^r\frac{r\,dr}{[2\,r + (e-1)\,r^2/r_p - (e+1)\,r_p]^{1/2}} =
\sqrt{G\,M}\,\,t.
\end{displaymath}](img795.png) |
(279) |
Consider an elliptical orbit characterized by
. Let us write
![\begin{displaymath}
r = \frac{r_p}{1-e}\,(1-e\,\cos E),
\end{displaymath}](img797.png) |
(280) |
where
is termed the elliptic anomaly. In fact,
is an angle which
varies between
and
. Moreover, the perihelion point corresponds to
, and the aphelion point to
. Now,
![\begin{displaymath}
dr = \frac{r_p}{1-e}\,e\,\sin E\,dE,
\end{displaymath}](img802.png) |
(281) |
whereas
Thus, Equation (279) reduces to
![\begin{displaymath}
\int_0^E (1-e\,\cos E)\,dE = \left(\frac{G\,M}{a^3}\right)^{1/2} t,
\end{displaymath}](img806.png) |
(283) |
where
. This equation can immediately be integrated to give
![\begin{displaymath}
E - e\,\sin E = 2\pi\left(\frac{t}{T}\right),
\end{displaymath}](img808.png) |
(284) |
where
is the orbital period. Equation (284),
which is known as Kepler's equation, is a transcendental equation
which does not possess a simple analytic solution. Fortunately, it is fairly straightforward to
solve numerically. For instance, using an iterative approach,
if
is the
th guess then
![\begin{displaymath}
E_{n+1} = 2\pi\left(\frac{t}{T}\right) + e\,\sin E_n.
\end{displaymath}](img811.png) |
(285) |
The above iteration scheme converges very rapidly (except in the limit
as
).
Equations (276) and (280) can be combined
to give
![\begin{displaymath}
\cos\theta = \frac{\cos E - e}{1-e\,\cos E}.
\end{displaymath}](img812.png) |
(286) |
Thus,
![\begin{displaymath}
1+\cos\theta = 2\,\cos^2(\theta/2) = \frac{2\,(1-e)\,\cos^2( E/2)}{1-e\,\cos E},
\end{displaymath}](img813.png) |
(287) |
and
![\begin{displaymath}
1-\cos\theta = 2\,\sin^2(\theta/2) = \frac{2\,(1+e)\,\sin^2 (E/2)}{1-e\,\cos E}.
\end{displaymath}](img814.png) |
(288) |
The previous two equations imply that
![\begin{displaymath}
\tan (\theta/2) = \left(\frac{1+e}{1-e}\right)^{1/2}\tan (E/2).
\end{displaymath}](img815.png) |
(289) |
We conclude that, in the case of an elliptical orbit, the solution of the Kepler problem reduces to the solution of the following three equations:
Here,
and
. Incidentally, it is clear that if
then
, and
. In other words, the motion is periodic with period
.
For the case of a parabolic orbit, characterized by
, similar analysis to
the above yields:
Here,
is termed the parabolic anomaly, and varies between
and
, with the perihelion point corresponding to
. Note that Equation (293) is a cubic equation,
possessing a single real root,
which can, in principle, be solved analytically. However, a numerical
solution is generally more convenient.
Finally, for the case of a hyperbolic orbit, characterized by
,
we obtain:
Here,
is termed the hyperbolic anomaly, and varies between
and
, with the perihelion point corresponding to
. Moreover,
. As in the elliptical
case, Equation (296) is a transcendental equation which is most easily solved numerically.
Next: Motion in a General
Up: Planetary Motion
Previous: Orbital Energies
Richard Fitzpatrick
2011-03-31