Kepler's equation
It was derived by Johannes Kepler in 1609 in Chapter 60 of his Astronomia nova,[1][2] and in book V of his Epitome of Copernican Astronomy (1621) Kepler proposed an iterative solution to the equation.
[3][4] This equation and its solution, however, first appeared in a 9th-century work by Habash al-Hasib al-Marwazi, which dealt with problems of parallax.
[5][6][7][8] The equation has played an important role in the history of both physics and mathematics, particularly classical celestial mechanics.
is useful to compute the position of a point moving in a Keplerian orbit.
As for instance, if the body passes the periastron at coordinates
, then to find out the position of the body at any time, you first calculate the mean anomaly
Numerical analysis and series expansions are generally required to evaluate
The standard Kepler equation is used for elliptic orbits (
slightly above 1 results in a hyperbolic orbit with a turning angle of just under 180 degrees.
goes to infinity, the orbit becomes a straight line of infinite length.
The Radial Kepler equation for the case where the object does not have enough energy to escape is:
to 1: and then making the substitution The radial equation for when the object has enough energy to escape is:
When the energy is exactly the minimum amount needed to escape, then the time is simply proportional to the distance to the power 3/2.
Confusion over the solvability of Kepler's equation has persisted in the literature for four centuries.
[9] Kepler himself expressed doubt at the possibility of finding a general solution: I am sufficiently satisfied that it [Kepler's equation] cannot be solved a priori, on account of the different nature of the arc and the sine.
But if I am mistaken, and any one shall point out the way to me, he will be in my eyes the great Apollonius.Fourier series expansion (with respect to
: Evaluating this yields: These series can be reproduced in Mathematica with the InverseSeries operation.
Therefore, this solution is a formal definition of the inverse Kepler equation.
Indeed, the derivative goes to zero at an infinite set of complex numbers when
This means that the radius of convergence of the Maclaurin series is
was found by Karl Stumpff in 1968,[14] but its significance wasn't recognized.
[15][clarification needed] One can also write a Maclaurin series in
) for the case in which the object does not have enough energy to escape can similarly be written as: Evaluating this yields: To obtain this result using Mathematica: For most applications, the inverse problem can be computed numerically by finding the root of the function: This can be done iteratively via Newton's method: Note that
This iteration is repeated until desired accuracy is obtained (e.g. when
Numerous works developed accurate (but also more complex) start guesses.
, which is in the denominator of Newton's method, can get close to zero, making derivative-based methods such as Newton-Raphson, secant, or regula falsi numerically unstable.
In that case, the bisection method will provide guaranteed convergence, particularly since the solution can be bounded in a small initial interval.
On modern computers, it is possible to achieve 4 or 5 digits of accuracy in 17 to 18 iterations.
[17] A similar approach can be used for the hyperbolic form of Kepler's equation.
on the right yields a simple fixed-point iteration algorithm for evaluating
