The 'eccentric anomaly' 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,, at time, then to find out the position of the body at any time, you first calculate the mean anomaly from the time and themean motion by the formula, then solve the Kepler equation above to get, then get the coordinates relative to the central gravitational body from:
There are several forms of Kepler's equation. Each form is associated with a specific type of orbit. The standard Kepler equation is used for elliptic orbits (). The hyperbolic Kepler equation is used for hyperbolic trajectories (). The radial Kepler equation is used for linear (radial) trajectories ().Barker's equation is used for parabolic trajectories (for which). With the parabolic orbit, unlike the elliptical or hyperbolic orbits, it is possible to solve Barker's equation and find aclosed-form expression for the position as a function of time.
When, the orbit is circular. Increasing causes the circle to become elliptical. When, there are four possibilities:
a parabolic trajectory,
a trajectory that goes back and forth along a line segment from the centre of attraction to a point at some distance away,
a trajectory going in or out along an infinite ray emanating from the centre of attraction, with its speed going to zero with distance
or a trajectory along a ray, but with speed not going to zero with distance.
A value of slightly above 1 results in a hyperbolic orbit with a turning angle of just under 180 degrees. Further increases reduce the turning angle, and as goes to infinity, the orbit becomes a straight line of infinite length.
where is the hyperbolic eccentric anomaly.This equation is derived by redefining M to be thesquare root of −1 times the right-hand side of the elliptical equation:
(in which is now imaginary) and then replacing by.
The Radial Kepler equation for the case where the object does not have enough energy to escape is:
where is proportional to time and is proportional to the distance from the centre of attraction along the ray and attains the value 1 at the maximum distance. This equation is derived by multiplying Kepler's equation by 1/2 and setting 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.
Calculating for a given value of is straightforward. However, solving for when is given can be considerably more challenging. There is noclosed-form solution. Solving for is more or less equivalent to solving for thetrue anomaly, or the difference between the true anomaly and the mean anomaly, which is called the "Equation of the center".
One can write aninfinite series expression for the solution to Kepler's equation usingLagrange inversion, but the series does not converge for all combinations of and (see below).
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 greatApollonius.
The inverse Kepler equation is the solution of Kepler's equation for all real values of:
Evaluating this yields:
These series can be reproduced inMathematica with the InverseSeries operation.
InverseSeries[Series[M-Sin[M],{M,0,10}]]
InverseSeries[Series[M-eSin[M],{M,0,10}]]
These functions are simpleMaclaurin series. SuchTaylor series representations of transcendental functions are considered to be definitions of those functions. Therefore, this solution is a formal definition of the inverse Kepler equation. However, is not anentire function of at a given non-zero. Indeed, the derivative
goes to zero at aninfinite set of complex numbers when the nearest to zero being at and at these two points
(where inverse cosh is taken to be positive), and goes to infinity at these values of. This means that the radius of convergence of the Maclaurin series is and the series will not converge for values of larger than this. The series can also be used for the hyperbolic case, in which case the radius of convergence is The series for when converges when.
While this solution is the simplest in a certain mathematical sense,[which?], other solutions are preferable for most applications. Alternatively, Kepler's equation can be solved numerically.
One can also write a Maclaurin series in. This series does not converge when is larger than theLaplace limit (about 0.66), regardless of the value of (unless is a multiple of2π), but it converges for all if is less than the Laplace limit. The coefficients in the series, other than the first (which is simply), depend on in a periodic way with period2π.
Note that and are in units of radians in this computation. This iteration is repeated until desired accuracy is obtained (e.g. when < desired accuracy). For most elliptical orbits an initial value of is sufficient. For orbits with, a initial value of can be used. Numerous works developed accurate (but also more complex) start guesses.[16] If is identically 1, then the derivative of, 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, thebisection 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.[18]: 66–67 In the case of a parabolic trajectory,Barker's equation is used.
A related method starts by noting that. Repeatedly substituting the expression on the right for the on the right yields a simplefixed-point iteration algorithm for evaluating. This method is identical to Kepler's 1621 solution.[4] Inpseudocode:
^It is often claimed that Kepler's equation "cannot be solvedanalytically"; see for examplehere. Other authors claim that it cannot be solved at all; see for example Madabushi V. K. Chari; Sheppard Joel Salon;Numerical Methods in Electromagnetism, Academic Press, San Diego, CA, USA, 2000,ISBN0-12-615760-X,p. 659
^Boyd, John P. (2007). "Rootfinding for a transcendental equation without a first guess: Polynomialization of Kepler's equation through Chebyshev polynomial equation of the sine".Applied Numerical Mathematics.57 (1):12–18.doi:10.1016/j.apnum.2005.11.010.
Conway, Bruce A. (1986). "An improved algorithm due to Laguerre for the solution of Kepler's equation".24th Aerospace Sciences Meeting.doi:10.2514/6.1986-84.
Boyd, John P. (2007). "Rootfinding for a transcendental equation without a first guess: Polynomialization of Kepler's equation through Chebyshev polynomial equation of the sine".Applied Numerical Mathematics.57 (1):12–18.doi:10.1016/j.apnum.2005.11.010.