# Kepler's equation

Kepler's equation solutions for five different eccentricities between 0 and 1

In orbital mechanics, Kepler's equation relates various geometric properties of the orbit of a body subject to a central force.

It was first 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] The equation has played an important role in the history of both physics and mathematics, particularly classical celestial mechanics.

## Equation

Kepler's equation is

 ${\displaystyle M=E-e\sin E}$

where M is the mean anomaly, E is the eccentric anomaly, and e is the eccentricity.

The 'eccentric anomaly' E 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 x = a(1 − e), y = 0, at time t = t0, then to find out the position of the body at any time, you first calculate the mean anomaly M from the time and the mean motion n by the formula M = n(tt0), then solve the Kepler equation above to get E, then get the coordinates from:

 ${\displaystyle {\begin{array}{lcl}x&=&a(\cos E-e)\\y&=&b\sin E\end{array}}}$

where a is the semi-major axis, b the semi-minor axis.

Kepler's equation is a transcendental equation because sine is a transcendental function, meaning it cannot be solved for E algebraically. Numerical analysis and series expansions are generally required to evaluate E.

## Alternate forms

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 (0 ≤ e < 1). The hyperbolic Kepler equation is used for hyperbolic trajectories (e ≫ 1). The radial Kepler equation is used for linear (radial) trajectories (e = 1). Barker's equation is used for parabolic trajectories (e = 1).

When e = 0, the orbit is circular. Increasing e causes the circle to become elliptical. When e = 1, there are three possibilities:

• a parabolic trajectory,
• a trajectory going in or out along an infinite ray emanating from the centre of attraction,
• or a trajectory that goes back and forth along a line segment from the centre of attraction to a point at some distance away.

A slight increase in e above 1 results in a hyperbolic orbit with a turning angle of just under 180 degrees. Further increases reduce the turning angle, and as e goes to infinity, the orbit becomes a straight line of infinite length.

### Hyperbolic Kepler equation

The Hyperbolic Kepler equation is:

 ${\displaystyle M=e\sinh H-H}$

where H is the hyperbolic eccentric anomaly. This equation is derived by redefining M to be the square root of −1 times the right-hand side of the elliptical equation:

${\displaystyle M=i\left(E-e\sin E\right)}$

(in which E is now imaginary) and then replacing E by iH.

 ${\displaystyle t(x)=\sin ^{-1}({\sqrt {x}})-{\sqrt {x(1-x)}}}$

where t is proportional to time and x is proportional to the distance from the centre of attraction along the ray. This equation is derived by multiplying Kepler's equation by 1/2 and setting e to 1:

${\displaystyle t(x)={\frac {1}{2}}\left[E-\sin(E)\right].}$

and then making the substitution

${\displaystyle E=2\sin ^{-1}({\sqrt {x}}).}$

## Inverse problem

Calculating M for a given value of E is straightforward. However, solving for E when M is given can be considerably more challenging. There is no closed-form solution.

One can write an infinite series expression for the solution to Kepler's equation using Lagrange inversion, but the series does not converge for all combinations of e and M (see below).

Confusion over the solvability of Kepler's equation has persisted in the literature for four centuries.[5] Kepler himself expressed doubt at the possibility of finding a general solution.

### Inverse Kepler equation

The inverse Kepler equation is the solution of Kepler's equation for all real values of ${\displaystyle e}$ :

${\displaystyle E={\begin{cases}\displaystyle \sum _{n=1}^{\infty }{\frac {M^{\frac {n}{3}}}{n!}}\lim _{\theta \to 0^{+}}\!{\Bigg (}{\frac {\mathrm {d} ^{\,n-1}}{\mathrm {d} \theta ^{\,n-1}}}{\bigg (}{\bigg (}{\frac {\theta }{\sqrt[{3}]{\theta -\sin(\theta )}}}{\bigg )}^{\!\!\!n}{\bigg )}{\Bigg )},&e=1\\\displaystyle \sum _{n=1}^{\infty }{\frac {M^{n}}{n!}}\lim _{\theta \to 0^{+}}\!{\Bigg (}{\frac {\mathrm {d} ^{\,n-1}}{\mathrm {d} \theta ^{\,n-1}}}{\bigg (}{\Big (}{\frac {\theta }{\theta -e\sin(\theta )}}{\Big )}^{\!n}{\bigg )}{\Bigg )},&e\neq 1\end{cases}}}$

Evaluating this yields:

${\displaystyle E={\begin{cases}\displaystyle s+{\frac {1}{60}}s^{3}+{\frac {1}{1400}}s^{5}+{\frac {1}{25200}}s^{7}+{\frac {43}{17248000}}s^{9}+{\frac {1213}{7207200000}}s^{11}+{\frac {151439}{12713500800000}}s^{13}+\cdots {\text{ with }}s=(6M)^{1/3},&e=1\\\\\displaystyle {\frac {1}{1-e}}M-{\frac {e}{(1-e)^{4}}}{\frac {M^{3}}{3!}}+{\frac {(9e^{2}+e)}{(1-e)^{7}}}{\frac {M^{5}}{5!}}-{\frac {(225e^{3}+54e^{2}+e)}{(1-e)^{10}}}{\frac {M^{7}}{7!}}+{\frac {(11025e^{4}+4131e^{3}+243e^{2}+e)}{(1-e)^{13}}}{\frac {M^{9}}{9!}}+\cdots ,&e\neq 1\end{cases}}}$

These series can be reproduced in Mathematica with the InverseSeries operation.

InverseSeries[Series[M - Sin[M], {M, 0, 10}]]
InverseSeries[Series[M - e Sin[M], {M, 0, 10}]]

These functions are simple Maclaurin series. Such Taylor 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, E is not an entire function of M at a given non-zero e. The derivative

${\displaystyle dM/dE=1-e\cos E}$

goes to zero at an infinite set of complex numbers when e<1. There are solutions at ${\displaystyle E=\pm i\cosh ^{-1}(1/e),}$  and at those values

${\displaystyle M=E-e\sin E=\pm i\left(\cosh ^{-1}(1/e)-{\sqrt {1-e^{2}}}\right)}$

(where inverse cosh is taken to be positive), and dE/dM goes to infinity at these points. This means that the radius of convergence of the Maclaurin series is ${\displaystyle \cosh ^{-1}(1/e)-{\sqrt {1-e^{2}}}}$  and the series will not converge for values of M larger than this. The series can also be used for the hyperbolic case, in which case the radius of convergence is ${\displaystyle \cos ^{-1}(1/e)-{\sqrt {e^{2}-1}}.}$  The series for when e = 1 converges when m < 2π.

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.

The solution for e ≠ 1 was found by Karl Stumpff in 1968,[7] but its significance wasn't recognized.[8][clarification needed]

One can also write a Maclaurin series in e. This series does not converge when e is larger than the Laplace limit (about 0.66), regardless of the value of M (unless M is a multiple of ), but it converges for all M if e is less than the Laplace limit. The coefficients in the series, other than the first (which is simply M), depend on M in a periodic way with period .

The inverse radial Kepler equation (e = 1) can also be written as:

${\displaystyle x(t)=\sum _{n=1}^{\infty }\left[\lim _{r\to 0^{+}}\left({\frac {t^{{\frac {2}{3}}n}}{n!}}{\frac {\mathrm {d} ^{\,n-1}}{\mathrm {d} r^{\,n-1}}}\!\left(r^{n}\left({\frac {3}{2}}{\Big (}\sin ^{-1}({\sqrt {r}})-{\sqrt {r-r^{2}}}{\Big )}\right)^{\!-{\frac {2}{3}}n}\right)\right)\right]}$

Evaluating this yields:

${\displaystyle x(t)=p-{\frac {1}{5}}p^{2}-{\frac {3}{175}}p^{3}-{\frac {23}{7875}}p^{4}-{\frac {1894}{3931875}}p^{5}-{\frac {3293}{21896875}}p^{6}-{\frac {2418092}{62077640625}}p^{7}-\ \cdots \ {\bigg |}{p=\left({\tfrac {3}{2}}t\right)^{2/3}}}$

To obtain this result using Mathematica:

InverseSeries[Series[ArcSin[Sqrt[t]] - Sqrt[(1 - t) t], {t, 0, 15}]]

## Numerical approximation of inverse problem

For most applications, the inverse problem can be computed numerically by finding the root of the function:

${\displaystyle f(E)=E-e\sin(E)-M(t)}$

This can be done iteratively via Newton's method:

${\displaystyle E_{n+1}=E_{n}-{\frac {f(E_{n})}{f'(E_{n})}}=E_{n}-{\frac {E_{n}-e\sin(E_{n})-M(t)}{1-e\cos(E_{n})}}}$

Note that E and M are in units of radians in this computation. This iteration is repeated until desired accuracy is obtained (e.g. when f(E) < desired accuracy). For most elliptical orbits an initial value of E0 = M(t) is sufficient. For orbits with e > 0.8, an initial value of E0 = π should be used. A similar approach can be used for the hyperbolic form of Kepler's equation.[9]:66–67 In the case of a parabolic trajectory, Barker's equation is used.

### Fixed-point iteration

A related method starts by noting that ${\displaystyle E=M+e\sin {E}}$ . Repeatedly substituting the expression on the right for the ${\displaystyle E}$  on the right yields a simple fixed-point iteration algorithm for evaluating ${\displaystyle E(e,M)}$ . This method is identical to Kepler's 1621 solution.[4]

function E(e,M,n)
E = M
for k = 1 to n
E = M + e*sin E
next k
return E


The number of iterations, ${\displaystyle n}$ , depends on the value of ${\displaystyle e}$ . The hyperbolic form similarly has ${\displaystyle H=e\sinh H-M}$ .

This method is related to the Newton's method solution above in that

${\displaystyle E_{n+1}=E_{n}-{\frac {E_{n}-e\sin(E_{n})-M(t)}{1-e\cos(E_{n})}}=E_{n}+{\frac {(M+e\sin {E_{n}}-E_{n})(1+e\cos {E_{n}})}{1-e^{2}(\cos {E_{n}})^{2}}}}$

To first order in the small quantities ${\displaystyle M-E_{n}}$  and ${\displaystyle e}$ ,

${\displaystyle E_{n+1}\approx M+e\sin {E_{n}}}$ .