Kepler's equation is

where ${\displaystyle M}$  is the mean anomaly, ${\displaystyle E}$  is the eccentric anomaly, and ${\displaystyle e}$  is the eccentricity.

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

where ${\displaystyle a}$  is the semi-major axis, ${\displaystyle b}$  the semi-minor axis.

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

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 (${\displaystyle 0\leq e<1}$ ). The hyperbolic Kepler equation is used for hyperbolic trajectories (${\displaystyle e>1}$ ). The radial Kepler equation is used for linear (radial) trajectories (${\displaystyle e=1}$ ). Barker's equation is used for parabolic trajectories (${\displaystyle e=1}$ ).

When ${\displaystyle e=0}$ , the orbit is circular. Increasing ${\displaystyle e}$  causes the circle to become elliptical. When ${\displaystyle e=1}$ , 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 ${\displaystyle e}$  above 1 results in a hyperbolic orbit with a turning angle of just under 180 degrees. Further increases reduce the turning angle, and as ${\displaystyle e}$  goes to infinity, the orbit becomes a straight line of infinite length.

Hyperbolic Kepler equation edit

The Hyperbolic Kepler equation is:

where ${\displaystyle H}$  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 ${\displaystyle E}$  is now imaginary) and then replacing ${\displaystyle E}$  by ${\displaystyle iH}$ .

Radial Kepler equation edit

The Radial Kepler equation is:

where ${\displaystyle t}$  is proportional to time and ${\displaystyle x}$  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 ${\displaystyle e}$  to 1:

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

and then making the substitution

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

Calculating ${\displaystyle M}$  for a given value of ${\displaystyle E}$  is straightforward. However, solving for ${\displaystyle E}$  when ${\displaystyle M}$  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 ${\displaystyle e}$  and ${\displaystyle M}$  (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 great Apollonius.

— Johannes Kepler^[10]

Fourier series expansion (with respect to ${\displaystyle M}$ ) using Bessel functions is ^{[11][12] [13]}

${\displaystyle E=M+\sum _{m=1}^{\infty }{\frac {2}{m}}J_{m}(me)\sin(mM),\quad e\leq 1,\quad M\in [-\pi ,\pi ].}$ 

With respect to ${\displaystyle e}$ , it is a Kapteyn series.

Inverse Kepler equation edit

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, ${\displaystyle E}$  is not an entire function of ${\displaystyle M}$  at a given non-zero ${\displaystyle e}$ . Indeed, the derivative

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

goes to zero at an infinite set of complex numbers when ${\displaystyle e<1,}$  the nearest to zero being at ${\displaystyle E=\pm i\cosh ^{-1}(1/e),}$  and at these two points

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

(where inverse cosh is taken to be positive), and ${\displaystyle \mathrm {d} E/\mathrm {d} M}$  goes to infinity at these values of ${\displaystyle M}$ . 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 ${\displaystyle M}$  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 ${\displaystyle e=1}$  converges when ${\displaystyle M<2\pi }$ .

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 ${\displaystyle e\neq 1}$  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 ${\displaystyle e}$ . This series does not converge when ${\displaystyle e}$  is larger than the Laplace limit (about 0.66), regardless of the value of ${\displaystyle M}$  (unless ${\displaystyle M}$  is a multiple of 2π), but it converges for all ${\displaystyle M}$  if ${\displaystyle e}$  is less than the Laplace limit. The coefficients in the series, other than the first (which is simply ${\displaystyle M}$ ), depend on ${\displaystyle M}$  in a periodic way with period 2π.

Inverse radial Kepler equation edit

The inverse radial Kepler equation (${\displaystyle e=1}$ ) 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:

${\displaystyle x(t)=p-{\frac {1}{5}}p^{2}-{\frac {3}{175}}p^{3}-{\frac {23}{7875}}p^{4}-{\frac {1894}{3031875}}p^{5}-{\frac {3293}{21896875}}p^{6}-{\frac {2418092}{62077640625}}p^{7}-\ \cdots \ {\bigg |}{p=\left({\tfrac {3}{2}}t\right)^{2/3}}}$ 

To obtain this result using Mathematica:

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

Newton's method edit

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 ${\displaystyle E}$  and ${\displaystyle M}$  are in units of radians in this computation. This iteration is repeated until desired accuracy is obtained (e.g. when ${\displaystyle f(E)}$  < desired accuracy). For most elliptical orbits an initial value of ${\displaystyle E_{0}=M(t)}$  is sufficient. For orbits with ${\displaystyle e>0.8}$ , a initial value of ${\displaystyle E_{0}=\pi }$  can be used. Numerous works developed accurate (but also more complex) start guesses.^[16] If ${\displaystyle e}$  is identically 1, then the derivative of ${\displaystyle f}$ , 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.^{[18]: 66–67} In the case of a parabolic trajectory, Barker's equation is used.

Fixed-point iteration edit

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 ${\displaystyle e}$ ,

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

• Kepler's laws of planetary motion
• Kepler problem
• Kepler problem in general relativity
• Radial trajectory

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• Kepler's Equation at Wolfram Mathworld