# Kepler's equation

(Redirected from Kepler's Equation)

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, and in book V of his Epitome of Copernican Astronomy (1621) Kepler proposed an iterative solution to the equation. The equation has played an important role in the history of both physics and mathematics, particularly classical celestial mechanics.

## Equation

Kepler's equation is

$M=E-e\sin E$

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 $t=t_{0}$ , 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 $M=n(t-t_{0})$ , then solve the Kepler equation above to get $E$ , then get the coordinates from:

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

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 ($0\leq e<1$ ). 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:

$M=e\sinh H-H$

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:

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

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

### Radial Kepler equation

The Radial Kepler equation is:

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

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:

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

and then making the substitution

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

## 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. 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

Fourier series expansion (with respect to $M$ ) using Bessel functions is 

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

With respect to $e$ , it is a Kapteyn series.

### Inverse Kepler equation

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

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

Evaluating this yields:

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

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

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

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

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

(where inverse cosh is taken to be positive), and $\mathrm {d} E/\mathrm {d} M$  goes to infinity at these points. This means that the radius of convergence of the Maclaurin series is $\cosh ^{-1}(1/e)-{\sqrt {1-e^{2}}}$  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 $\cos ^{-1}(1/e)-{\sqrt {e^{2}-1}}.$  The series for when $e=1$  converges when $M<2\pi$ .

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 $e\neq 1$  was found by Karl Stumpff in 1968, but its significance wasn't recognized.[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 .

### Inverse radial Kepler equation

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

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

Evaluating this yields:

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

To obtain this result using Mathematica:

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

## Numerical approximation of inverse problem

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

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

This can be done iteratively via Newton's method:

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

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 $E_{0}=M(t)$  is sufficient. For orbits with $e>0.8$ , an initial value of $E_{0}=\pi$  should be used. If $e$  is identically 1, then the derivative of $f$ , 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. A similar approach can be used for the hyperbolic form of Kepler's equation.: 66–67  In the case of a parabolic trajectory, Barker's equation is used.

### Fixed-point iteration

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

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

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

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

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

$E_{n+1}\approx M+e\sin {E_{n}}$ .