Kepler's equation

File:Kepler_equation_solutions.svg

Kepler's equation is

{{Equation box 1

|indent =:

|equation = $M\; =\; E\; -\; e\; \backslash sin\; E$

|border colour = #50C878

|background colour = #ECFCF4}}

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:

{{Equation box 1

|indent =:

|equation = $\backslash begin\{array\}\{lcl\}$

x & = & a (\cos E - e) \\

y & = & b \sin E

\end{array}

|border colour = #50C878

|background colour = #ECFCF4}}

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, and it cannot be solved for $E$ algebraically. Numerical analysis and series expansions are generally required to evaluate $E$.

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 (). 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 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 $e$ 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 $e$ goes to infinity, the orbit becomes a straight line of infinite length.

The Hyperbolic Kepler equation is:

{{Equation box 1

|indent =:

|equation = $M\; =\; e\; \backslash sinh\; H\; -\; H$

|border colour|background colour }}

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:

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

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

The Radial Kepler equation for the case where the object does not have enough energy to escape is:

{{Equation box 1

|indent =:

|equation = $t(x)\; =\backslash pm\backslash biggr[\; \backslash sin^\{-1\}(\; \backslash sqrt\{\; x\; \}\; )\; -\; \backslash sqrt\{\; x\; (\; 1\; -\; x\; )\; \}\; \backslash biggr]$

|border colour|background colour}}

where $t$ is proportional to time and $x$ 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 $e$ to 1:

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

and then making the substitution

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

The radial equation for when the object has enough energy to escape is:

{{Equation box 1

|indent =:

|equation = $t(x)\; =\; \backslash pm\; \backslash biggr[\backslash sinh^\{-1\}(\; \backslash sqrt\{\; x\; \}\; )\; -\; \backslash sqrt\{\; x\; (\; 1\; +\; x\; )\; \}\; \backslash biggr]$

|border colour|background colour}}

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 $M$ 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. Solving for $E$ is more or less equivalent to solving for the true anomaly, or the difference between the true anomaly and the mean anomaly, which is called the "Equation of the center".

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.It is often claimed that Kepler's equation "cannot be solved analytically"; see for example [http://www.jgiesen.de/kepler/kepler.html here]. 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, {{ISBN|0-12-615760-X}}, [https://books.google.com/books?id=pEoti_sUxtoC&pg=PA659 p. 659] Kepler himself expressed doubt at the possibility of finding a general solution:

{{blockquote|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[https://www.e-rara.ch/zut/content/pageview/162862 "Mihi ſufficit credere, ſolvi a priori non poſſe, propter arcus & ſinus ετερογενειαν. Erranti mihi, quicumque viam monſtraverit, is erit mihi magnus Apollonius."] {{Cite journal |title=Kepler's Problem |first=Asaph |last=Hall |author-link=Asaph Hall |url=https://archive.org/details/jstor-2635832 |date=May 1883 |journal=Annals of Mathematics |volume=10 |issue=3 |pages=65–66 |doi=10.2307/2635832|jstor=2635832 }}}}

Fourier series expansion (with respect to $M$) using Bessel functions is {{cite book|last1=Fitzpatrick|first1=Philip Matthew|year=1970|title=Principles of celestial mechanics|isbn=0-12-257950-X|publisher=Academic Press}}{{cite journal | last1 = Colwell | first1 = Peter | title = Bessel Functions and Kepler's Equation | journal = The American Mathematical Monthly | date = January 1992 | volume = 99 | issue = 1 | pages = 45–48 | issn = 0002-9890 | doi = 10.2307/2324547| jstor = 2324547 }}

{{cite journal |first1=John P. |last1=Boyd

|title=Rootfinding for a transcendental equation without a first guess: Polynomialization of Kepler's equation through Chebyshev polynomial equation of the sine

|doi=10.1016/j.apnum.2005.11.010 |year=2007 |volume=57 |number=1

|journal=Applied Numerical Mathematics |pages=12–18 }}

:$$

E = M + \sum_{m=1}^\infty \frac{2}{m} J_m(me) \sin(mM), \quad e\le 1, \quad M \in [-\pi,\pi].

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

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 \ne 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 = ( 6 M )^{1/3}

, & e = 1\\

\\

\displaystyle

\frac{1}{1-e} M

- \frac{e}{(1-e)^4 } \frac{M^3}{3!}

+ \frac{(9 e^2 + e)}{(1-e)^7 } \frac{M^5}{5!}

- \frac{(225 e^3 + 54 e^2 + e) }{(1-e)^{10} } \frac{M^7}{7!}

+ \frac{ (11025 e^4 + 4131 e^3 + 243 e^2 + e) }{(1-e)^{13} } \frac{M^9}{9!}+ \cdots

, & e \ne 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 $e$. Indeed, the derivative

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

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

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

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

While this solution is the simplest in a certain mathematical sense,{{which|date=August 2017}}, other solutions are preferable for most applications. Alternatively, Kepler's equation can be solved numerically.

The solution for $e\; \backslash ne\; 1$ was found by Karl Stumpff in 1968,{{Cite journal |last=Stumpff |first=Karl |date=1 June 1968 |title=On The application of Lie-series to the problems of celestial mechanics |url=https://ntrs.nasa.gov/search.jsp?R=19680017171 |id=NASA Technical Note D-4460}} but its significance wasn't recognized.{{Cite book |last=Colwell |first=Peter |year=1993 |title=Solving Kepler's Equation Over Three Centuries |isbn=0-943396-40-9 |publisher=Willmann–Bell |page=43}}{{clarify|date=August 2017}}

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 {{math|2π}}), 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 {{math|2π}}.

The inverse radial Kepler equation ($e\; =\; 1$) for the case in which the object does not have enough energy to escape can similarly 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:

:$x(t)\; =\; p\; -\; \backslash frac\{1\}\{5\}\; p^2\; -\; \backslash frac\{3\}\{175\}p^3$

- \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}]]

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 $e\; >\; 0.8$, a initial value of $E\_0\; =\; \backslash pi$ can be used. Numerous works developed accurate (but also more complex) start guesses.{{cite journal | last1=Odell | first1=A. W. | last2=Gooding | first2=R. H. | title=Procedures for solving Kepler's equation | journal=Celestial Mechanics | publisher=Springer Science and Business Media LLC | volume=38 | issue=4 | year=1986 | issn=1572-9478 | doi=10.1007/bf01238923 | pages=307–334 | bibcode=1986CeMec..38..307O| s2cid=120179781 }} 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.{{cite web |last1=Keister |first1=Adrian |title=The Numerical Analysis of Finding the Height of a Circular Segment |url=https://www.winemantech.com/blog/the-numerical-analysis-of-finding-the-height-of-a-circular-segment |website=Wineman Technology |publisher=Wineman Technology, Inc. |access-date=28 December 2019}} A similar approach can be used for the hyperbolic form of Kepler's equation.{{cite book |last1=Pfleger |first1=Thomas |first2=Oliver |last2=Montenbruck |title=Astronomy on the Personal Computer |date=1998 |publisher=Springer |location=Berlin, Heidelberg |isbn=978-3-662-03349-4 |edition= Third |url=https://archive.org/details/springer_10.1007-978-3-662-03349-4 }}{{rp|66–67}} In the case of a parabolic trajectory, Barker's equation is used.

A related method starts by noting that $E\; =\; M\; +\; e\; \backslash sin\{E\}$. 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. In pseudocode:

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 $H\; =\; \backslash sinh^\{-1\}\backslash left(\backslash frac\{H+M\}\{e\}\backslash right)$.

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 $e$,

:$E\_\{n+1\}\; \backslash approx\; M\; +\; e\; \backslash sin\{E\_\{n\}\}$.

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

{{Reflist}}

{{commons category|Kepler's equation}}

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• [http://mathworld.wolfram.com/KeplersEquation.html Kepler's Equation at Wolfram Mathworld]

{{Johannes Kepler}}

Category:Eponymous equations of physics

Category:Johannes Kepler

Category:Orbits