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Kepler's equation

From Wikipedia, the free encyclopedia
Orbital mechanics term
For specific applications of Kepler's equation, seeKepler's laws of planetary motion.
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Inorbital mechanics,Kepler's equation relates various geometric properties of the orbit of a body subject to acentral force.

It was derived byJohannes Kepler in 1609 in Chapter 60 of hisAstronomia nova,[1][2] and in book V of hisEpitome of Copernican Astronomy (1621) Kepler proposed an iterative solution to the equation.[3][4] This equation and its solution, however, first appeared in a 9th-century work byHabash al-Hasib al-Marwazi, which dealt with problems of parallax.[5][6][7][8] The equation has played an important role in the history of both physics and mathematics, particularly classicalcelestial mechanics.

Equation

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

Kepler's equation is

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

whereM{\displaystyle M} is themean anomaly,E{\displaystyle E} is theeccentric anomaly, ande{\displaystyle e} is theeccentricity.

The 'eccentric anomaly'E{\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 coordinatesx=a(1e){\displaystyle x=a(1-e)},y=0{\displaystyle y=0}, at timet=t0{\displaystyle t=t_{0}}, then to find out the position of the body at any time, you first calculate the mean anomalyM{\displaystyle M} from the time and themean motionn{\displaystyle n} by the formulaM=n(tt0){\displaystyle M=n(t-t_{0})}, then solve the Kepler equation above to getE{\displaystyle E}, then get the coordinates relative to the central gravitational body from:

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

wherea{\displaystyle a} is thesemi-major axis,b{\displaystyle b} thesemi-minor axis.

Kepler's equation is atranscendental equation becausesine is atranscendental function, and it cannot be solved forE{\displaystyle E}algebraically.Numerical analysis andseries expansions are generally required to evaluateE{\displaystyle E}.

Alternate forms

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

Whene=0{\displaystyle e=0}, the orbit is circular. Increasinge{\displaystyle e} causes the circle to become elliptical. Whene=1{\displaystyle 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 ofe{\displaystyle 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 ase{\displaystyle e} goes to infinity, the orbit becomes a straight line of infinite length.

Hyperbolic Kepler equation

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The Hyperbolic Kepler equation is:

M=esinh(H)H{\displaystyle M=e\sinh(H)-H}

whereH{\displaystyle H} 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:

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

(in whichE{\displaystyle E} is now imaginary) and then replacingE{\displaystyle E} byiH{\displaystyle iH}.

Radial Kepler equations

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

t(x)=±[sin1(x)x(1x)]{\displaystyle t(x)=\pm {\biggr [}\sin ^{-1}({\sqrt {x}})-{\sqrt {x(1-x)}}{\biggr ]}}

wheret{\displaystyle t} is proportional to time andx{\displaystyle 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 settinge{\displaystyle e} to 1:

t(x)=12[EsinE].{\displaystyle t(x)={\frac {1}{2}}\left[E-\sin E\right].}

and then making the substitution

E=2sin1(x).{\displaystyle E=2\sin ^{-1}({\sqrt {x}}).}

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

t(x)=±[sinh1(x)x(1+x)]{\displaystyle t(x)=\pm {\biggr [}\sinh ^{-1}({\sqrt {x}})-{\sqrt {x(1+x)}}{\biggr ]}}

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.

Inverse problem

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CalculatingM{\displaystyle M} for a given value ofE{\displaystyle E} is straightforward. However, solving forE{\displaystyle E} whenM{\displaystyle M} is given can be considerably more challenging. There is noclosed-form solution. Solving forE{\displaystyle E} 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 ofe{\displaystyle e} andM{\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 greatApollonius.

— Johannes Kepler[10]

Fourier series expansion (with respect toM{\displaystyle M}) usingBessel functions is[11][12][13]

E=M+m=12mJm(me)sin(mM),e1,M[π,π].{\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 toe{\displaystyle e}, it is aKapteyn series.

Inverse Kepler equation

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

E={n=1Mn3n!limθ0+(dn1dθn1((θθsin(θ)3)n)),e=1n=1Mnn!limθ0+(dn1dθn1((θθesin(θ))n)),e1{\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:

E={s+160s3+11400s5+125200s7+4317248000s9+12137207200000s11+15143912713500800000s13+ with s=(6M)1/3,e=111eMe(1e)4M33!+(9e2+e)(1e)7M55!(225e3+54e2+e)(1e)10M77!+(11025e4+4131e3+243e2+e)(1e)13M99!+,e1{\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 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,E{\displaystyle E} is not anentire function ofM{\displaystyle M} at a given non-zeroe{\displaystyle e}. Indeed, the derivative

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

goes to zero at aninfinite set of complex numbers whene<1,{\displaystyle e<1,} the nearest to zero being atE=±icosh1(1/e),{\displaystyle E=\pm i\cosh ^{-1}(1/e),} and at these two points

M=EesinE=±i(cosh1(1/e)1e2){\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), anddE/dM{\displaystyle \mathrm {d} E/\mathrm {d} M} goes to infinity at these values ofM{\displaystyle M}. This means that the radius of convergence of the Maclaurin series iscosh1(1/e)1e2{\displaystyle \cosh ^{-1}(1/e)-{\sqrt {1-e^{2}}}} and the series will not converge for values ofM{\displaystyle M} larger than this. The series can also be used for the hyperbolic case, in which case the radius of convergence iscos1(1/e)e21.{\displaystyle \cos ^{-1}(1/e)-{\sqrt {e^{2}-1}}.} The series for whene=1{\displaystyle e=1} converges whenM<2π{\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 fore1{\displaystyle e\neq 1} was found byKarl Stumpff in 1968,[14] but its significance wasn't recognized.[15][clarification needed]

One can also write a Maclaurin series ine{\displaystyle e}. This series does not converge whene{\displaystyle e} is larger than theLaplace limit (about 0.66), regardless of the value ofM{\displaystyle M} (unlessM{\displaystyle M} is a multiple of), but it converges for allM{\displaystyle M} ife{\displaystyle e} is less than the Laplace limit. The coefficients in the series, other than the first (which is simplyM{\displaystyle M}), depend onM{\displaystyle M} in a periodic way with period.

Inverse radial Kepler equation

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The inverse radial Kepler equation (e=1{\displaystyle e=1}) for the case in which the object does not have enough energy to escape can similarly be written as:

x(t)=n=1[limr0+(t23nn!dn1drn1(rn(32(sin1(r)rr2))23n))]{\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:

x(t)=p15p23175p3237875p418943031875p5329321896875p6241809262077640625p7  |p=(32t)2/3{\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 usingMathematica:

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

Numerical approximation of inverse problem

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Newton's method

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

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

This can be done iteratively viaNewton's method:

En+1=Enf(En)f(En)=EnEnesin(En)M(t)1ecos(En){\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 thatE{\displaystyle E} andM{\displaystyle M} are in units of radians in this computation. This iteration is repeated until desired accuracy is obtained (e.g. whenf(E){\displaystyle f(E)} < desired accuracy). For most elliptical orbits an initial value ofE0=M(t){\displaystyle E_{0}=M(t)} is sufficient. For orbits withe>0.8{\displaystyle e>0.8}, a initial value ofE0=π{\displaystyle E_{0}=\pi } can be used. Numerous works developed accurate (but also more complex) start guesses.[16] Ife{\displaystyle e} is identically 1, then the derivative off{\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, 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.

Fixed-point iteration

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A related method starts by noting thatE=M+esinE{\displaystyle E=M+e\sin {E}}. Repeatedly substituting the expression on the right for theE{\displaystyle E} on the right yields a simplefixed-point iteration algorithm for evaluatingE(e,M){\displaystyle E(e,M)}. This method is identical to Kepler's 1621 solution.[4] Inpseudocode:

functionE(e,M,n)E=Mfork=1tonE=M+e*sinEnextkreturnE

The number of iterations,n{\displaystyle n}, depends on the value ofe{\displaystyle e}. The hyperbolic form similarly hasH=sinh1(H+Me){\displaystyle H=\sinh ^{-1}\left({\frac {H+M}{e}}\right)}.

This method is related to theNewton's method solution above in that

En+1=EnEnesin(En)M(t)1ecos(En)=En+(M+esinEnEn)(1+ecosEn)1e2(cosEn)2{\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 quantitiesMEn{\displaystyle M-E_{n}} ande{\displaystyle e},

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

See also

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References

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  1. ^Kepler, Johannes (1609)."LX. Methodus, ex hac Physica, hoc est genuina & verissima hypothesi, extruendi utramque partem æquationis, & distantias genuinas: quorum utrumque simul per vicariam fieri hactenus non potuit. argumentum falsæ hypotheseos".Astronomia Nova Aitiologētos, Seu Physica Coelestis, tradita commentariis De Motibus Stellæ Martis, Ex observationibus G. V. Tychonis Brahe (in Latin). pp. 299–300.
  2. ^Aaboe, Asger (2001).Episodes from the Early History of Astronomy. Springer. pp. 146–147.ISBN 978-0-387-95136-2.
  3. ^Kepler, Johannes (1621). "Libri V. Pars altera.".Epitome astronomiæ Copernicanæ usitatâ formâ Quæstionum & Responsionum conscripta, inq; VII. Libros digesta, quorum tres hi priores sunt de Doctrina Sphæricâ (in Latin). pp. 695–696.
  4. ^abSwerdlow, Noel M. (2000)."Kepler's Iterative Solution to Kepler's Equation".Journal for the History of Astronomy.31 (4):339–341.Bibcode:2000JHA....31..339S.doi:10.1177/002182860003100404.S2CID 116599258.
  5. ^Colwell, Peter (1993).Solving Kepler's Equation Over Three Centuries. Willmann-Bell. p. 4.ISBN 978-0-943396-40-8.
  6. ^Dutka, J. (1997-07-01)."A note on "Kepler's equation"".Archive for History of Exact Sciences.51 (1):59–65.Bibcode:1997AHES...51...59D.doi:10.1007/BF00376451.S2CID 122568981.
  7. ^North, John (2008-07-15).Cosmos: An Illustrated History of Astronomy and Cosmology. University of Chicago Press.ISBN 978-0-226-59441-5.
  8. ^Livingston, John W. (2017-12-14).The Rise of Science in Islam and the West: From Shared Heritage to Parting of The Ways, 8th to 19th Centuries. Routledge.ISBN 978-1-351-58926-0.
  9. ^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,ISBN 0-12-615760-X,p. 659
  10. ^"Mihi ſufficit credere, ſolvi a priori non poſſe, propter arcus & ſinus ετερογενειαν. Erranti mihi, quicumque viam monſtraverit, is erit mihi magnus Apollonius."Hall, Asaph (May 1883)."Kepler's Problem".Annals of Mathematics.10 (3):65–66.doi:10.2307/2635832.JSTOR 2635832.
  11. ^Fitzpatrick, Philip Matthew (1970).Principles of celestial mechanics. Academic Press.ISBN 0-12-257950-X.
  12. ^Colwell, Peter (January 1992). "Bessel Functions and Kepler's Equation".The American Mathematical Monthly.99 (1):45–48.doi:10.2307/2324547.ISSN 0002-9890.JSTOR 2324547.
  13. ^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.
  14. ^Stumpff, Karl (1 June 1968)."On The application of Lie-series to the problems of celestial mechanics". NASA Technical Note D-4460.{{cite journal}}:Cite journal requires|journal= (help)
  15. ^Colwell, Peter (1993).Solving Kepler's Equation Over Three Centuries. Willmann–Bell. p. 43.ISBN 0-943396-40-9.
  16. ^Odell, A. W.; Gooding, R. H. (1986). "Procedures for solving Kepler's equation".Celestial Mechanics.38 (4). Springer Science and Business Media LLC:307–334.Bibcode:1986CeMec..38..307O.doi:10.1007/bf01238923.ISSN 1572-9478.S2CID 120179781.
  17. ^Keister, Adrian."The Numerical Analysis of Finding the Height of a Circular Segment".Wineman Technology. Wineman Technology, Inc. Retrieved28 December 2019.
  18. ^Pfleger, Thomas; Montenbruck, Oliver (1998).Astronomy on the Personal Computer (Third ed.). Berlin, Heidelberg: Springer.ISBN 978-3-662-03349-4.

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