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MathGroup Archive 2001

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Re: kepler equation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31241] Re: kepler equation
  • From: "Hermann Meier" <hermann.ernst.meier at swissonline.ch>
  • Date: Tue, 23 Oct 2001 04:53:19 -0400 (EDT)
  • References: <9qrd3f$c17$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Kepler's equation: mean anomaly (input) and eccentric anomaly (output) are
both in radians:

keplerEquation[meanAnomaly_,eccentricity_] :=
FixedPoint[meanAnomaly+eccentricity*Sin[#]&,meanAnomaly]

The following solution may also be used (for moderate values of eccentricity
only; quite slow) (Taff, Celestial Mechanics, p. 34)

kep[M_,e_] :=
   (M+2*NSum[
   1/m*(((e*m)^m*BesselJ[m, Sqrt[e^2*m^2]])/(e^2*m^2)^(-(-m/2)))*
   Sin[m*M],
 {m,1,Infinity}, Method -> Fit])

M = 199.7855*Degree; e = 0.047837;

keplerEquation[M,e] / Degree // InputForm or kep[M,e]/Degree // InputForm
both give 198.89778838060755

Hermann Meier
hermann.ernst.meier at swisspnline.ch



"ammar" <ammarsak at emirates.net.ae> wrote in message
news:9qrd3f$c17$1 at smc.vnet.net...
>
> Dear sir
> would you send to me some source files of Kepler equation based on
mathematica
> or any related  topics
> thanks
> Ammar Jamil Sakaji
> P.O.Box 45247
> Abu Dhabi - UAE
>
>



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