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Two-Bodies Problem REAL SOLUTION

  1. Dec 26, 2011 #1
    hi everybody,

    I've recently got down to celestial mechanics and I can't remember how I used to derive the time dependent vector of motion r(t). Any (mathematical) help would be appreciated.

    PS: To avoid misundertandings, I don't need the classical r(θ) plane curve orbits!
     
  2. jcsd
  3. Dec 26, 2011 #2

    mathman

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  4. Dec 26, 2011 #3

    D H

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    There is no solution to r(t) in the elementary functions. You will need to find θ(t) (which also does not have a solution in the elementary functions) and then use that r(θ) that you don't want to use.

    The standard way to find θ(t) is to first determine the mean anomaly M(t). This is pretty easy; the mean anomaly is simply a linear function of time. Then solve for the eccentric anomaly by via Kepler's equation [itex]M(t) = E(t) - e\sin(E(t))[/itex]. Kepler's equation is a transcendental function. The inverse Kepler function does cannot be expressed in the elementary functions. Once you have E(t), solve for θ(t) via [itex]\tan \frac {\theta(t)} 2 = \sqrt{ \frac {1+e}{1-e} } \tan \frac {E(t)} 2[/itex].
     
  5. Dec 26, 2011 #4
    @mathman do you recognize a derivation of r(t) there? because I don't..

    @D H I'll put your suggestion to test as soon as possible! I have a question though, what's the point of referring to the solution as not compiled with elementary functions?
     
  6. Dec 26, 2011 #5

    D H

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    It means that there isn't a nice closed-form solution.
     
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