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Trigonometry related to Kepler's Equation

  1. Jul 3, 2012 #1
    In deriving Kepler's Equation, there's a little trig work that gets at nu (see figure), which is the true anomaly. I'm curious how the third equation makes it into the 4th (last) equation. It's been a very long time since I've played with trig. nu of course is the atan of the last equation.
     

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  2. jcsd
  3. Jul 3, 2012 #2

    You should write down, with LaTeX, directly the equations here so that it'll be possible to check the notation and give, eventually, and answer, otherwise

    it is way too cumbersome.

    DonAntonio
     
  4. Jul 3, 2012 #3
    See <http://rip94550.wordpress.com/2011/05/02/elliptical-orbits-%E2%80%93-deriving-keplers-equation/>. [Broken] Attached is the part I was looking for.
     

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    Last edited by a moderator: May 6, 2017
  5. Jul 4, 2012 #4

    D H

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    You're using the wrong equation as the source. Use the first two. From the second, compute [itex]1+\cos\nu[/itex]. Next compute [itex]sin\nu/(1+\cos\nu)[/itex]. This is just [itex]\tan(\nu/2)[/itex] per the half angle formula. Use the same formula to simplify the right hand side.
     
  6. Jul 4, 2012 #5
    What source are you referring to? The link or one of the equations in the attachment? The link is 10 pages long.
     
    Last edited: Jul 4, 2012
  7. Jul 4, 2012 #6

    D H

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    I'm referring to the attachment in your first post.
     
  8. Jul 5, 2012 #7
    Ah, that was produced by a CM "pdf book" on the web. I'm satisfied with the longer attachment. I may contact the author of the pdf.
     
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