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Doubt about the polar equation of a Kepler orbit

  1. Jul 23, 2010 #1
    Good morning,

    I have a doubt about the differentiation of the polar equation of an orbit:
    [tex]r=\frac{p}{1+e\cos\nu}[/tex]
    It represents the relative position of a planet with respect to the central body.
    Here, p is the parameter, e is the eccentricity and r is the radius of the planet measured from the focus (where the central body is located).
    If I differentiate it, I obtain the following result:
    [tex]\dot{r}=\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{p}{1+e\cos\nu} \right )[/tex]
    [tex]p \left(\frac{e\dot{\nu}\sin \nu}{(1+e\cos\nu)^2}\right) = \frac{h^2}{\mu} \left(\frac{e\dot{\nu}\sin \nu}{(1+e\cos\nu)^2}\right)[/tex]
    However, according to the text I'm reading, I should be getting this:
    [tex]\dot{r}=\sqrt{\frac{\mu}{p}}e\sin\nu[/tex]
    I'm not sure on how I could get this result.
    Any ideas?

    Thank you in advance.
     
  2. jcsd
  3. Jul 24, 2010 #2

    Filip Larsen

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    Gold Member

    From Kepler's second law you have the relationship [itex]r^2 \dot{\nu} = h[/itex] which you can insert into your equation to get rid of [itex]\dot{\nu}[/itex].
     
  4. Jul 24, 2010 #3
    OK, I tried it and it led me to the correct result.
     
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