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Kepler's First Law Mathematical Query

  1. Sep 29, 2010 #1
    Hello all! It's been a long summer, and I thought I'd warm up things by going over Kepler's Laws.

    I've been following the mathematical derivation in Introduction to Modern Astrophysics, and to be honest I am little stumped on a part of it:

    We have the derived definition of angular momentum as

    [tex]\vec{L} = \mu r^{2}\hat{r}\times\frac{d}{dt}\hat{r} [/tex]

    Then what happens is out of the blue, it seems, the author takes the cross product of the acceleration vector and angular momentum:

    [tex]\vec{a}\times\vec{L} = -\frac{GM}{r^{2}}\hat{r}\times \left(\mu r^{2}\hat{r}\times\frac{d}{dt}\hat{r} \right)[/tex]

    Ugh, to be honest I am stumped at the physical significance of that cross product. I can understand that the result is a [tex]\vec{v}\times\vec{L}[/tex] expression, but it just seems to have come out of nowhere and I don't know (physically) why we go through that process

    For reference it is on page 44 of the red paperback (second) edition. It is under Chapter 2: Celestial Mechanics.
  2. jcsd
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