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Orbits: explicit r(t) and theta(t)

  1. Oct 6, 2007 #1

    nrqed

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    I am sure this is a stupid question but can someone give me the equations for r(t) and theta(t) for an orbit an inverse square law (treating the central mass as being at rest). The books always seem to only give r(theta) i.e. the shape of the orbit but not the actual time evolution. Is it ever done explicitly?

    thanks
     
  2. jcsd
  3. Oct 6, 2007 #2

    D H

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    A more proper treatment is treating the center of mass as being at rest. If the orbiting body is of negligible mass, the central mass will essentially be at rest.

    You are most likely talking about the equation

    [tex]r=\frac{a(1-e^2)}{1+e\cos\theta}[/tex]

    where [itex]\theta[/itex] is the true anomaly. The time evolution in terms of [itex]\theta[/itex] does not have an explicit form.

    An explicit form for the time evolution of the mean anomaly does exist, and it is quite simple in form:

    [tex]M(t)=M(t_0) + (t-t_0)\dot M[/tex]

    The mean anomaly is related to the eccentric anomaly [itex]\psi[/itex] via Kepler's equation:

    [tex]M(t) = \psi(t) - e\sin\psi(t)[/tex]

    The eccentric anomaly is in turn related to the true anomaly by

    [tex]\sqrt{1-e}\tan\frac{\theta}2 = \sqrt{1+e}\tan \frac{\psi(t)}2[/tex]
     
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