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Radius of curvature of planetary motion

  1. Apr 16, 2014 #1
    Suppose we have a planet of mass m orbiting a larger one of mass M along an elliptical path. If we use polar coordinates with the origin placed on the planet of mass M (focus of the ellipse) then at the instant when the smaller planet is at the point of closest approach we have:

    [itex] \boldsymbol{v} = r \frac{d\theta}{dt} \boldsymbol{u_{\theta}} [/itex]

    [itex] \boldsymbol{a} = - r \: (\frac{d\theta}{dt})^{2} \: \boldsymbol{u_{r}} = - \frac{G M}{r^{2}} \boldsymbol{u_{r}} [/itex]

    From which we obtain:

    [itex] \boldsymbol{a} = - \frac{v^{2}}{r} \: \boldsymbol{u_{r}}= - \frac{G M}{r^{2}} \: \boldsymbol{u_{r}} [/itex]

    Now, if we calculate the curvature:

    [itex] k = \frac{\left\|\boldsymbol{a} \times \boldsymbol{v} \right\|}{v^{3}}= \frac{G M}{r^{2} v^{2}}[/itex]

    From this equation and the previous one it looks as if the radius of curvature at the point of closest approach is the distance of closest approach r itself, which is not true. I'd appreciate if you could clear this up.

    EDIT: nevermind, the second equation is wrong, [itex] \frac{d^{2}r}{dt^{2}} [/itex] is not zero.
     
    Last edited: Apr 16, 2014
  2. jcsd
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