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Hill Sphere Explanation

  1. Aug 30, 2010 #1
    The Hill sphere of a celestial object is the area in which it is the dominating force in the area.

    I was wondering how the formula was determined.

    If the mass of the smaller body is m, and it orbits a heavier body of mass M with a semi-major axis a and an eccentricity of e, then the radius r of the Hill sphere for the smaller body is, approximately:

    Equation: [tex]r \approx a(1-e)\sqrt[3]{m/3M}[/tex]

    I believe I understand how to determine eccentricity and the semi-major axis, I'm just searching for how the whole thing comes together.

    http://en.wikipedia.org/wiki/Hill_sphere" [Broken]

    The derivation on the wikipedia article didn't help much.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 2, 2010 #2
    The reason why it didn't help is that it's wrong. The derivation is for the Laplace Sphere, which is a related concept but distinct mathematically...

    as discussed here on Physforums before

    ...there's Astrodynamics lecture notes available online covering it, but I'll have to track them down.
    Last edited by a moderator: May 4, 2017
  4. Sep 2, 2010 #3
    I tracked down my old copies of the University of Wisconsin EMA 550 lectures and the Wikipedia entries. The equation for the Sphere of Influence aka the Hill sphere has been confused with the Lagrange Points L1 & L2. They're similar concepts in some features, so it's an understandable confusion. So it goes like so...

    [itex]R_{H} = a \left(\frac{M_{\mathrm{minor}}}{M_{\mathrm{major}}}\right)^{2/5}[/itex]

    ...and the L1 & L2 points are found (+/-) in line with the the minor body's semimajor axis around the major body via...

    [itex]R_L = a \left(\frac{M_{\mathrm{minor}}}{3M_{\mathrm{major}}}\right)^{1/3}[/itex]
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