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Gauss' Law Conclusion in Gravitational Fields

  1. Mar 9, 2012 #1
    I've been told that electric and gravitational fields have a lot in common, and both are practically analogous to each other. Also, the conclusions made through Gauss' Law apply just as well (analogously) to gravitational fields.

    One of Gauss' Law's predictions is for a solid (conducting) charged body with a a cavity in it. According to it, no charge can exist inside the body...the only charge on it must exist on it's outer surface; that means no charge can exist on the inner surface. We get to that conclusion if we consider a Gaussian surface inside the body's material (not in the cavity).

    But, if we introduce a point charge in the cavity, a charge develops on the inner surface, so that if we take a Gaussian surface in the body's material again (not in the cavity), there's no net electric field inside the body, and therefore, no net flux through the surface (and therefore, no net charge within the surface).

    But, if we consider this for a gravitational field; we take a body with a cavity within it. For simplicity, I considered a spherical shell. I know that there's no gravitational field inside a shell. So, when I place a point mass inside the shell, why do we not observe any redistribution of the mass of the shell on the surface of the shell, so as to cancel out the gravitational field of the point mass (as we saw in the case of the point charge introduced in the cavity of the charged body). Or is there a tendency for a redistribution to take place.....but it is not observable (or not possible, for that matter), because we consider the body to be rigid??
  2. jcsd
  3. Mar 9, 2012 #2

    Doc Al

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    Staff: Mentor

    That's only true because the electrostatic field within a conductor must be zero. That goes beyond Gauss' law and has no parallel with gravity. (Gauss' law itself works just fine for both.)
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