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Intuitive explanation of gravity outside a sphere?

  1. Mar 7, 2004 #1
    Does anyone have an intuitive explanation why the gravitational force of a sphere on a distant point is the same as if all the sphere's mass were concentrated at the center? The usual integration over spherical shells goes through o.k. but the result is so simple it seems to me that maybe we could get there by insight instead of calculation...or at least part of the way there...
    Do we have the same result for a disk or a ring and a point in the same plane? Is the gravitational force of a ring on a point in the same plane equivalent to the force from the same mass concentrated at the center of the ring?
    I think this is true...so is there an intuitive way to get this result in the simpler case of a ring and a co-planar point, and does the spherical case then follow?
     
  2. jcsd
  3. Mar 7, 2004 #2

    HallsofIvy

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    Actually, that's only true if the density is uniform (unless by "distantpoint" the limit as distance goes to infinity). I'm not sure what you would consider "intuitive" but the simplest way to think about it is through symmetry. Take any point outside the uniform, spherical mass. Draw a line from the point to the center of the sphere (I'm going to call that the "center line"). For any point on the sphere, there is a "symmetrically opposite" point (equal distance on other side of the center line). The component of graviational force perpendicular to the center line of the two points are equal and opposite and so cancel. That is, the force is directed to the center of the sphere.

    The "very distant" (limit as distance goes to infinity) case is even more "intuitive": any figure, of any density, seen from far enough away looks like a point!
     
  4. Mar 7, 2004 #3

    Doc Al

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    I don't get your reasoning, Halls. Your argument shows that the force acts along the center line, but does not imply (as far as I can see) that the magnitude of the force is the same as if the mass were at the center.

    I can't think of a simple intuitive explanation. Of course, you can apply Gauss's theorem to gravity, but that requires explanation (for those unfamiliar with the Poisson equation for gravity).
     
  5. Mar 8, 2004 #4
    HallsofIvy's response misses the point of my question, but it brings up another aspect...Is the theorem true for any sperically symmetric distribution? The analogy with Gauss' theorem suggests that it is.
    But still no insight about why the force operates from the center. I'm not necessarily looking for a complete proof...just a little improvement over watching the result fall out of an integral.
     
  6. Mar 8, 2004 #5

    Doc Al

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    Yes.
     
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