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Gravitation: Sphere is same as point mass?

  1. Jun 29, 2008 #1
    Newton's law of universal gravitation is only valid for point masses. Is it just chance that a person on the earth experiences the same force of gravity that he would if he were on a shell of no mass and the same radius of earth with a single point at the shell's center which had the mass of the earth?

    Does anyone know of a proof that these two situations are equivalent? I've been trying to do it using spherical coordinates and I get stumped at,
    [tex]g=2G \pi \delta \int _0^{\pi }\int _0^r\frac{\rho ^2\sin\phi \text{cos}\left\frac{\phi }{2}\right}{r^2+\rho ^2-2\rho r \text{cos}\phi }d\rho d\phi [/tex].
    Where [itex]\delta[/itex] is the average density and [itex]r[/itex] is the radius.

    If anyone is interested in how I got to this, I'd be happy to explain it. If the two listed situations are identical, then the double integral should be [itex]\frac{2}{3}[/itex].

    Thanks in advance!
     
  2. jcsd
  3. Jun 29, 2008 #2
  4. Jun 29, 2008 #3
    In that picture it is claimed that the distance between the test mass and every point on the shaded band dM is s. Either I'm not getting something, or that distance is different at every point along the band.
     
  5. Jun 29, 2008 #4

    D H

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    Your integral looks a bit strange. I'm not saying its wrong.

    If you do it right you will get the result that any body with a spherical mass distribution behaves gravitationally exactly like a point mass for any point outside the body. Newton proved this using the shell theorem. Gauss derived the result in a different way using Gauss' law.
     
  6. Jun 29, 2008 #5

    rbj

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    2nd what D H says. if you are outside the sphere and consider the gravitational field vector (as a function of position relative to the center of the sphere), there is no reason (because of symmetry) to think that the direction of the field vector would be different in the two cases. so the direction of the field is the same. and Gauss's Theorem takes care of the magnitude between the two cases (again, for outside the sphere).
     
  7. Jun 29, 2008 #6
    I uploaded the word document with all my work on it http://www.mediafire.com/?3nm93snmxgd, look at that to see how I got to that integral. Btw, when I plug it into mathematica it gives me,
    [tex]g=\frac{4}{15}G\delta\pi r(1+4\log4)[/tex].

    If someone could look through my paper to see what I did wrong, that'd be great. I hate when I do something I feel is valid but get a different answer than the accepted answer. It's only 2 pages and most of the first isn't necessary to read. Thanks!
     
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