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Newton's Law of Gravitation

  1. Feb 2, 2008 #1
    So I'm reading "Mathematical Physics" by Donald H.

    Menzel, and I don't buy the following derivation from

    section 2.12

    The purpose of the derivation is derive the potential

    energy at a point Po which is a distance Ro from the

    center of a sphere of uniform density.

    First they derive the amount an infinitesimal amount of

    mass P(dv) located on the shell of the sphere contributes

    to the potential energy at a point Po.

    [tex]dV=-\frac{G \rho}{R}dxdydz[/tex]

    The following expression is given to compute the

    potential energy contribution of the infinitesimal piece

    of mas dV at P(dV) on a point Po.

    [tex]V=-G\rho \int_r^{r+dr}\int_{\theata

    =0}^{\pi}\int_{\phi=0}^{2\pi}\frac{r^2sin^2( \theta ) dr

    d \theata d \phi }{R}[/tex]
    r is the distance from the center of the sphere to P(dV)
    Ro is the distance from P to the center of the sphere
    R is the distance from P(dV) to Po

    A change in variables is derived by doing implicit

    differentiation with r contant on the law of cosines:

    [tex]R^2=R_o^2+r^2-2R_o r \ cos( \theta )[/tex]

    which gives:

    [tex]R \ dR = R_o r sin( \theta ) d \theta [/tex]

    So far I agree but then they say that this implies:

    [tex]V=-G\rho \int_r^{r+dr}\int_{R_o-r}^{R_o+r}\int_{0}^{2\pi}\frac{r}{R_o}drdR d \phi[/tex]

    However, if I do the above subsitution I get an extra factor of [tex]sin( \theta )[/tex] left over.
  2. jcsd
  3. Feb 2, 2008 #2


    User Avatar
    Homework Helper

    in the integral, i think it should just be
    [tex]V= -G\rho \int \frac{1}{R} r^2 \sin\theta \, dr d\theta d\phi[/tex]
    from change of variables
  4. Feb 2, 2008 #3
    That's what I started thinking last night after I posted. It should be easy enough to check.
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