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Gravity of a Flat Disc

  1. Jun 3, 2006 #1
    Hey guys, so I'm back in school after an 8-month break, and I'm feeling a bit rusty :rolleyes:

    So I've got a flat disc of radius A, with constant density p, in the z=0 plane. I want to calculate the gravitational field at any point up or down the z-axis.

    I integrated the potential over the disc got the correct potential function (which is given) of :

    [tex]\Phi(z)=-G\rho2\pi(\sqrt{a^2+z^2}-z)[/tex]
    So now I just take the negative derivative of this to get the Gravitational field, right?
    [tex]G(z)=-\nabla\Phi(z)[/tex]
    [tex] G(z)=-G\rho2\pi(\frac{z}{\sqrt{a^2+z^2}}-1)[/tex]

    But this means there is a force at z=0 when I think there should not be... :confused:

    [tex]How do I make a new line in Latex? \\ this doesn't seem to work? :@[/tex]
     
    Last edited: Jun 3, 2006
  2. jcsd
  3. Jun 3, 2006 #2

    Andrew Mason

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    What is the thickness of the disc? Where does that appear in your calculation?

    The force/unit mass at (0,0,z) from a ring element of the disc of radius r thickness h and width dr would be:

    [tex]dF = Gdm/s^2 = \frac{G\rho 2\pi r hdr}{r^2 + z^2}[/tex]

    assuming h to be small compared to z. Integrate that from r = 0 to r = A.

    AM
     
    Last edited: Jun 3, 2006
  4. Jun 3, 2006 #3

    arildno

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    The potential is not correct, since it is not symmetric about the disk.
    The proper potential is:
    [tex]\Phi(z)=-G2\pi\rho(\sqrt{a^{2}+z^{2}}-|z|)[/tex]
    yielding the proper force per unit mass along the z-axis (in the positive vertical direction) :
    [tex]f(z)=2\pi\rho{G}(\frac{z}{\sqrt{a^{2}+z^{2}}}-\frac{z}{|z|})[/tex]
    The limiting values as z goes to zero,
    [tex]\lim_{z\to{0}^{+}}f(z)=-2\pi\rho{G}, \lim_{z\to{0}^{-}}f(z)=2\pi\rho{G}[/tex]
    are the strengths of the force just outside the disk, on either side.
    There is a leap of discontinuity across the disk, where AT the origin, the force is, indeed 0.
     
    Last edited: Jun 3, 2006
  5. Jun 3, 2006 #4
    [tex]\frac{d|z|}{dz}=\frac{z}{|z|}[/tex]?
     
  6. Jun 3, 2006 #5

    arildno

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    Quite so. :smile:
    The derivative of the absolute value function is not defined at z=0.
     
  7. Jun 3, 2006 #6
    And so F cannot be defined at z=0? So we take that to mean there is no force there?
     
  8. Jun 3, 2006 #7

    arildno

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    No, it doesn't. It just means you have to consider the z=0 case separately.
     
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