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5.8.3: Plane Discs

  • Page ID
    8147
  • [ "article:topic", "authorname:tatumj" ]

    Refer to figure \(\text{V.2A}\). The potential at \(\text{P}\) from the elemental disc is

    \[dψ = -\frac{GδM}{\left( r^2 + z^2 \right)^{1/2}} = -\frac{2 \pi G σrδr}{\left( r^2 + z^2 \right)^{1/2}}. \label{5.8.10} \tag{5.8.10}\]

    The potential from the whole disc is therefore

    \[ψ = -2 \pi G σ \int_0^a \frac{r dr}{\left( r^2 + z^2 \right)^{1/2}}. \label{5.8.11} \tag{5.8.11}\]

    The integral is trivial after a brilliant substitution such as \(X = r^2 + z^2\) or \(r = z \tan θ\), and we arrive at

    \[ψ=-2 \pi G σ \left( \sqrt{z^2 + a^2} - z \right). \label{5.8.12} \tag{5.8.12}\]

    This increases to zero as \(z → ∞\). We can also write this as

    \[ψ = -\frac{2\pi Gm}{\pi a^2} \cdot \left[ z \left( 1 + \frac{a^2}{z^2} \right)^{1/2} - z \right] , \label{5.8.13} \tag{5.8.13}\]

    and, if you expand this binomially, you see that for large \(z\) it becomes, as expected, \(−Gm/z^2\) .