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# 5.8.3: Plane Discs

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

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, $$−Gm/z^2$$ .