5.8.3: Plane Discs

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, $−Gm/z$.