5.4.7: Solid Cylinder

We do this not because it has any particular relevance to celestial mechanics, but because it is easy to do. We imagine a solid cylinder, density $ρ$, radius $a$, length $l$. We seek to calculate the field at a point $\text{P}$ on the axis, at a distance $h$ from one end of the cylinder (figure $\text{V.8}$).

$\text{FIGURE V.8}$

The field at $\text{P}$ from an elemental disc of thickness $δz$ a distance z below P is (from Equation 5.4.9)

$$δg = Gρδzω. \label{5.4.19} \tag{5.4.19}$$

Here $ω$ is the solid angle subtended at $\text{P}$ by the disc, which is $2 \pi \left[ 1 - \frac{z}{\left( z^2 + a^2 \right)^{1/2}} \right]$. Thus the field at $\text{P}$ from the entire cylinder is

$$g = 2 \pi G ρ \int_h^{l+h} \left[ 1 - \frac{z}{(z^2 + a^2)^{1/2}} \right] dz, \label{5.4.20} \tag{5.4.20}$$

or $$g = 2 \pi G ρ \left( l - \sqrt{(l+h)^2 + a^2} + \sqrt{h^2+ a^2} \right), \label{5.4.21} \tag{5.4.21}$$

or $$g = 2\pi G ρ ( l - r_2 + r_1 ). \label{5.4.22} \tag{5.4.22}$$

It might also be of interest to express $g$ in terms of the height $y (= \frac{1}{2}l + h)$ of the point $\text{P}$ above the mid-point of the cylinder. Instead of Equation $\ref{5.4.21}$, we then have

$$g = 2 \pi G ρ \left( l - \sqrt{(y + \frac{1}{2}l)^2 + a^2} + \sqrt{(y - \frac{1}{2}l)^2 + a^2} \right) . \label{5.4.23} \tag{5.4.23}$$

If the point $\text{P}$ is inside the cylinder,at a distance $h$ below the upper end of the cylinder, the limits of integration in Equation $\ref{5.4.20}$ are $h$ and $l − h$, and the distance $y$ is $\frac{1}{2}l − h$. In terms of $y$ the gravitational field at $\text{P}$ is then

$$g = 2 \pi G ρ \left( 2y - \sqrt{(y + \frac{1}{2}l)^2 + a^2} + \sqrt{(y-\frac{1}{2}l)^2 + a^2} \right). \label{5.4.24} \tag{5.4.24}$$

In the graph below I have assumed, by way of example, that $l$ and $a$ are both 1, and I have plotted $g$ in units of $2 \pi Gρ$ (counting $g$ as positive when it is directed downwards) from $y = −1$ to $y = + 1$. The portion inside the cylinder $(- \frac{1}{2} ≤ y ≤ \frac{1}{2}l)$, represented by Equation $\ref{5.4.24}$, is almost, but not quite, linear. The field at the centre of the cylinder is, of course, zero.

Below, I draw the same graph, but for a thin disc, with $a = 1$ and $l = 0.1$. We see how it is that the field reaches a maximum immediately above or below the disc, but is zero at the centre of the disc.