5.8.6: Rods
- Page ID
- 8150
Refer to figure \(\text{V.5}\). The potential at \(\text{P}\) due to the element \(δx\) is \(-\frac{Gλδx}{r} = − Gλ \sec θδθ\). The total potential at \(\text{P}\) is therefore
\[ψ = - G λ \int_α^β \sec θdθ = - Gλ \ln \left[ \frac{\sec β + \tan β}{\sec α + \tan α} \right] . \label{5.8.15} \tag{5.8.15}\]
\(\text{FIGURE V.24}\)
Refer now to figure \(\text{V.24}\), in which \(A = 90^\circ + α\) and \(B = 90^\circ − β\).
\[\frac{\sec β + \tan β}{\sec α + \tan α} = \frac{\cos α (1+ \sin β)}{\cos β (1 + \sin α)} = \frac{\sin A (1+ \cos B)}{\sin B(1-\cos A)} = \frac{2 \sin \frac{1}{2}A \cos \frac{1}{2} A . 2 \cos^2 \frac{1}{2} B}{2\sin \frac{1}{2} B \cos \frac{1}{2} B . 2 \sin^2 \frac{1}{2} A} = \cot \frac{1}{2} A \cot \frac{1}{2} B = \sqrt{\frac{s(s-r_2)}{(s-r_1)(s-2l)}}\cdot \sqrt{\frac{s(s-r_1)}{(s-2l)(s-r_2)}},\]
where \(s = \frac{1}{2} (r_1 + r_2 + 2l)\). (You may want to refer here to the formulas on pp. 37 and 38 of Chapter 2.)
Hence \[ψ = - Gλ \ln \left[ \frac{r_1 + r_2 + 2l}{r_1 + r_2 -2l} \right]. \label{5.8.16} \tag{5.8.16}\]
If \(r_1\) and \(r_2\) are very large compared with \(l\), they are nearly equal, so let’s put \(r_1 + r_2 = 2r\) and write equation 5.8.17 as
\[ψ = -\frac{Gm}{2l} \ln \left[ \frac{2r \left( 1 + \frac{2l}{2r} \right)}{2r \left( 1 - \frac{2l}{2r} \right)} \right] = -\frac{Gm}{2l} \left[ \ln \left(1 + \frac{l}{r} \right) - \ln \left( 1 - \frac{l}{r} \right) \right] .\]
Maclaurin expand the logarithms, and you will see that, at large distances from the rod, the potential is, expected, \(−Gm/r\).
Let us return to the near vicinity of the rod and to equation \(\ref{5.8.16}\). We see that if we move around the rod in such a manner that we keep \(r_1 + r_2\) constant and equal to \(2a\), say − that is to say if we move around the rod in an ellipse (see our definition of an ellipse in Chapter 2, Section 2.3) − the potential is constant. In other words the equipotentials are confocal ellipses, with the foci at the ends of the rod. Equation \(\ref{5.8.16}\) can be written
\[ψ = - Gλ \ln \left( \frac{a+l}{a-l} \right) . \label{5.8.17} \tag{5.8.17}\]
For a given potential \(ψ\), the equipotential is an ellipse of major axis
\[2a = 2l \left( \frac{e^{ψ/(Gλ)}+1}{e^{ψ/(Gλ)}-1} \right), \label{5.8.20} \tag{5.8.20}\]
where \(2l\) is the length of the rod. This knowledge is useful if you are exploring space and you encounter an alien spacecraft or an asteroid in the form of a uniform rod of length \(2l\).