21.7: Inverse Cube Attractive Force

A particle moves in a field such that the attractive force on it varies inversely as the cube of the distance from a centre of attraction. What is the shape of the path? How does the angle $\theta$ vary with time?

Let’s suppose that the radial acceleration is $a(r) = -k^3/r^3 = - k^3u^3$. (I want the coefficient of $1/r^3$ to be negative, so that the force is attractive, which is why I have written the coefficient as $-k^2$. Besides, the dimensions of $k$ are then L²T⁻¹, which are the same as those of $h$, the angular momentum per unit mass, which helps to make the algebra simple.) The differential equation to the path (Equation 21.6.10) is then $h^2 u^2 \frac{d^2u}{d \theta^2} + h^2u^3 = k^2 u^3$ or

$$h^2 \frac{d^2u}{d \theta^2} + h^2u = k^3 u. \label{eq:21.7.11}$$

That is,

$$\frac{d^2u}{d \theta ^2} = \frac{k^2-h^2}{h^2} u. \label{eq:21.7.12}$$

The form of the motion evidently depends on whether $k^2 > h^2$ (a strongly attractive force, or a small angular momentum), or if $k^2 < h^2$ (a weak force, or a large angular momentum.) If we start the particle rolling with just the right amount of angular momentum $( k^2 = h^2 )$, there will evidently be zero radial acceleration, and the particle will move in a circle.

Before integrating Equation $\ref{eq:21.7.12}$, let us look at the equivalent potential. For $a(r) = -k^2/r^3$, the potential in the inertial frame is $\Omega = -\frac{1}{2} k^2/r^2$ provided we take the potential at infinity to be zero. The equivalent potential is then (see equation 21.2.5)

$$\Omega ' = - \frac{k^2}{2r^2} + \frac{h^2}{2r^2} . \label{eq:21.7.13}$$

We see that, if $k^2 = h^2$, the potential is zero and independent of distance. If $h^2 < k^2$, the equivalent potential is negative, increasing to zero as $r$ → ∞, and the particle accelerates towards the centre of attraction. If $h^2 > k^2$, the potential is positive, decreasing to zero as $r$ → ∞, and the particle accelerates away from the centre of attraction. This sounds like a contradiction, but what is happening is $h^2 > k^2$, that means that the particle has initially been given a large angular momentum, and, in the corotating frame, the centrifugal force is larger than the attractive force.

If $h^2 < k^2$, the equation of motion (Equation $\ref{eq:21.7.12}$) is

$$\frac{d^2u}{d \theta^2} = c^2 u , \tag{21.7.14}\label{eq:21.7.14}$$

where

$$c^2 = \frac{k^2 -h^2}{h^2}. \tag{21.7.15}\label{eq:21.7.15}$$

The general solution is

$$u = Ae^{c \theta} + Be^{-c \theta} \tag{21.7.16}\label{eq:21.7.16}$$

If the initial conditions are that at $t = 0, r = r_0, u = u_0, \frac{du}{d \theta} = 0$ (this last condition means that the particle was launched in a direction at right angles to the radius vector, this solution becomes

$$u = y_0 \cos h c \theta. \tag{21.7.17}\label{eq:21.7.17}$$

That is,

$$r = r_0 \sec h c \theta\tag{21.7.18}\label{eq:21.7.18}$$

I have drawn this below for $c = 0.1$; that is, for $k \approx 1.05h$ And for $c = 0.5$; that is $k \approx 1.22h$, for a smaller angular momentum.

We also need to consider the case $h^2 > k^2$, in which case the general solution is of the form $u = A \cos c \theta + B \sin c \theta$. Alas, I haven’t had the energy to do this yet. Perhaps some viewer can beat me to it, and let me know.