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# 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

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

That is,

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

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 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)

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

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 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.