# 21.7: Inverse Cube Attractive Force

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

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.

This page titled 21.7: Inverse Cube Attractive Force is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.