An interesting trajectory

We can see this as a classical central-force problem, and thus apply the Binet's equation to solve this problem. Before we begin solving the problem, a brief overview of the derivation of the equation is as follows: we know that particles moving under a central force experience no change in angular momentum, since the central force only acts in the radial component of the acceleration of the particle [for the full mathematical proof, look here ]. So we can let $h$ be the specific angular momentum of the particle (which is the angular momentum of the particle divided by its mass $m$ ), where $h$ is a constant. Now, if we let $u = \frac{1}{r}$ be the reciprocal of the distance from the origin, then we can write the central force in terms of $u$ using Binet's formula: $F(u) = -mh^2u^2\left(\dfrac{\mathrm{d}^{2}u}{\mathrm{d}\theta^2}+u\right)$ [for the full mathematical proof, look here ].

So first, we need to determine what $\dfrac{\mathrm{d}^{2}u}{\mathrm{d}\theta^2}$ is.

$r^2 = R^2 \cos (2\theta) \\ \dfrac{1}{u^2} = R^2 \cos (2\theta) \\ \dfrac{\mathrm{d}}{\mathrm{d}\theta}(\dfrac{1}{u^2}) = \dfrac{\mathrm{d}}{\mathrm{d}\theta}[R^2 \cos (2\theta)] \\ \dfrac{1}{u^3}\dfrac{\mathrm{d}u}{\mathrm{d}\theta} = R^2 \sin (2\theta) \\ \dfrac{\mathrm{d}u}{\mathrm{d}\theta} = u^3 R^2 \sin (2\theta) \\ \begin{aligned} \dfrac{\mathrm{d}^{2}u}{\mathrm{d}\theta^2} &= R^2u^3[2 \cos (2\theta)] + R^2 \sin (2\theta) (3u^2 \dfrac{\mathrm{d}u}{\mathrm{d}\theta}) \\ &= 2u^3[R^2 \cos (2\theta)]+ 3u^5R^2 \sin (2\theta) (\dfrac{1}{u^3} \dfrac{\mathrm{d}u}{\mathrm{d}\theta}) \\ &= 2u^3(\dfrac{1}{u^2})+ 3u^5R^2 \sin (2\theta) [R^2 \sin (2\theta)] \\ &= 2u + 3u^5R^4 \sin^2 (2\theta) \\ &= 2u + 3u^5R^4 [1 - \cos^2 (2\theta)] \\ &= 2u + 3u^5R^4 - 3u^5R^4 \cos^2 (2\theta) \\ &= 2u + 3u^5R^4 - 3u^5[R^2 \cos (2\theta)]^2 \\ &= 2u + 3u^5R^4 - 3u^5(\dfrac{1}{u^2})^2 \\ &= 2u + 3u^5R^4 - 3u \\ &= 3u^5R^4 - u \end{aligned}$

So now we substitute this back into the Binet's equation to obtain

$\begin{aligned} F(u) &= -mh^2u^2\left(\dfrac{\mathrm{d}^{2}u}{\mathrm{d}\theta^2}+u\right) \\ &= -mh^2u^2(3u^5R^4 - u + u) \\ &= -mh^2u^2(3u^5R^4) \\ &= -3R^4mh^2u^7 . \end{aligned}$

Since $r = \dfrac{1}{u}$ , then we can substitute this back into the expression above to get

$\begin{aligned} F(r) &= F(\dfrac{1}{u}) \\ &= -3R^4mh^2(\dfrac{1}{u})^7 \\ &= \dfrac{-3R^4mh^2}{u^7} \\ &= \dfrac{C}{u^7} , \end{aligned}$ where $C = -3R^4mh^2$ is a constant.

Therefore, $n = 7$ .

In planar polar coordinates, the radial and transverse components of acceleration are $\ddot{r} - r\dot\theta^2$ and $r\ddot\theta + 2r\dot{r}\dot\theta = \tfrac{1}{r}\tfrac{d}{dt}(r^2\dot\theta)$ . Thus, for a particle of mass $m$ in an attractive central field of magnitude $F(r)=\tfrac{C}{r^n}$ , we have $\ddot{r} - r\dot\theta^2 = -\tfrac{C}{m}r^{-n} \qquad \qquad h \; = \; r^2\dot\theta$ where $h$ is constant, the angular momentum per unit mass.

If we change variables, and consider $u = r^{-1}$ , then $\begin{array}{rcl} \dot{r} & = & -u^{-2}\frac{du}{d\theta}\dot\theta \; = \; -h\frac{du}{d\theta} \\ \ddot{r} & = & -h\frac{d^2u}{d\theta^2}\dot\theta \; = \; -h^2u^2\frac{d^2u}{d\theta^2} \end{array}$ so that the differential equation becomes $\begin{array}{rcl} -h^2u^2\frac{d^2u}{d\theta^2} - h^2u^3 & = & -\tfrac{C}{m}u^n \\ \frac{d^2u}{d\theta^2} + u & = & \tfrac{C}{mh^2}u^{n-2} \end{array}$ The trajectory $r^2 = R^2\cos2\theta$ leads to $\begin{array}{rcl} u & = & R^{-1}(\sec2\theta)^{\frac12} \\ \frac{du}{d\theta} & = & R^{-1}\tfrac{1}{2}(\sec2\theta)^{-\frac12} \times2\sec2\theta\tan2\theta \\ & = & R^{-1}(\sec2\theta)^{\frac12}\tan2\theta \\ \frac{d^2u}{d\theta^2} & = & R^{-1}(\sec2\theta)^{\frac12}\tan^22\theta + 2R^{-1}(\sec2\theta)^{\frac12}\sec^22\theta \\ & = & R^{-1}\big[\tan^22\theta + 2\sec^22\theta\big](\sec2\theta)^{\frac12} \\ \frac{d^2u}{d\theta^2} + u & = & R^{-1}\big[\tan^22\theta + 1 + 2\sec^22\theta\big](\sec2\theta)^{\frac12} \\ & = & \tfrac{3}{R}(\sec2\theta)^{\frac52} \; = \; 3R^4u^5 \end{array}$ and so this trajectory is possible provided that $n=7$ .

Reference: See for example David Morin's book on Classical Mechanics.

First note that the force law $F(r) = C/r^n , \ C < 0,$ implies that the potential is $V(r) = - \frac{C}{nr^{n-1}} \equiv \frac{B}{r^{n-1}} \ \text{such that } F(r) = -\frac{dV}{dr} \equiv - V'(r).$

For central forces, acceleration is only along the radial direction. The tangential part is zero. The radial and tangential accelerations of a point mass in polar coordinates are $a_r = \ddot{r} - r \dot{\theta}^2; \qquad a_\theta = 2\dot{r}\dot{\theta} + r\ddot{\theta} = 0.$ The second equation results in the conservation of angular momentum $L$ and it allows us to express angular velocity in terms of $L$ : $L = mr^2\dot \theta = \text{Const.} \quad \Rightarrow \dot{\theta} = \frac{L}{mr^2}$ The Newton's law can now be expressed as $m a_r = m \ddot{r} - \frac{L}{mr^3} = F(r) = - V'(r).$ This equation can be integrated once (after multiplication with $\dot{r}$ ) to yield conservation energy equation: $\frac{1}{2} m \dot{r}^2 + \frac{L}{2mr^2} + V(r) = E.$

Noting that $\dot{r} = \frac{dr}{d\theta} \dot{\theta} = \frac{dr}{d\theta} \frac{L}{mr^2},$ the conservation energy equation can be transformed into $\left(\frac{1}{r^2} \frac{dr}{d\theta}\right)^2 + \frac{1}{r^2}= \frac{2mE}{L^2} - \frac{2mV(r)}{L^2}.$ Given $r^2(\theta) = R^2 \cos(2\theta)$ , and noting that $r \frac{dr}{d\theta} = - R^2 \sin (2\theta),$ and substituting this into the last differential equation, we get $\frac{R^4}{r^6} = \frac{2mE}{L^2} + \frac{2mV(r)}{L^2}.$ Comparing the equations, it is evident that $E = 0, \text{ and } n = 7.$

Simplest solution from elementry concepts