The Binet equation

The radial component of Newton's second law was show in an earlier post to be \begin{align} F &= m\left(\ddot r - r\dot\theta^2\right)\,. \end{align} Given some radial force, $\vec F = F(r,\theta)\hat r$, solving this differential equation gives a description of the movement of a particle subject to the force.

Assuming a non-vanishing angular velocity and radial coordinate ($\dot\theta\neq 0$, $r\neq 0$) the radial coordinate $r$ can be assumed to be a function of the angular coordinate $\theta$. This enables us to do the substition $u(\theta) = \frac{1}{r(\theta)}$. The derivative of $u$ w.r.t. $t$ is \begin{align} \frac{\mathrm du}{\mathrm dt} &= \frac{\mathrm du}{\mathrm d\theta}\frac{\mathrm d\theta}{\mathrm d t} \end{align} so that the derivative w.r.t $\theta$ becomes \begin{align} u'(\theta) = \frac{\mathrm du}{\mathrm d\theta} &= \frac{1}{\dot\theta} \frac{\mathrm du}{\mathrm dt}\\ &= \frac{1}{\dot\theta}\frac{\mathrm d}{\mathrm d t}\left(\frac{1}{r}\right)\\ &= - \frac{\dot r}{r^2\dot \theta}\\ &= - \frac{\dot r}{h}\,, \end{align} where $h = r^2\dot\theta$ is the specific angular momentum, which, as shown before, is a constant when the force is purely radial.

The derivative w.r.t. $t$ of this is \begin{align} \frac{\mathrm d}{\mathrm d t}\left(\frac{\mathrm d u}{\mathrm d \theta}\right) &= \frac{\mathrm d \theta}{\mathrm d t} \frac{\mathrm d}{\mathrm d\theta}\left(\frac{\mathrm d u}{\mathrm d \theta}\right) = \frac{\mathrm d \theta}{\mathrm d t} \frac{\mathrm d^2 u}{\mathrm d \theta^2}\,, \end{align} so that \begin{align} u''(\theta) = \frac{\mathrm d^2 u}{\mathrm d \theta^2} &= \frac{1}{\dot \theta} \frac{\mathrm d}{\mathrm d t}\left(-\frac{\dot r}{h}\right)\\ &= - \frac{\ddot r}{h \dot\theta}\,. \end{align}

Note that $\ddot r = -h\dot\theta u'' = - h^2 u^2 u''$, and that $r\dot \theta^2 = \frac{h^2}{r^3} = h^2 u^3$. Newton's second law becomes \begin{align} F &= m\left(-h^2 u^2 u'' - h^2 u^3\right)\\ &= - m h^2 u^2 \left(u'' + u\right)\,. \end{align} This result is the Binet equation, which is a restatement of Newton's second law that in some cases proves more tractable that the forms presented earlier.

As an example, consider a force satisfying an inverse square law (such as gravity), i.e., \begin{align} \vec F &= - \frac{f}{r^2}\hat r = - f u^2 \hat r\,, \end{align} where $f>0$ for an attractive force. Plugging $F=-f u^2$ into the left-hand side of the Binet equation gives \begin{align} - f u^2 &= - m h^2 u^2\left(u'' + u\right) \end{align} or \begin{align} \frac{f}{m h^2} &=u'' + u\,. \end{align}

This is a linear differential equation, the homogeneous version of which is $u''+u=0$ with complete solution \begin{align} u_h(\theta) = c_1 \cos(\theta) + c_2 \sin(\theta)\,. \end{align} The inhomogeneous equation has a particular solution \begin{align} u_p(\theta) = \frac{f}{m h^2}\,, \end{align} ans so the complete solution to the original equation is \begin{align} u(\theta) &= c_1\cos(\theta) + c_2 \sin(\theta) + \frac{f}{m h^2}\,. \end{align}

Returning to the original coordinates gives \begin{align} r(\theta) &= \frac{1}{c_1\cos(\theta) + c_2\sin(\theta) + \frac{f}{m h^2}}\\ &= \frac{1}{c \sin(\theta-\theta_0) + \frac{f}{m h^2}}\,, \end{align} where $\theta_0 = -\arctan\left(\frac{c_1}{c_2}\right)$ and $c = \sqrt{c_1^2 + c_2^2}$.

This is the general polar form of a conic section with one focus at the origin. For simplicity, and without loss of generality, we orient the coordinate system such that $\theta_0=0$, and the function becomes \begin{align} r(\theta) &= \frac{1}{c \sin(\theta) + \frac{f}{mh^2}}\,, \end{align} which is symmetric around the $y$-axis. Depending on the values of the constants the trajectory described by this function may be elliptic, parabolic or hyperbolic.

If $c<\frac{f}{mh^2}$ the denominator in the fraction is positive for any $\theta$, and thus the value of $r$ remains finite; the particle follows a closed elliptic orbit. This is Kepler's first law of planetary motion: planets move in elliptical orbits with the Sun in one of the foci.

If $c = \frac{f}{mh^2}$ the particle follows a parabolic trajectory. The denominator in $r(\theta)$ vanishes exactly as $\theta=-\frac{\pi}{2}$, while $r>0$ for all other values of $\theta$. This implies that the trajectory does not intersect the negative $y$-axis; it does however intersect twice any other line passing through the origin.

If $c>\frac{f}{m h^2}$ the trajectory is hyperbolic. Values of $\theta$ for which $r<0$ and $r>0$, respectively, define the two separate branches of the hyperbola. The two values of $\theta$ for which the denominator in $r(\theta)$ vanishes define the lines approached asymptotically by the hyperbolic branches.