The Orbit Equation and Binet's Substitution

The Geometric Shift

Imagine tracking a planet. Newton's laws tell us how the distance \(r\) changes with time \(t\). However, the "orbit" is a static curve in space. We want to know \(r\) as a function of the angle \(\theta\), not time. The difficulty is that time and angle are linked by the angular velocity. If we could express the radial acceleration directly in terms of \(\theta\), we could solve for the shape of the path without worrying about how fast the object moves along it. This is the core idea: we are changing the independent variable from \(t\) to \(\theta\), and we are changing the dependent variable from \(r\) to \(u = 1/r\). This inversion is crucial because the central force law often becomes linear in \(u\), making the mathematics tractable.

Deriving Binet's Formula

To perform the change of variables, we use the chain rule. Let \(u(\theta) = 1/r(\theta)\). The radial velocity is \(\dot{r} = \frac{dr}{d\theta}\dot{\theta}\). Since angular momentum \(\ell = m r^2 \dot{\theta}\) is conserved, we have \(\dot{\theta} = \frac{\ell}{m r^2} = \frac{\ell u^2}{m}\). Substituting this into the expression for \(\dot{r}\):

Now we need the radial acceleration \(\ddot{r}\). Differentiating \(\dot{r}\) with respect to time again, and using \(\frac{d}{dt} = \dot{\theta} \frac{d}{d\theta} = \frac{\ell u^2}{m} \frac{d}{d\theta}\), we obtain the famous Binet formula:

Notice that the radial acceleration is now expressed entirely in terms of \(u\) and its derivatives with respect to \(\theta\). This is the key step that decouples the geometry from the dynamics of the time parameter.

The Orbit Differential Equation

Newton's second law in the radial direction for a central force \(F(r)\) is \(m \ddot{r} = F(r) + m r \dot{\theta}^2\). The term \(m r \dot{\theta}^2\) is the centrifugal force. Substituting \(\dot{\theta} = \frac{\ell u^2}{m}\) and \(r = 1/u\), we get \(m r \dot{\theta}^2 = m \frac{1}{u} \left( \frac{\ell u^2}{m} \right)^2 = \frac{\ell^2 u}{m}\). Now substitute Binet's formula for \(\ddot{r}\):

Rearranging to isolate the second derivative, we arrive at the general orbit equation:

This equation is universal. For any central force, if you can write the force as a function of \(u\), you can solve this differential equation to find the shape of the orbit. The left-hand side is always the same; the right-hand side encodes the specific physics of the force.

Application to Inverse-Square Law

For the gravitational or electrostatic force, \(F(r) = -\frac{k}{r^2} = -k u^2\), where \(k > 0\) is the strength of the attraction. Substituting this into the right-hand side of the orbit equation:

The \(u^2\) terms cancel perfectly, leaving a constant. The differential equation becomes:

This is a linear second-order differential equation with constant coefficients. The homogeneous solution is \(A \cos \theta + B \sin \theta\), and the particular solution is the constant \(\frac{mk}{\ell^2}\). By choosing the zero of \(\theta\) at the periapsis (closest approach), we set \(B=0\) and \(A = \frac{mk}{\ell^2} e\), where \(e\) is the eccentricity. The general solution is:

Inverting this gives the conic section equation \(r(\theta) = \frac{\ell^2/mk}{1 + e \cos \theta}\). The eccentricity \(e\) is determined by the total energy \(E\) of the system, linking the geometric shape directly to the dynamical state.