Ellipses and the Two-Body Problem

Summer research 2024: the three-body problem · Part 3 of 11

A weekly research log from my summer project at the University of Edinburgh, supervised by Prof. Jenni Smillie.

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Before tackling three bodies, this post revisits ellipses and the Kepler problem.

Ellipses

Semi-major and semi-minor axes

I discussed the parametrisation of ellipses in an earlier post. An ellipse can be written as

\[\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1,\]

with semi-major axis \(a\), semi-minor axis \(b\) and eccentricity \(e\), where

\[b^2=a^2(1-e^2).\]

Polar equations

Measured from the centre, at polar angle \(\theta_{\text{c}}\), the radius is

\[R= \frac{a \sqrt{1-e^2}}{\left(1-e^2 \cos^2\theta_{\text{c}}\right)^{1/2}} .\]

Measured from a focus, at polar angle \(\theta_{\text{f}}\), it is

\[r=\frac{a(1-e^2)}{1+e \cos\theta_{\text{f}}} = \frac{p}{1+e \cos\theta_{\text{f}}},\]

where \(p\) is the semi-latus rectum, which appeared in my post on Kepler’s third law.

The two-body problem

I have written about the reduction to relative motion and about two bodies falling straight towards each other. Here I focus on the shape of the orbits. Throughout, \(\mu = GM\) and \(\mathbf{r}\) is the relative position, which obeys \(\ddot{\mathbf{r}} = -\mu\,\mathbf{r}/r^3\).

Energy

The energy per unit (reduced) mass is conserved:

\[\frac{1}{2}v^2 - \frac{\mu}{r} = E .\]

Suppose \(E<0\) and write \(E=- \alpha^2\). Since \(\frac12 v^2 \ge 0\),

\[\frac{\mu}{r} \ge \alpha^2 \quad\Longrightarrow\quad r \leq \frac{\mu}{\alpha^2} = 2a, \qquad \text{where } a=\frac{\mu}{2 \alpha^2}.\]

So negative energy implies a bounded orbit. (In fact \(a\) turns out to be the semi-major axis.)

Angular momentum and the Laplace–Runge–Lenz vector

As shown in my post on Noether’s theorem, the angular momentum per unit mass, \(\mathbf{C} = \mathbf{r}\times\dot{\mathbf{r}}\), is conserved.

If \(\mathbf{C}=\mathbf{0}\), the motion is purely radial: the bodies move along a straight line, and collide if they are approaching. If \(\mathbf{C} \neq \mathbf{0}\), then \(\mathbf{r} \cdot \mathbf{C}=0\), so the motion lies in the fixed plane perpendicular to \(\mathbf{C}\).

Next, differentiate the unit vector \(\mathbf{e}_r = \mathbf{r}/r\):

\[\begin{aligned} \frac{d\mathbf{e}_r}{dt}=\frac{d}{d t}\left(\frac{\mathbf{r}}{r}\right) & =\frac{r \dot{\mathbf{r}}-\dot{r}\, \mathbf{r}}{r^2}=\frac{r^2 \dot{\mathbf{r}}-r \dot{r}\, \mathbf{r}}{r^3} \\ & =\frac{(\mathbf{r} \cdot \mathbf{r}) \dot{\mathbf{r}}-(\mathbf{r} \cdot \dot{\mathbf{r}}) \mathbf{r}}{r^3}=\frac{\mathbf{r} \times(\dot{\mathbf{r}} \times \mathbf{r})}{r^3}=\frac{\mathbf{C} \times \mathbf{r}}{r^3}, \end{aligned}\]

using \(r\dot r = \mathbf{r}\cdot\dot{\mathbf{r}}\) and \(\dot{\mathbf{r}}\times\mathbf{r} = -\mathbf{C}\). Using the equation of motion, \(\mathbf{r}/r^3 = -\ddot{\mathbf{r}}/\mu\), and since \(\mathbf{C}\) is constant,

\[\frac{d\mathbf{e}_r}{dt}=-\frac{\mathbf{C} \times \ddot{\mathbf{r}}}{\mu}=-\frac{1}{\mu} \frac{d}{d t}(\mathbf{C} \times \dot{\mathbf{r}})=\frac{1}{\mu} \frac{d}{d t}(\mathbf{v} \times \mathbf{C}) .\]

Integrating, there is a constant vector \(\mathbf{e}\) such that

\[\mu\left(\mathbf{e}_r+\mathbf{e}\right)=\mathbf{v} \times \mathbf{C} = \dot{\mathbf{r}} \times(\mathbf{r} \times \dot{\mathbf{r}})=v^2\, \mathbf{r}-r \dot{r}\, \dot{\mathbf{r}} .\]

Since \(\mathbf{C}\) is perpendicular to both \(\mathbf{e}_r\) and \(\mathbf{v}\times\mathbf{C}\), taking the dot product with \(\mathbf{C}\) gives \(\mathbf{e} \cdot \mathbf{C}=0\): the vector \(\mathbf{e}\) lies in the plane of the motion.

The conserved vector \(\mathbf{e}\) is the eccentricity vector, a rescaling of the Laplace–Runge–Lenz vector.

Kepler’s first law

Now take the dot product of the last equation with \(\mathbf{r}\). On the right, \((\mathbf{v}\times\mathbf{C})\cdot\mathbf{r} = \mathbf{C}\cdot(\mathbf{r}\times\mathbf{v}) = C^2\). On the left, writing \(\theta\) for the angle between \(\mathbf{r}\) and \(\mathbf{e}\), we get \(\mu(r + e\, r\cos\theta)\). Hence

\[r = \frac{C^2/\mu}{1 + e\cos\theta},\]

which is exactly the polar equation of a conic with a focus at the origin, semi-latus rectum \(p = C^2/\mu\) and eccentricity \(e = \lvert\mathbf{e}\rvert\). Bound orbits (\(e<1\)) are ellipses: this is Kepler’s first law. The eccentricity vector points from the focus towards the point of closest approach.

The second and third laws are derived in my post on Kepler’s third law.

A hidden symmetry

By Noether’s theorem, a conserved quantity signals a symmetry, but for the Laplace–Runge–Lenz vector the symmetry is hidden. For bound orbits, the Kepler problem is mathematically equivalent to free motion on the surface of a four-dimensional (hyper)sphere, so it is symmetric under rotations in four dimensions, a larger symmetry group than the obvious rotations of three-dimensional space.

Progress

The project is proving more demanding than I expected, and working through the literature in detail takes time, but writing up each topic here is a good way to organise my understanding.

Credits

This post is based on the book Integrable Systems in Celestial Mechanics by Diarmuid Ó’Mathúna. The remark about the four-dimensional sphere is based on Wikipedia.

Changes on 25 October 2026

What was wrong. Nothing in the results, but the bound \(r \le 2a\) was stated without its derivation, and the derivation of the eccentricity vector stopped before reaching the orbit equation.

What changed. Both are now complete, with \(\mu = GM\) throughout.