Kepler Orbits and the Runge-Lenz Vector

The Geometry of Closed Orbits

Imagine a particle moving in a plane under a central force. We already know from the previous lecture that the shape of the path is a conic section. However, knowing the shape is not enough; we need to know where the "focus" of that conic section lies relative to the particle's motion. In a generic central force field, the orbit might slowly rotate or "precess," meaning the point of closest approach moves around the center over time. This would look like a flower petal pattern.

For the specific case of gravity or electrostatics, where the force drops off exactly as the inverse square of the distance, something special happens. The orbit does not precess. The ellipse (or hyperbola) stays fixed in space. This implies that there is a vector quantity that always points from the center of force to the perihelion, and its magnitude remains constant. This vector is the Laplace–Runge–Lenz vector. It acts as a "compass needle" for the orbit, locking the major axis in place. This is a rare symmetry in classical mechanics, and it is the key to understanding why Kepler's laws hold so precisely for ideal two-body systems.

Defining the Conserved Vector

Let us define the Laplace–Runge–Lenz vector \(\mathbf{A}\) for a particle of mass \(m\) moving under the potential \(V(r) = -k/r\). The vector is constructed from the linear momentum \(\mathbf{p}\), the angular momentum \(\mathbf{L}\), and the position vector \(\mathbf{r}\):

Here, \(k\) is the force constant (e.g., \(G M m\) for gravity), and \(\hat{\mathbf{r}}\) is the unit vector pointing from the center of force to the particle. Both terms lie in the plane of motion: \(\mathbf{p} \times \mathbf{L}\) is perpendicular to \(\mathbf{L}\), which is itself perpendicular to that plane, and \(\hat{\mathbf{r}}\) lies in it. The combination is carefully chosen so that its time derivative vanishes when the force is inverse-square.

It is crucial to note that \(\mathbf{A}\) is perpendicular to the angular momentum \(\mathbf{L}\). We can verify this by taking the dot product:

Since \(\mathbf{A}\) is perpendicular to \(\mathbf{L}\), it must lie entirely within the plane of the orbit. This confirms that \(\mathbf{A}\) points in a specific direction within the orbital plane, which we will identify as the direction of the perihelion.

Direction and the Orbit Equation

Take the dot product of \(\mathbf{A}\) with \(\mathbf{r}\). By the scalar triple product, \((\mathbf{p} \times \mathbf{L}) \cdot \mathbf{r} = \mathbf{p} \cdot (\mathbf{L} \times \mathbf{r})\), and \(\mathbf{L} \times \mathbf{r} = (\mathbf{r} \times \mathbf{p}) \times \mathbf{r} = \mathbf{p}\, r^2 - \mathbf{r}\,(\mathbf{p} \cdot \mathbf{r})\), so

Write \(\mathbf{A} \cdot \mathbf{r} = |\mathbf{A}|\, r \cos\theta\), with \(\theta\) measured from the direction of \(\mathbf{A}\), and solve for \(r\):

This is the conic of the previous lecture with eccentricity \(e = |\mathbf{A}|/(mk)\), that is \(|\mathbf{A}| = mke\). Because \(r\) is smallest at \(\theta = 0\), \(\mathbf{A}\) points from the centre of force to the perihelion. For a circular orbit \(e = 0\) and \(\mathbf{A} = 0\): a circle has no distinguished perihelion.

Magnitude and Energy

Squaring the definition, with \((\mathbf{p} \times \mathbf{L})^2 = p^2 L^2\) (because \(\mathbf{p} \perp \mathbf{L}\)) and \(\hat{\mathbf{r}} \cdot (\mathbf{p} \times \mathbf{L}) = L^2/r\) from above:

so \(|\mathbf{A}| = mk\sqrt{1 + 2EL^2/(mk^2)} = mke\). Bound orbits (\(E < 0\)) have \(e < 1\) and are ellipses; \(E = 0\) gives the parabola \(e = 1\), and \(E > 0\) the hyperbolae \(e > 1\).