The Laplace vector

The conservation of energy and agular momentum were already discussed in earlier sections of these notes. They give us four first integrals of the motion (since the angular momentum has three components). However, there is also another quantity in celestial mechanics which is conserved in the unperturbed Kepler motion.This quantity is the Laplace vector, which is defined as

$${\bf L}= {\bf v}\times {\bf A}-\frac{\mu}{r} {\bf r}.$$

From this we can compute its derivative

$$\dot{{\bf L}} = \dot{{\bf v}} \times {\bf A}+ {{\bf v}} \ti... ... +\frac{\mu}{r^2} \dot{r} {\bf r}-\frac{\mu}{r} \dot{{\bf r}}$$

and using the usual substitutions we get

$$\dot{{\bf L}} = \left(-\frac{\mu}{r}{\bf r} + {\bf F}\right... ... F}) +\frac{\mu}{r^2} \dot{r} {\bf r} -\frac{\mu}{r} {\bf v}$$

and therfore

$$\dot{{\bf L}} = \frac{\mu}{r^3}\left[ ({\bf r}\cdot{\bf v})... ...\bf F}) + \frac{\mu}{r^2}\dot{r}{\bf r}-\frac{\mu}{r}{\bf v}$$

This quantity seems to be rather messy, however, a little vector algebra son yields

$$\dot{{\bf L}} = {\bf F}\times{\bf A}+ {{\bf v}} \times ({\bf r}\times {\bf F}).$$

Now, if ${\bf F}= 0$, the right hand side of this last expression vanishes, i.e. the Laplace vector is conserved, yielding three more first integrals.

Before continuing we should give some geometric interpretation of these first integrals. Let us start with the angular momentum. In the unperturbed case, the objects move in a plane, and both ${\bf r}$ and ${\bf v}$ lie in this plane. Since ${\bf A}$ is perpendicular to both ${\bf r}$ and ${\bf v}$, ${\bf A}$ is perpendicular to the plane of motion. Since any plane through the origin in $\rm I \hskip -3pt R^3$ is uniquely defined by its normal, the angular momentum defines the plane of the motion. If the angular momentum is not conserved it means that the motion is not planar any more. A little analysis also give s a geometric interpretation of the energy, namely the energy is directly proportional to the area of the ellipse.

Since $L$ is a constant vector, we can compute its components at any time. We can also use a special coordinate system which makes the computation easy. We compute ${\bf L}$ at the point in time when the particle as at pericenter, in a coordinate system which is in the plane of the orbit, the $x$ axis aligned with the direction of pericenter. Hence,

$${\bf r}= (r, 0, 0), \;\;\;\;\ {\bf v}= (0, v, 0)$$

Then

$${\bf A}= {\bf r}\times {\bf v}= (0, 0, rv)$$

and

$${\bf v}\times {\bf A}= (rv^2, 0, 0)$$

and

$${\bf L}= (rv^2-\mu, 0, 0)$$

We conclude that ${\bf L}$ points in the direction of the pericenter. We still have to compute the length of ${\bf L}$, which will be the subject of the next section.