Perturbations in the pericenter

In this section we consider the potential

$$U(r) = \frac{\mu}{r}\left(1+\frac{\epsilon}{r^2}\right).$$

This potential differs from the potential of the Kepler motion by the term

$$\frac{\epsilon\mu}{r^3},$$

in which we will assume that $\epsilon$ is small. However, the potential has still the spherical symmetry of the Kepler motion.

This potential leads to the new equation of motion

$$\ddot{{\bf r}} = -\frac{\mu}{r^3}{\bf r}- \frac{3\epsilon\mu}{r^5}{\bf r}.$$

Hence, the perturbing force is

$${\bf F}= - \frac{3\epsilon\mu}{r^5}{\bf r}$$

Clearly, ${\bf F}$ is a central force ( ${\bf r}\times {\bf F}= 0$) and we have

$$\dot{{\bf A}} = 0$$

and

$$\dot{{\bf L}} = {\bf F}\times {\bf A}$$

In particular, since ${\bf A}$ is constant, the force ${\bf F}$ does not change the plane of the orbit.

We now compute $\dot{{\bf L}}$ in a convenient coordinate system. Assume that the pericenter is aligned with the $x$ axis at the instant for which we do the computations. Then

$${\bf A}= (0,0,A)$$

where $A$ is the scalar angular momentum of the orbit (conserved). We also have

$${\bf L}= (\mu e, 0, 0).$$

and also

$${\bf F}= -\frac{3\epsilon\mu}{r^5} (x, y, 0)$$

Direct computation of cross products yields

$$\dot{{\bf L}} = -\frac{3\epsilon\mu A}{r^5} (y, -x, 0) .$$ (29)

Now this is a difficult differential equation for $\L$, since $r$ of course meeds to observe the equation of motion. On the other hand since $\epsilon$ is small and $r^5$ is typically rather large, we see that ${\bf L}$ will change relatively slowly, compared with the change of location of the body. Since the change of ${\bf L}$ is small, it is sufficient to know its change over a whole orbit instead of its instantaneous change. In other words we average the perturbation over the entire orbit.

In simple terms we have a differential equation of the form

$$\dot{f}(t) = g(t, f(t)),$$

where $f$ is a first integral of the unperturbed motion (e.g., an orbital element which is constant for an orbit). When $g = 0$, i.e., there is no perturbation, $f$ is a constant, $f_0$. In principle, $g(t, f)$ should be computed using the perturbed orbit on which $f(t)$ is not constant. This would make the computation very difficult. Instead, since the changes in $f$ are small for a single orbit, we take the approximation

$$\dot{f}(t) = g(t, f_0),$$

We then get

$$f(t) = \int_{0}^{t} g(\tau, f_0) \,d \tau$$

The right hand side can be computed, with some effort.

We also observe the following. Since $g(t)$ is function evaluated on a periodic orbit, it is also periodic, with the orbital period. We can define decomposition

$$g(t) = g_0 + \tilde{g}(t)$$

where $g_0$ is the average of $g(t)$ for an orbit, and $\tilde{g}(t)$ is whatever left of $g(t)$. In other words,

$$g_0 = \frac{1}{T}\int_0^T g(t) \,d t,$$

and

$$\tilde{g}(t) = g(t) - g_0.$$

Clearly, we must have that

$$\frac{1}{T}\int_0^T \tilde{g}(t) \,d t = 0.$$

Then, we have

$$f(t) = g_0 t + \int_{0}^{t} \tilde{g}(\tau) \,d \tau.$$

The first term on the right hand side gives a linear trend in time, as so-called secular perturbation in $f$. The second term gives purely periodic, so-called short-period perturbations. We are usually much more interested in the secular perturbations than in the short-period ones. Note, that instead of time, we can use a linear function of time, for instance the mean anomaly $M$ to compute $g_0$. In other words, we also have

$$g_0 = \frac{1}{2\pi}\int_0^{2\pi} g(M) \,d M.$$

In the following section we will apply this idea to the Laplace vector.