Can one approximate the semi-major axis of an orbit as the average orbital distance for eccentric orbits? Kepler's 3rd Law (regarding the relation between orbital period and the semi-major axis of an orbit) applies to all elliptical orbits. But as I understand it, the only reason it is safe to use the approximation that the semi-major axis of an orbit is approximately equal to the average orbital distance is because the planets of our solar system are generally not very eccentric. 



As an example, the Earth is about 1 AU away from the Sun and has a very low orbital eccentricity, so one can say that the semi-major axis of the Earth's orbit is also approximately 1 AU in length. Since P^2 (yr) = a^3 (AU), one can deduce that the orbit of Earth should take about a year. But, what if the Earth had a higher eccentricity - If the orbital eccentricity of the Earth were higher, would it still be wise to approximate the orbital distance as the semi-major axis of an orbit?
The maths says that the semi-major axis is not a good measure of average distance for high eccentricity (elliptical) orbits.

There are basically two ways to measure this : (1) an average over the entire orbit on a purely geometric basis, and (2) the average over time.  These give quite different results - qualitatively different.

Average distance on geometric basis.

In a way this is the most obvious one.  The mathematics is :

$$R_g = \frac a {2\pi} \int_0^{2\pi}r\,d\theta$$

And we use for an ellipse :

$$r = \frac \rho {1+\epsilon\,cos\theta}$$

The result is (eventually :-) ) :

$$R_g = a\,\sqrt{1-\epsilon^2} \tag{1}$$

Average over time.

This is a tricker calculation, but in fact it's the most "human" average.  We want :

$$R_t = \frac 1 T \int_0^T r\,dt$$

The maths on this is tedious and not at all useful, but the result ( with a little help from Wolframalpha.com ) is :

$$R_t = a\,\frac 2 {\pi} \frac{2+\epsilon^2}{(1-\epsilon^2)^{\frac 5 2}}tan^{-1}\left( \sqrt{\frac {1-\epsilon}{1+\epsilon}} \right) \tag{2}$$

Now by any definition that's a messy equation, but the main point is to note the key difference between the time averaged expression (2) and the geometric one (1).

The geometric expression gets smaller as eccentricity gets larger.

The time averaged one gets larger as eccentricity gets larger (a slight simplification).

In fact for both $\frac R a$ starts off at $1.000$ but when $\epsilon\to 1$ then  $R_g\to 0$ and $R_t\to \infty$ !

Why ?

The geometric expression doesn't allow for the orbital speed of the object.  As you increase eccentricity an orbits ends up with having a faster close approach, so less time close up to it's primary.  As it moves away from the primary, the orbital speed also decreases, so it spends much more time at large distances than at short distances.