The website Swinburne University of Technology (AU) gives the definition of semi-major axis as:

...half of the longest diameter of an ellipse.

This website give the definition of semi-major axis as:

Half of the major axis is called the semi-major axis, and the semi-major axis is also the average sun-planet distance.

But the average sun-planet distance should be the average of the semi-major and semi-minor axes. I've also seen the mean orbital distance described as the semi-major axis in many other places, not just the one cited, so that seems to be a fairly consistent definition, too. But both seem to be inconsistent; how could it be both?

And it also seems to be that the semi-major axis ought to be the same as the aphelion distance, and the semi-minor axis ought to be the same perihelion.

Obviously, the confusion must be mine, but I'm not sure where I'm getting things wrong.

Edit: It occurred to me shortly after posting this and shutting down my computer for the day that (as I'm sure 90% of this SE community knows) the semi-major and semi-minor axes are measured to the center of the ellipse, whereas the aphelion and perihelion are measured to the Sun. I think I'm starting to understand how the semi-major axis can be both the mean orbital radius with respect to the sun and the largest distance from the center of the ellipse.

  • 1
    $\begingroup$ docplayer.net/docs-images/61/46032371/images/2-0.png is the closest thing I could find to a good image. The periapsis and apoapsis add up to the major axis, so half the major axis, aka the semimajor axis, is the average of those two distances, and, it turns out, also the average distance (by symmetry) $\endgroup$
    – user21
    Oct 17 '18 at 4:41
  • $\begingroup$ There are several different ways to define the mean orbital distance, depending on how the averaging is performed. The semimajor axis is an important one, because of its relation to the orbital period, but Wikipedia mentions 2 other options. Unfortunately, I can't find a formula for the mean orbital distance averaging over the distance traveled along the orbit, but I suspect it's not simple, due to the elliptic integral in the ellipse arc length formula. $\endgroup$
    – PM 2Ring
    Oct 17 '18 at 5:00
  • $\begingroup$ @barrycarter Sure, the semimajor axis is the mean of the periapsis & apoapsis distances. But what are you implying by "and, it turns out, also the average distance"? It's certainly not the average distance by time, although it is the average by eccentric anomaly. $\endgroup$
    – PM 2Ring
    Oct 17 '18 at 5:06
  • 1
    $\begingroup$ @PM2Ring Damn, I was hoping no one would call me out on that one. When I typed it, I was thinking the "top half" and "bottom half" of an elliptical orbit is symmetric, which it is, but that's not enough to guarantee the time averaged distance is the length of the semimajor axis. It still might be true (enough stuff might cancel out), but it's not necessarily true. I retract that portion of my statement. $\endgroup$
    – user21
    Oct 17 '18 at 19:22

I think your confusion is just an incorrect (though not uncommon!) visualization of a planetary orbit. The parent star (usually) doesn't lie at the center of the ellipse, but at one of the foci; only for circular orbits are the foci at the same location as the center of the ellipse. Therefore, for non-circular orbits, the aphelion distance is greater than the semi-major axis, and the average of the semi-major and semi-minor axes is the average distance to the center of the ellipse - not the star.

Mathematically, note that the perihelion and aphelion distances are related to the semi-major axis $a$ and eccentricity $e$ by $$R_p=a(1-e),\quad R_a=a(1+e)$$ As a quick check, adding these together and dividing by 2 yields $a$. In the case of a circular orbit, where $e=0$, we have $R_p=R_a=a$.

  • $\begingroup$ Gee willikers, if only someone could draw a diagram :) $\endgroup$
    – user21
    Oct 17 '18 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.