<p>I <em>think</em> 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 <em>foci</em>; 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.</p>

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

CC BY-SA 4.0
2018-10-17T00:15:44.717Z
2018-10-17T00:32:59.067Z
2018-10-17T00:32:59.067Z
HDE 226868
<p>The website <a href="https://astronomy.swin.edu.au/cosmos/S/Semi-major+Axis" rel="nofollow noreferrer">Swinburne University of Technology (AU)</a> gives the definition of semi-major axis as:</p>

<blockquote>
  <p>...half of the longest diameter of an ellipse.</p>
</blockquote>

<p><a href="http://astro.wsu.edu/worthey/astro/html/lec-ellipse.html" rel="nofollow noreferrer">This website</a> give the definition of semi-major axis as:</p>

<blockquote>
  <p>Half of the major axis is called the semi-major axis, and the semi-major axis is also the average sun-planet distance.</p>
</blockquote>

<p>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?</p>

<p>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.</p>

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

<p><strong>Edit:</strong> 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.</p>

CC BY-SA 4.0
2018-10-17T05:00:37.470Z
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](https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes#Average_distance) 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.
PM 2Ring
CC BY-SA 4.0
2018-10-17T05:06:02.163Z
@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.
PM 2Ring
<orbital-elements>
Semi-major axis, vs aphelion, vs mean orbital distance?
BillDOe
# Semi-major axis, vs aphelion, vs mean orbital distance?

(Asked by: username_0 on 2018-10-17)

The website [Swinburne University of Technology (AU)](https://astronomy.swin.edu.au/cosmos/S/Semi-major+Axis) gives the definition of semi-major axis as:

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

[This website](http://astro.wsu.edu/worthey/astro/html/lec-ellipse.html) 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.

**Tags:** orbital-elements
> **Comments:**
> **username_1**: 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](https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes#Average_distance) 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.  
> **username_1**: @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.  

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## 1 answers

### Answer by username_2 on 2018-10-17. Score: 3 | Accepted answer

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 $R\_p=R\_a=a$.


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