Historically, $1~\text{AU}$ is defined as $1/2(r_{min}+r_{max})$, but the problem is that this is not the average radius if Earth orbit would be transformed into circle, with Sun at the center and Earth flying around with $r=\text{const}$. If we want to extrapolate true average radius of Earth "circle orbit to be", then we need to compare ellipse orbit and circle orbit areas, $\pi ab=\pi r^2$, from which we can deduce average distance from center to circle orbit: $$\overline r = \sqrt {ab} \tag 1$$,- which obviously is geometric mean of ellipse semi-major and semi-minor axis.

Now given Earth orbit axis parameters $$ a \approx 149~597~500~\text{km} \\ b\approx 149~576~567 ~\text{km} \tag 2$$

we can substitute them into (1) getting geometric mean shift from the current AU definition by $$1 AU - \overline r \approx 10~837~\text{km} \tag 3,$$

which means that if Sun would be in the center of circle and Earth had an ideal circle orbit,- then this orbit radius would be by over $10~\text{million meters}$ smaller than it is currently defined as 1AU, "average distance to Earth".

Question: Why it was chosen average distance from Earth to Sun be simply orbit semimajor-axis even if circle formed from it area $\pi a^2 \neq \pi ab,$ current ellipse area and semi-minor axis is completely ignored when defining AU ?

Residual question is about pluses/minuses of having my proposed AU definition instead, which truly averages Earth orbit.

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    $\begingroup$ Related: astronomy.stackexchange.com/a/47107/16685 $\endgroup$
    – PM 2Ring
    Commented Jun 11 at 19:57
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    $\begingroup$ Also see en.wikipedia.org/wiki/Gaussian_gravitational_constant $\endgroup$
    – PM 2Ring
    Commented Jun 11 at 20:00
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    $\begingroup$ It's far more important that the astronomical unit be a consistent distance that jives with previous values than it be continually changing year after year. If a rogue planet passes through the solar system, altering the Earths orbit and casting it into the void of interstellar space, the last astronomers scrounging for pails of air on the frozen, lifeless world will say, "Yep, still 149 597 870 700 m, why do you ask?" $\endgroup$
    – notovny
    Commented Jun 11 at 23:40
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    $\begingroup$ Where did you get "a" and "b" from? These are also averages, approximation, or arbitrarily pulled from an instant. Earthorbit is constantly changing, so the AU as a physical entity isn't really that useful today. It was more useful when we didn't know the scale of the solar system. $\endgroup$ Commented Jun 12 at 0:31
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    $\begingroup$ Ok. But as James says, it makes sense to define the mean radius as the (mean ellipse) semi-major axis because of its important connection to the period. We normally use a in ellipse-based orbit calculations. The total ellipse area isn't that important; OTOH, the rate of change of the area is proportional to the angular momentum. $\endgroup$
    – PM 2Ring
    Commented Jun 13 at 8:06

1 Answer 1


Because the orbital period depends on the semimajor axis. Two orbits of small bodies around the sun with the same semimajor axis will have the same orbital period.

So the "size" of the orbit is best described by this parameter, rather than the area of the ellipe (or the curved length, or the average distance of the planet over time, or any other alternative ways of defining the AU.)

  • $\begingroup$ So in principle, AU definition is binded to the orbital period of Earth, makes sense. $\endgroup$ Commented Jun 12 at 7:49
  • $\begingroup$ Everything is right. But in response, it would be worth emphasizing that such a definition of AU simplifies the application of Kepler's third law. $\endgroup$
    – Serge3leo
    Commented Jun 12 at 8:51
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    $\begingroup$ @AgniusVasiliauskas Yes, the old defintion of AU was "the radius of an unperturbed circular Newtonian orbit about the Sun of a particle having infinitesimal mass, moving with an angular frequency of 0.017 202 098 95 radians per day (or one orbit in 365.256 898 3 days)". $\endgroup$
    – James K
    Commented Jun 12 at 16:50

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