<p>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.</p>
<p>So the &quot;size&quot; 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.)</p>

CC BY-SA 4.0
2024-06-12T07:49:36.037Z
So in principle, AU definition is binded to the orbital period of Earth, makes sense.
Agnius Vasiliauskas
CC BY-SA 4.0
2024-06-12T08:51:59.957Z
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**.
Serge3leo
CC BY-SA 4.0
2024-06-12T16:50:40.697Z
@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)".
James K
CC BY-SA 4.0
2024-06-11T21:02:52.857Z
2024-06-11T21:02:52.857Z
James K
<p>Historically, <span class="math-container">$1~\text{AU}$</span> is defined as <span class="math-container">$1/2(r_{min}+r_{max})$</span>, 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 <span class="math-container">$r=\text{const}$</span>. If we want to extrapolate true average radius of Earth &quot;<em>circle orbit to be</em>&quot;, then we need to compare ellipse orbit and circle orbit areas, <span class="math-container">$\pi ab=\pi r^2$</span>, from which we can deduce average distance from center to circle orbit: <span class="math-container">$$\overline r = \sqrt {ab} \tag 1$$</span>,- which obviously is <strong>geometric mean</strong> of ellipse semi-major and semi-minor axis.</p>
<p>Now given Earth orbit axis parameters <span class="math-container">$$ a \approx 149~597~500~\text{km} \\ b\approx 149~576~567 ~\text{km} \tag 2$$</span></p>
<p>we can substitute them into (1) getting geometric mean shift from the current AU definition by <span class="math-container">$$1 AU - \overline r \approx 10~837~\text{km} \tag 3,$$</span></p>
<p>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 <span class="math-container">$10~\text{million meters}$</span> smaller than it is currently defined as 1AU, &quot;average distance to Earth&quot;.</p>
<p>Question: Why it was chosen average distance from Earth to Sun be simply orbit <strong>semimajor-axis</strong> even if circle formed from it area <span class="math-container">$\pi a^2 \neq \pi ab,$</span> current ellipse area and semi-minor axis is <em>completely</em> ignored when defining AU ?</p>
<p>Residual question is about pluses/minuses of having my proposed AU definition instead, which truly averages Earth orbit.</p>

CC BY-SA 4.0
2024-06-11T19:57:17.207Z
Related: https://astronomy.stackexchange.com/a/47107/16685
PM 2Ring
CC BY-SA 4.0
2024-06-11T20:00:03.123Z
Also see https://en.wikipedia.org/wiki/Gaussian_gravitational_constant
PM 2Ring
CC BY-SA 4.0
2024-06-11T23:40:12.887Z
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?"
notovny
CC BY-SA 4.0
2024-06-12T00:31:19.487Z
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.
Greg Miller
CC BY-SA 4.0
2024-06-12T01:39:51.533Z
See https://en.wikipedia.org/wiki/Astronomical_unit
Walter
CC BY-SA 4.0
2024-06-12T07:32:05.983Z
@GregMiller They are extrapolated from perihelion and aphelion, as $a=0.5(r_{min}+r_{max}),~b=\sqrt{r_{min} \cdot r_{max}}$. Perihelion and aphelion data is taken from NASA Earth facts sheet.
Agnius Vasiliauskas
CC BY-SA 4.0
2024-06-13T02:49:26.777Z
This fact sheet? https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html That has 6 digit mean perihelion & aphelion data (& only 3 digits for the eccentricity), so you can't get 9 digit ellipse axes from it. But as Greg says, the Earth's orbit is dynamic. Its actual shape is only approximately an ellipse, and its parameters vary over time.
PM 2Ring
CC BY-SA 4.0
2024-06-13T02:54:34.613Z
As I've said elsewhere on this site, the Earth's orbit around the Sun is only approximately a Keplerian ellipse. A better approximation considers the EMB, the Earth-Moon barycentre (their centre of mass), as the entity orbiting the Sun, and then we look at the orbit of the Earth and Moon around that barycentre. I have some info on the variation in perihelion & aphelion: https://astronomy.stackexchange.com/a/49605/16685 & on the variation in the lunar orbit: https://astronomy.stackexchange.com/a/55112/16685
PM 2Ring
CC BY-SA 4.0
2024-06-13T06:18:49.167Z
@PM2Ring As I understand it 6 significant digits of perihelion and aphelion data limits ellipse axis accuracy to the 6 digits or less, but it does not mean, that you can't extrapolate it at all, just that it will be a "crude estimate". This question is principal and not about accuracy, but more about which method should be employed to calculate 1 AU or in other words - How to average Earth orbit's radius and Why this averaging method was chosen (or is best).
Agnius Vasiliauskas
CC BY-SA 4.0
2024-06-13T08:06:13.413Z
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](https://en.wikipedia.org/wiki/Areal_velocity) is proportional to the angular momentum.
PM 2Ring
<the-sun><distances><definition><earths-orbit>
Distance from Earth to Sun averaging method in AU definition
Agnius Vasiliauskas
# Distance from Earth to Sun averaging method in AU definition

(Asked by: username_0 on 2024-06-11)

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.

**Tags:** the-sun, distances, definition, earths-orbit

> **Comments:**
> **username_1**: Related: https://astronomy.stackexchange.com/a/47107/16685  
> **username_1**: Also see https://en.wikipedia.org/wiki/Gaussian_gravitational_constant  
> **username_2**: 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?"  
> **username_3**: 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.  
> **username_4**: See https://en.wikipedia.org/wiki/Astronomical_unit  
> **username_0**: @GregMiller They are extrapolated from perihelion and aphelion, as $a=0.5(r_{min}+r_{max}),~b=\sqrt{r_{min} \cdot r_{max}}$. Perihelion and aphelion data is taken from NASA Earth facts sheet.  
> **username_1**: This fact sheet? https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html That has 6 digit mean perihelion & aphelion data (& only 3 digits for the eccentricity), so you can't get 9 digit ellipse axes from it. But as Greg says, the Earth's orbit is dynamic. Its actual shape is only approximately an ellipse, and its parameters vary over time.  
> **username_1**: As I've said elsewhere on this site, the Earth's orbit around the Sun is only approximately a Keplerian ellipse. A better approximation considers the EMB, the Earth-Moon barycentre (their centre of mass), as the entity orbiting the Sun, and then we look at the orbit of the Earth and Moon around that barycentre. I have some info on the variation in perihelion & aphelion: https://astronomy.stackexchange.com/a/49605/16685 & on the variation in the lunar orbit: https://astronomy.stackexchange.com/a/55112/16685  
> **username_0**: @PM2Ring As I understand it 6 significant digits of perihelion and aphelion data limits ellipse axis accuracy to the 6 digits or less, but it does not mean, that you can't extrapolate it at all, just that it will be a "crude estimate". This question is principal and not about accuracy, but more about which method should be employed to calculate 1 AU or in other words - How to average Earth orbit's radius and Why this averaging method was chosen (or is best).  
> **username_1**: 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](https://en.wikipedia.org/wiki/Areal_velocity) is proportional to the angular momentum.  

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## 1 answers

### Answer by username_5 on 2024-06-11. Score: 6 | Accepted answer

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.)

> **Comments:**
> **username_0**: So in principle, AU definition is binded to the orbital period of Earth, makes sense.  
> **username_6**: 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**.  
> **username_5**: @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)".  

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