I am using the Earth-Sun distance to teach kids the usefulness of trigonometry. This site is helpful.

But one of the more engaged asked what the most accurate measurement is. After some research, I found that elapsed time measurements of reflected radar returns seems to be the way we do it now. By measuring the distance to another planet this way and combining the parallax method, we can calculate accurate distance to the sun. But is this THE most accurate way to measure the AU? What is the error margin? If the AU were changing by .0000001 percent annually, could we detect it?

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    $\begingroup$ The Astronomical Unit is no longer measured, it's defined. It is exactly 149 597 870 700 meters. $\endgroup$
    – notovny
    Commented Oct 17, 2021 at 13:42
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    $\begingroup$ If it was defined, why wasn't it rounded to 1.5E11? :D $\endgroup$
    – johnDanger
    Commented Oct 19, 2021 at 17:54
  • $\begingroup$ @johnDanger because it wasn't that long when they changed from measuring it to defining it. $\endgroup$
    – masher
    Commented Oct 20, 2021 at 6:57
  • $\begingroup$ @masher, it surely wasn't 149 597 870 700 meters though either? I wonder if they rounded it to such purely based on measurement uncertainties or if there was some sort of practical consideration. $\endgroup$
    – johnDanger
    Commented Dec 6, 2021 at 18:16

3 Answers 3


The astronomical unit was defined to be exactly 149 597 870 700 meters in 2012, so it is no longer a measured value. In fact, it never was a measured value. It was instead a computed value, and it was computed rather strangely.

Prior to 2012, the astronomical unit was defined as the distance between the center of the Sun at which a tiny particle in an unperturbed circular orbit about the Sun would yield a value of exactly 0.0172020985 for the Gaussian gravitational constant. That value corresponds to the distance at which a tiny particle in an unperturbed circular orbit about the Sun would have an orbital angular velocity of 0.0172020985 radians per solar day. That value corresponds to an orbital period of 365.256898 days. That value, now called a Gaussian year, was based on measurements available to Gauss of the length of a sidereal year. (The currently accepted value of the sidereal year is 365.256363004 days of 86400 seconds each.)

This outdated value in the length of a sidereal year was one of the reasons the astronomical unit was given a defined value in 2012. There were other reasons. One is that that definition made the concept of the astronomical unit a bit (more than a bit?) counter-intuitive. With that definition, uncertainty in the computed value of astronomical unit depended on the ability of solar system astronomers to estimate the Sun's gravitational parameter (conceptually, the product of the Newtonian gravitational constant and the Sun's mass; in practice, a quantity that could be estimated directly as a consequence of models used to generate ephemerides) and the ability to measure time.

The ability to measure time (currently about one part in $10^{16}$) has far outpaced the ability to estimate the Sun's gravitational parameter (currently less than one part in $10^{10}$). This means that if the 2012 change had not happened, the uncertainty in the astronomical unit would depend on the ability estimate the Sun's gravitational parameter, about three parts in $10^{11}$, or about 0.000000003%. That's a lot better than the 0.0000001% precision asked about in the question.

  • $\begingroup$ Thank you. So the AU was defined as the radius for a hypothetical orbit of a hypothetical particle that required 365.256898 days to complete that orbit. To make matters more ridiculous, this particle would have the mass of the earth so that it could obey celestial mechanics. But then what is the current method for measuring the distance to the sun with an accuracy of 1 part in 3 x 10^11? $\endgroup$ Commented Oct 17, 2021 at 21:17
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    $\begingroup$ @aquagremlin The middle part is incorrect ("To make matters more ridiculous, this particle would have the mass of the earth so that it could obey celestial mechanics."). The particle has negligible mass. Keep in mind that the ratio of the Earth's mass to that of the Sun is about $3\times10^-6$, about twice the error in the Gaussian year. Also keep in mind that telescopes and timekeeping weren't all that precise in the mid 19th century. Gauss's value was very good considering the timeframe. $\endgroup$ Commented Oct 17, 2021 at 21:38
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    $\begingroup$ Developers of ephemerides beginning in the late 19th century elected to use the Gaussian gravitational constant as a given, even though timekeeping had progressed a bit. The point of ephemerides at that point was to know where to point telescopes at specific points in time. (That remains a goal, but milliarcsecond astronomy has made that requirement considerably more strict.) $\endgroup$ Commented Oct 17, 2021 at 21:43
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    $\begingroup$ Developing a good ephemeris model is extremely complicated, and has been getting ever more complicated as time goes on. JPL has been doing this since near the start of the space age. For example, JPL created its DE101 ephemeris model in 1961. (It wasn't published until 1983.) Note well: One of the authors of that 1983 paper has his name on the 2021 paper to which I linked in a previous comment. People study this concept in graduate school and then spend their entire career refining the concepts. $\endgroup$ Commented Oct 17, 2021 at 22:25
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    $\begingroup$ @MarkFoskey The Earth's mass is not quite that small. The older definition of the astronomical unit explicitly assumed a particle of negligible mass. In other words, a little tiny rock rather than the Earth. $\endgroup$ Commented Oct 19, 2021 at 1:10

As mentioned in a comment, the AU is no longer measured, it is defined, as 149 597 870 700 metres, or exactly the distance that light travels in a vacuum in $499 \frac{9823}{2053373}$seconds.

But I suppose your question is about the accuracy of the measurement of the distance between the Earth and Sun.

As noted in the linked question, the measured value of the Earth's Semimajor axis is 149,597,887,500 m, A measured difference on the order of 0.00001% from the defined AU. This is found by radar measurements of the planets, combined with a form of Kepler's third law, that relates the distances to the planets to their orbital periods. (Kepler's third law isn't actually precise enough for this, you need a Newtonian n-body model of the solar system) The basic idea is that if you know the relative positions of the Earth and Venus you can model their orbits, and this allows you to calculate the size of their orbits, and hence the Astronomical Unit (in the old definition)

  • $\begingroup$ Im sorry but I could not decode the answer from your statement. If we currently know the distance to the sun is 149,597,887,500 m, is it accurate to 1 meter, 10 meters? 100 meters? 1 kilometer? Where can I read more about how that measurement's accuracy is determined? $\endgroup$ Commented Oct 17, 2021 at 18:00
  • $\begingroup$ I read this astronomy.stackexchange.com/questions/35045/… but it did not enlighten me either. $\endgroup$ Commented Oct 17, 2021 at 18:05
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    $\begingroup$ @aquagremlin That question was misinformed in the sense that the questioner thought we could directly measure the distance to the Sun via radar. We cannot. I suspect we'll be able to ping Ultima Thule with radar before with can ping the Sun. For example, see Why can we not measure the distance to the Sun directly using Radar? at Physics.SE. $\endgroup$ Commented Oct 17, 2021 at 19:01
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    $\begingroup$ Unsourced answers are detrimental to Stack Exchange. How do future readers know the author didn't just make some or all of this up? Where can they confirm the numerical values written here? Where did they come from? This could be a great Quora answer but in Stack Exchange answers we should support our assertions with authoritative sources. Our goal as answer authors is not to show that we know the answer ourselves, it is to provide future readers with high quality answers, not just unsupported assertions. $\endgroup$
    – uhoh
    Commented Oct 17, 2021 at 22:34
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    $\begingroup$ @aquagremlin To give you a clear answer to your first comment: we don't know that the distance to the sun is 149597887500 m. What we know, because we've defined it that way, is that the unit of length called "astronomical unit" a.k.a. "AU" is exactly 149597887500 m. But that unit of length (AU) is not exactly the same as the distance to the sun. The actual distance to the sun at any given moment might be, say, 1.0001... AU or 0.997... AU or so on. $\endgroup$
    – David Z
    Commented Oct 19, 2021 at 0:22

As Wikipedia says, the Astronomical unit is "roughly the distance from Earth to the Sun".

 The astronomical unit was originally conceived as the average of Earth's aphelion and perihelion; however, since 2012 it has been defined as exactly $149\,597\,870\,700$ m.

From the JPL Solar System Dynamics site:

The average Sun-Earth distance is not an exact quantity because the orbit of the Earth about the Sun is not exactly elliptical due to changing perturbations by other planets and because general relativity slightly modifies the elliptical solutions obtained from Newton's theory of gravity.

The astronomical unit (AU) is a convenient unit for approximating orbits around the Sun. According to Kepler's 3rd law, the orbital period $T$ of a planet in an orbit with semi-major axis $a$ is given by $$T^2=a^3$$ where $T$ is in years, and $a$ is in AU. However, that equation neglects the mass of the planet. That's fine for a rough approximation, since the Sun is far more massive than any planet (even the Sun / Jupiter mass ratio is over 1000). Such a system, which treats the mass of the orbiting body as negligible, is called a one-body system.

Fortunately, Newton discovered that a two-body gravitational system could be "reduced" to a pair of one-body systems, where we treat one of the bodies as being fixed at the focus of the orbit. The Newtonian version of the previous equation for a two-body system is $$a^3=\frac{G(m_1+m_2)}{4\pi^2}T^2$$ where $m_1$ and $m_2$ are the masses of the two bodies, and $G$ is the universal gravitational constant.

Using Newtonian mechanics, astronomers were able to calculate the relative distances and masses of bodies in the Solar System. The AU was a convenient way to express those relative distances. Although they were certainly curious to know the true size of the AU, they didn't need to know it in order to predict the positions of bodies on the celestial sphere.

It was realised that by making very careful parallax observations that the length of the AU could be estimated. The transit of Venus is an ideal phenomenon for this because the amount of parallax is relatively large, if you make several simultaneous measurements from widely-separated points on the Earth. Of course, this requires that you can accurately calculate the distance between your observation posts, and can determine the timings of the transit with sufficient precision.

The AU is a convenient "measuring rod" for the Solar System. Although we say it's the mean distance from the Earth to the Sun, that does not imply that it's the time-averaged distance, even in an idealised two-body model of the Sun & Earth.

From Wikipedia Semi-major and semi-minor axes:

It is often said that the semi-major axis is the "average" distance between the primary focus of the ellipse and the orbiting body. This is not quite accurate, because it depends on what the average is taken over.

  • averaging the distance over the eccentric anomaly indeed results in the semi-major axis.

  • averaging over the true anomaly (the true orbital angle, measured at the focus) results in the semi-minor axis, $b=a{\sqrt {1-e^{2}}}$

  • averaging over the mean anomaly (the fraction of the orbital period that has elapsed since pericentre, expressed as an angle) gives the time-average, $a\left(1+{\frac {e^{2}}{2}}\right)$.

In the above equations, $e$ is the eccentricity of the ellipse. For Earth's orbit, $e\approx 0.0167$, so all of those averages are roughly equal.

Of course, the Earth's true motion relative to the Sun is far more complicated than a simple two-body system. Firstly, the Moon's mass is relatively large: the Moon / Earth mass ratio is $\approx 0.0123$, and the mean distance from the centre of the Earth to the Earth-Moon barycentre is $\approx 4\,670$ km. So with just those three bodies, we no longer have a simple mean Sun-Earth distance.

And then we need to include perturbations from the other Solar System bodies. FWIW, the analytical theory of the Moon's motion is quite complex: "The number of terms needed to express the Moon's position with the accuracy sought at the beginning of the twentieth century was over 1400".

The best modern ephemeris calculations do not use perturbed Newtonian ellipses. Instead, they integrate the equations of motion of the major masses in the Solar System. The Jet Propulsion Laboratory Development Ephemeris takes into account the 343 most massive asteroids in its calculations. You can freely access their data, spanning from 9999 BC to 9999 AD, via the Horizons system.

Here's a plot of the Sun-Earth distance, generated using Horizons, for the period 2020-Apr-3 to 2022-Apr-3. The mean (time-averaged) distance over that period is $\approx 149\,619\,270$ km.

Sun-Earth distance

If you want to generate your own plots, here's a live link to my Sage / Python script. There are some instructions on using the script in this Space Exploration answer.

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    $\begingroup$ Thank you for your insight into this problem. It is interesting how our methods change over time as we reach for ever greater precision. The qualitative differences of the solutions are remarkable. They also highlight a recurring theme - as we reach ever greater precision and resolve our answer ever more, we find intrinsic variations that render our answer nebulous. $\endgroup$ Commented Oct 18, 2021 at 10:42
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    $\begingroup$ I think the important point here (that seems to be missing from other answers so far) is that the exact length of the AU is much less important than its use as a relative value. So we can say that for instance the orbit of Mars is (per Google) is 1.524 AU, without knowing the exact length of the AU. $\endgroup$
    – jamesqf
    Commented Oct 18, 2021 at 16:51
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    $\begingroup$ @jamesqf I also wanted to point out that the semi-major axis $a$ is only the mean distance between the orbiting body & its primary for a simple elliptical orbit, and even in that case it's just the mean of the minimum & maximum distances between the body centres, not the time-averaged distance. FWIW, the time-averaged mean Sun-Earth distance is ~21,000 km greater than the mean of its perihelion & aphelion distances. $\endgroup$
    – PM 2Ring
    Commented Oct 18, 2021 at 23:14
  • $\begingroup$ @PM 2Ring: You know, I'd never thought of that :-) But once I do, it seems obvious - comets being an extreme case. $\endgroup$
    – jamesqf
    Commented Oct 18, 2021 at 23:28
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    $\begingroup$ @jamesqf it's much like light years (or PS), which you'll rarely see to more than 1 or 2 decimal places because when you are dealing with distances where light years become useful there rarely is a measurement precise enough for it to matter beyond that. In practice, much like AU, its usefulness comes from being able to quickly get a rough understanding of how far something really far away is. $\endgroup$
    – eps
    Commented Oct 19, 2021 at 14:40

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