Talk:Astronomical unit

"ua" vs. "au"

Since the NIST Guide in the links uses "ua" as the symbol for this unit, and the IAU Style Manual in the links uses "au" as the symbol for this unit, why does this article use a different symbol, "AU"?

Why doesn't it at least mention those other symbols? Gene Nygaard 03:35, 17 Dec 2004 (UTC)

In practice, "AU" or "au" is universal. Using "ua" would cause no end of confusion. The IAU is the authority for astronomy.

As for whether to use AU or au, it seems that Astronomy magazine uses "AU" and Sky and Telescope uses "a.u." (!) So as they say, "go figure". -- Curps 21:41, 26 Dec 2004 (UTC)

One reason why it may be suitable for ua (and why the above guide uses ua) is the international country code for Australia is au, causing confusion. But, as said previously, AU is most used. So that's how it stands.

Taylor 08:27, 3 August 2005 (UTC)

"Ua" is the international code for Ukraine. So using "ua" does not solve the problem, which luckily never existed in the first place. -- Heptor _talk 18:54, 29 April 2013 (UTC)

removed Francophobic comments re: the Bureau des Poids et Mesures, and replaced it with the more neutral explanation. — Preceding unsigned comment added by [[User:{{{1}}}|{{{1}}}]] ([[User talk:{{{1}}}|talk]] • [[Special:Contributions/{{{1}}}|contribs]])

The IAU style manual of 1989 pdf file given presently as ref 3 says in Section 5.15, pS24, says "The symbol for the astronomical unit is au". So i'm correcting the lead, which is wrong. La version francophone n'est pas apparement acceptee par l'UAI. Boud (talk) 22:14, 20 February 2011 (UTC)

It looks like the French translation has crept back in. I removed it again. Heptor _talk 21:44, 29 April 2013 (UTC)

I added a 'citation needed' for the comment that units are usually capitalized only for those that are named after a specific person. This is most assuredly not a universal practice, almost all electrical units for example are not capitalized despite being named after someone, like coulomb and ampere. I would say most acronyms are capitalized, and AU is an acronym. Nerfer (talk) 15:54, 18 April 2011 (UTC)

The 2012 IAU RESOLUTION B2 on the re-definition of the astronomical unit of length, recommends "that the unique symbol “au” be used for the astronomical unit." It looks like we've got a difference between the IAU (2012) recommendation and the BIPM (2006) report. I suspect the BIPM and the IAU will sort this out, but for the moment we probably should let both stand with the most recent IAU recommendation getting some priority. --SteveMcCluskey (talk) 17:43, 11 May 2013 (UTC)

estimates of AU (old discussion)

I came across a web-page that show the best AU is different from 149 597 870 691 ± 30 metres.

The address list below. And the contents show between two lines. http://www.scenta.co.uk/tcaep/nonxml/science/constant/details/Astronomical%20Unit.htm

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Science
Constants

Name	Astronomical Unit
Symbol	AU
Value	1.495 978 706 60 ± 0.000 000 000 20 × 1011 m
Category	Astrophysical
Comments	The astronomical unit is defined as the mean distance between the Earth and the Sun. The mean distance between the centre of mass of the Sun and the centre of mass of the Earth-Moon system is 1.000 0023 AU. The best measurement of the astronomical unit comes from phase-modulated continuous-wave radio signals reflected off other planets. It was continuous-wave signals returned from the Viking probes on Mars that allowed the determination of the astronomical unit to 20m.

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YuWong yuwong36@yahoo.com

Do you know how to read scientific notation? The values are the same, except the one at your link is 30 m less, and it’s ±20 instead of 30. It’s an imperceptible difference, really (unless you do it for a living). —Frungi 04:57, 25 February 2006 (UTC)

How did they do it in ancient times

How did people measure the distance of the sun from the earth accurately in ancient times? Does anyone know? —The preceding unsigned comment was added by 81.179.39.220 (talk • contribs) 13:51, 19 January 2006 (UTC).

In ancient times, people didn’t believe in the solar system. Except for Galileo, which was why they ganged up on him. —Frungi 04:14, 25 February 2006 (UTC)

Well, not really. Ancient Greeks did know how to calculate the distance of the Sun from the Earth (they already knew that the Earth revolves around the Sun). They also knew that the Earth was round and they knew how to calculate its diameter. Aristarchus of Samos, who lived circa 310 BC - 230 BC used a method for determining the Earth-Sun distance, which had to do with measuring the angle in the sky between the Sun and the Moon when it is precisely in its first quarter phase. This was a particularly difficult thing to measure accuratly at that time. You would also have to know the Earth-Moon distance (which he did) to perform the calculation. Even though his measures weren't extremely accurate, the model was correct. And it's quite impressive that a man, some 2200 years ago, without telescopes, instruments or even a calculator, actually managed to find out how far away the Sun is. Leschatz 15:17, 5 March 2006 (UTC)

Message regarding "acceptance" of the value I edited here:

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The value I wrote in for

"AU = 149 597 870 696.0 +/- 0.146"

is from Russia, figured from their ephemeris called "EPM2004" (of 2004). The newest NASA value (figured using their ephemeris called "DE410" of 2003) is

"AU = 149 597 870 697.4 +/- 0.3"

which is do to work headed by Myles Standish of the JPL. Both values were presented at the IAU Colloquium 196. In the transcript of the discussions between the Russians and Myles Standish at that meeting, the Russians tell him his value has a uncertainty of "1.4" ( and thus by implication, that their value is better). He said "perhaps the extra digit they have is significant, I am not sure".

 Thus I choose to use the Russian value for the AU.
                       Sincerely, EGB 

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The Americans have more spacecrafts in deep space and better radio ranging technology. The Russians have first class mathematicians though, nobody would argue with that. In any case, deciding which is the better value is something even the IAU didn't want to do. DonPMitchell (talk) 04:06, 28 May 2010 (UTC)

astronomical units

how long does it take to travel one astronomical unit in measures of time (minutes, hours, days, months, etc.) ??????????????? —The preceding unsigned comment was added by 71.168.192.207 (talk) 00:19, 6 March 2007 (UTC).

>> Depends how fast you're moving. (Darktachyon 15:24, 5 May 2007 (UTC))

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conversion

The conversion of AU to angstroms is entirely unnecessary. (Darktachyon 15:24, 5 May 2007 (UTC))

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Proposed WikiProject

Right now the content related to the various articles relating to measurement seems to be rather indifferently handled. This is not good, because at least 45 or so are of a great deal of importance to Wikipedia, and are even regarded as Vital articles. On that basis, I am proposing a new project at Wikipedia:WikiProject Council/Proposals#Measurement to work with these articles, and the others that relate to the concepts of measurement. Any and all input in the proposed project, including indications of willingness to contribute to its work, would be greatly appreciated. Thank you for your attention. John Carter 20:58, 2 May 2007 (UTC)

Definition of AU

The JPL appears to define the AU as the distance travelled by light in 499.004 783 806(10) s... I don't know the subject well enough to wish to amend the article, but could someone more knowledgeable check this out? Thanks! Physchim62 (talk) 09:50, 29 May 2007 (UTC)

That's true, because what is actually measured is the light travel time. In fact, this comes down to the same thing as defining it as a number of metres, because the metre is defined as the distance travelled by light in 299792458 seconds 1/299792458 of a second. I don't think it's an important enough distinction to be worth mentioning here. Cosmo0 16:32, 19 September 2007 (UTC)

Yeah, um, no: the meter isn't defined as light's traverse of nearly 300 million seconds, turning it into molasses; it's its traverse 'cross one-nearly-three-hundred-millionth of one second. Wink. —Preceding unsigned comment added by 69.148.169.120 (talk) 03:58, 28 May 2008 (UTC)

Oops. Cosmo0 (talk) 09:49, 28 May 2008 (UTC)

The above JPL time constant in the link source is now displayed as 499.004 783 836 s, derived by dividing the AU distance by the speed of light. The site now states: "Last Updated: 2012-Oct-24." A 3 m margin of error in the AU would correspond to a 0.010 μs (3/299 792 458) margin of error for time. — Glenn L (talk) 11:44, 8 November 2012 (UTC)

Definition

I changed the intro to make the first sentence a little clearer to the lay reader and moved the bit about an AU being slightly less than the semi-major axis of the earth's orbit to below the full definition. I think it would confuse some readers if the first sentence of the article defined an AU as being “almost equal to” something else. Cosmo0 10:18, 22 September 2007 (UTC)

million vs billion

I noticed that the distance is written as almost 150 billion kilometers, but it says just after that it is 150 million. I am not an expert, and so will not make an edit one way or the other, but would someone check the value?Llama (talk) 01:53, 14 April 2008 (UTC)

It is written as 150 million km or 150 billion meters - but be careful - in the U.S. a "billion" is 10^9 but in England it is 10^12 - so best not to use the term. Use "mega", "giga" and so on. Anyway it is OK and consistent as it is.Carrionluggage (talk) 03:15, 14 April 2008 (UTC)

Is there any way we can make this less confusing somehow? It does seem to catch a lot of people out, judging by the number of times it gets 'corrected'. Do we even need to give the value in both metres and kilometres, given that converting from one to the other is not exactly difficult? Cosmo0 (talk) 20:48, 15 April 2008 (UTC)

I agree, there is no need to have both. In my eyes, meters should be prefered over kilometers as they are the SI unit for distance.--Fogeltje (talk) 06:38, 16 April 2008 (UTC)

Meters might be the SI unit, but to me it sounds a bit ridiculous because it gives the impression that the distance is known (and constant) to an accuracy that warrants the use of "meters". Even for distances on earth, in the order of kilometers, one would never use "meters"... JH-man (talk) 07:34, 16 April 2008 (UTC)

Gotta agree with JH; meter may be a human-scale distance unit, but we ought to go with larger scales, kilometers at least. This after all being astronomy, and not dustin' crops, boy. —Preceding unsigned comment added by 69.148.169.120 (talk) 04:05, 28 May 2008 (UTC)

Looks like there's an error in the Examples section

Kind of a somewhat minor nitpick, I suppose, but consider the following two lines:

1. 50,000 AU: possible closest estimate of the "Outer Oort Cloud" limits (1.0 ly)
2. 100,000 AU: possible farthest estimate of the "Outer Oort Cloud" limits (1.6 ly).

It doesn't appear to me that it can be mathematically true that 50,000 AU equals 1.0 ly (to one significant digit) if 100,000 AU = 1.6 ly (to one significant digit). Basic multiplication would suggest that either 50,000 AU is approx 0.8 ly (maybe .7 or .9 due to rounding), or 100,000 AU is about 2.0 ly (I could see it being 1.9 ly due to rounding). Anyhow, later in the article, we see: 1 light-year ≈ 63,241 AU so that gives us 50,000 ≈ .8 ly. —Preceding unsigned comment added by 216.136.68.194 (talk) 18:11, 23 June 2008 (UTC)

Fixed. Cosmo0 (talk) 20:00, 23 June 2008 (UTC)

Timeline

There is an inconsistency in the timeline of the history section. It says that the richer and cassini performed the first estimate of the AU, yet below it says Horrocks did it 10 years earlier... This also desperately needs citations!! —Preceding unsigned comment added by Lzkelley (talk • contribs) 21:39, 25 January 2009 (UTC)

Thanks for pointing that out. The confusion seems to arise because Horrocks attempted to estimate the AU, but his method was flawed and he got the wrong answer. I guess whether you still consider that an estimate or not is a moot point, but I've revised the text to say that "The value of the AU was first estimated with reasonable accuracy by Jean Richer and Giovanni Domenico Cassini in 1672" (no bolding in main article). I didn't have time to look for citations, but I agree it needs some. Cosmo0 (talk) 12:38, 27 January 2009 (UTC)

What's that (6) in the figure?

I've checked the reference, and it says the value has an uncertainty of 6 metres, so plus/minus 3 metres? But I had no clue what the (6) meant until I downloaded the reference pdf, then searched it, then found the table on page 4!

I've no idea how to use this:

• {{val|1.49597870691|(6)|e=11|ul=m}}

Could someone educate me? Is it being used correctly here?

I see that the rounded figure is given in kilometres, then the accurate figure in metres. Definite room for confusion there - I made the mistake of copy & pasting the accurate figure and missed the metres unit - oops! HarryAlffa (talk) 21:03, 3 May 2009 (UTC)

OK! First time use of these template thingys! Above gives a significance of the figure, not an uncertainty. I took away the parenthesis around the (6), which makes it +-6, should it be +-3? HarryAlffa (talk) 21:22, 3 May 2009 (UTC)

I think ±6 is correct, at least according to the source cited. NIST, however, give the value as 1.49597870691(6)×10¹¹ m, while claiming to get it from the same source. So either they're using a bizarre notation or one of them is wrong. Apparently this means the same thing in the notation they use, although it's ambiguous IMO.

Be careful with the template, though: how you wrote it implied an error of 6×10¹¹ m! I've reformatted it as 149597870691±6 m, which seems to be the most sensible way to fit in the small error.

Yes! The very same occured to me before bedtime. I thought that maybe a word in the template authors ear to put the +-val after the exponent! HarryAlffa (talk) 13:33, 5 May 2009 (UTC)

Or simply use the parentheses convention rather than the misleading +/-… Physchim62 (talk) 06:01, 6 May 2009 (UTC)

And yes, the use of both metres and kilometres in the introduction has been the source of much confusion in the past!

Cosmo0 (talk) 13:07, 5 May 2009 (UTC)

Units

On the sidebar of the page, it shows this:

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SI units
149.60×10^6 km 149.60×10^9 m
Astronomical units
4.8481 × 10−6 pc 15.813×10^−6 ly
US customary / Imperial units
92.956×10^6 mi 490.81×10^9 ft
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I don't know how to edit it, but this should be:

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SI units
1.4960×10^8 km 1.4960×10^11 m
Astronomical units
4.8481 × 10^−6 pc 15.813×10^−6 ly
US customary / Imperial units
9.2956×10^7 mi 4.9081×10^11 ft
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This can be found on any site explaining significant figures, including wikipedia:
http://en.wikipedia.org/wiki/Significant_figures —Preceding unsigned comment added by 24.76.182.175 (talk) 22:46, 7 March 2010 (UTC)

But see Engineering notation.

I have no idea why this convention is used, but the box is generated by a site-wide template so it can't be changed here. Cosmo0 (talk) 21:39, 8 March 2010 (UTC)

Jabir ibn Aflah and Mu’ayyad al-Din al-’Urdi

The article claimed that these two astronomers realized that Ptolemy grossly underestimated the distance from the Earth to the Sun. I took this out because it is a misreading of the source, which is p. 237 of A History of Arabic Astronomy: Planetary Theories During the Golden Age of Islam, George Saliba, New York University Press, 1994, and states:

We have also seen above that Ptolemy had assumed the apparent size of the solar disk to be invariable at all geometric distances of the Sun, namely 0;31,20°. In addition to this approximation, there were also the faulty numbers associated with Ptolemy's grossly underestimated real geocentric distance of the Sun. As a result of that, at least two medieval astronomers, namely Jābir Ibn Aflaḥ (fl. 1150) and ʿUrḍī (d. 1266), argued against Ptolemy's order of the planets and, by using Ptolemy's own figures, defended an order in which Venus was to be placed above the Sun.

The source says that Ptolemy's figure for the earth-sun distance caused problems with his cosmological model, which Jābir Ibn Aflaḥ and ʿUrḍī reacted to. But, it does not say that they were aware that this figure was a gross underestimate. To further explain, let's look at what ʿUrḍī did. According to van Helden (p. 33, Measuring the Universe, University of Chicago Press, 1985, ISBN 0-226-84881-7):

Al-ʿUrḍī recalculated all the distances from parameters which he redetermined, making sure to add the diameter of each heavenly body to the calculated thickness of its sphere. He found that there was not enough room between Mercury and the Sun for the sphere of Venus. Venus therefore had to be above the Sun, so the order of the planets became Moon-Mercury-Sun-Venus-Mars-Jupiter-Saturn...Since his greatest solar distance was 1,266 e.r., about the same as Ptolemy's, Venus's greatest distance became 8,486 e.r., which was Mars's least distance.

So, al-ʿUrḍī rearranged Ptolemy's spheres, but did not significantly change the Earth-Sun distance.

As for Jābir Ibn Aflaḥ, Richard Lorch explains in his paper "The astronomy of Jābir Ibn Aflah" (doi:10.1111/j.1600-0498.1975.tb00315.x) that, according to Jābir, Ptolemy had determined a maximum solar parallax of 2′51″, and that if Venus and Mercury were inside the sphere of the Sun, they would have still larger parallaxes. In his treatise, Jābir therefore places Mercury and Venus above the Sun (see also [1]), but there is no sign that he greatly enlarged the Earth-Sun distance (if he had done this, there would be no need to move Mercury and Venus.)

Spacepotato (talk) 02:43, 29 May 2010 (UTC)

AU and m

NOTE: this discussion is now of historical interest only since the 2012 IAU definition fixes the AU at 1495977870700 meters exactly.2600:1000:B00C:57A5:7D9F:81E9:F7A3:51F0 (talk) 03:51, 18 October 2012 (UTC)

The article subsection Conversion factors states:

1AU = 149,597,870.700 ± 0.003 km

The IAU website states:

One AU is exactly 149,597,870.691 kilometres

The JPL website says:

AU = c * tauA = 1.49597870691 x 10¹¹ (± 3) m

The 2002 IERS documentation says (Table 1.1):

c�τ_A = 149597870691m ±3m [3] Astronomical unit in meters

The multi-volume Landolt-Bornstien compendium (p.4) says:

AU = 1.49597871464 × 10¹¹ m (no error bars)

Shirley & Fairbridge say (Table A8, p.49):

AU =1.4959787066 x 10¹¹ (no error bars)

Why the variety of values? Some of these numbers are close in value, but the IAU says their number is exact. Some sources quote no error bar. Is this an experimental number or a theoretical number based upon using Newton's laws with a particular gravitational constant?? Brews ohare (talk) 05:52, 23 July 2010 (UTC)

The IAU website is misleading: the current IAU best estimate (2009) is 149,597,870.700(3) km, as quoted in the article. The value of 149,597,870.691(3) km comes from the widely used DE-405 ephemeris (1998). Shirley & Fairbridge gives some discussion on other values.

The answer to your second question is a bit more subtle. For any given (modern) ephemeris, the astronomical unit is a fixed number (equivalent to a fixed value for the heliocentric gravitational constant): the ephemeris is a model of the solar system, and so internally exact. So the IAU writer wasn't actually wrong when s/he said that the AU is exactly 149,597,870.691 km: that's true for anyone using the DE-405 ephemeris (which is just about everyone). However, the people who compile ephemerides (the Jet Propulsion Laboratory in the case of DE-405) usually make an estimate of the uncertainty based on the residuals between the model and the available observations, and the ±3 m is JPL's estimate of the uncertainty. I say "estimate" because you cannot use a simple Gaussian treatment of measurement uncertainties – too many of the input data are correlated with one another – and there is some debate on the best way to make the estimate. The Russian team have produced an ephemeris with a claimed uncertainty in the AU of ±0.3 m, but the current consensus appears to be that they were too optimistic in their treatment of uncertainties, hence the return to ±3 m for the IAU best estimate. Physchim62 (talk) 11:47, 23 July 2010 (UTC)

It would appear that the article is rather lacking in its discussion of these points. Certainly something about the exact internally consistent model and the error in applicability of this model should be there. BTW, it looks a bit tough to determine the "experimental" error in applying a model that assumes an infinitesimal mass particle in an exactly circular orbit. Where in nature does one find this item? Brews ohare (talk) 14:05, 23 July 2010 (UTC)

Planetary masses are small compared to the Solar mass and, to calculate the perturbation due to finite planetary mass, you only need to know the mass ratios. Physchim62 (talk) 15:59, 23 July 2010 (UTC)

In short, if one selects a specific value for k, it would seem that there isn't any choice in the matter: the AU is determined by Newton's laws independent of any experimental input. On the other hand, if one takes k to be an experimental number, then the definition of the AU is not as stated, i.e. is not for a specific k-value. What is the case? Brews ohare (talk) 14:24, 23 July 2010 (UTC)

But of course! If you have a fixed value for k you have defined A, that's the whole point of a definition! But how do you compare that value with Earth-based units? You can't, unless you have a measurement both in astronomical units and in Earth-based units, so that you can calculate the conversion factor. Physchim62 (talk) 15:59, 23 July 2010 (UTC)