definition of 1 AU


by wg1337
Tags: definition
wg1337
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#1
Dec5-10, 01:34 PM
P: 11
Hi!
I have a homework at school and one question in it is to show how AU is defined.
So of course I took google and saw - 1 AU is the distance from Earth to Sun (or the radius of Earth orbit).
But distance from where? From the center of planet or from the surface?
I have seen some illustrations and all they seem to show the distance from center, but my physics book seems to show different.
Won't be first time when the book gives a wrong answer.
The same goes for parsec - http://coolcosmos.ipac.caltech.edu/c...ges/parsec.gif
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Matterwave
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#2
Dec5-10, 02:14 PM
P: 2,043
Well, because the distance from the Sun to the Earth is so much greater than the size of the Earth (~150million km, compared with ~7000km), it usually doesn't matter how you want to define it (for practical purposes). The AU is about 149,597,870.7km, which is close to the mean Earth-Sun distance. But the technical definition is a little more complicated.

I quote wikipedia:

The AU was originally defined as the length of the semi-major axis of the Earth's elliptical orbit around the Sun. In 1976 the International Astronomical Union revised the definition of the AU for greater precision, defining it as that length for which the Gaussian gravitational constant (k) takes the value 0.017 202 098 95 when the units of measurement are the astronomical units of length, mass and time.[5][6][7] An equivalent definition is 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,[2] or that length for which the heliocentric gravitational constant (the product GM☉) is equal to (0.017 202 098 95)2 AU3/d2. It is approximately equal to the mean Earth–Sun distance.
wg1337
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#3
Dec5-10, 02:18 PM
P: 11
Thought that elliptical orbit will cause problems with this, also found many websites that says it is from the center.
Thanks!

twofish-quant
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#4
Dec6-10, 04:48 AM
P: 6,863

definition of 1 AU


Also, one interesting thing about the definition is that it's deliberately unclear from the definition what the actual distance is in standard distance units (i.e. km). The reason for this is that we can do celestial mechanics to nine significant figures, but the conversion factor from AU to km is known only to six significant figures.
RocketSci5KN
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#5
Dec6-10, 06:37 PM
P: 157
For calculations to make sense, it should be defined as the average distance from the center of the sun to the center of the earth. The radius of the sun plus earth is 701,400 km, so you'd add this to 149.6 million km *if* it was surface to surface. But, I don't think that's the case.
twofish-quant
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#6
Dec6-10, 07:47 PM
P: 6,863
Quote Quote by RocketSci5KN View Post
For calculations to make sense, it should be defined as the average distance from the center of the sun to the center of the earth.
It's intentionally not defined that way since we don't know exactly what the distance is between the earth and the sun in km.

The radius of the sun plus earth is 701,400 km, so you'd add this to 149.6 million km *if* it was surface to surface. But, I don't think that's the case.
The curious thing is that the uncertainty in the conversion between km and AU is larger than this uncertainty.
D H
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#7
Dec7-10, 04:50 AM
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Quote Quote by twofish-quant View Post
Also, one interesting thing about the definition is that it's deliberately unclear from the definition what the actual distance is in standard distance units (i.e. km). The reason for this is that we can do celestial mechanics to nine significant figures, but the conversion factor from AU to km is known only to six significant figures.
This is just wrong. The conversion from AU to km is known to almost eleven significant digits, 1AU = 1495978707003 meters. The calculation is quite simple:

[tex]1\,\text{AU} = \left(\frac{GM_{\odot}\,(86400\,\text{s})^2}{k^2}\right)^{1/3}[/tex]

The sole source of uncertainty here is the heliocentric gravitational constant, and that is known to almost eleven significant digits, GM = 1.32712440041(10)1020 m3/s2 (TDB-compatible value).

Are you talking perhaps about the dispute between TDB versus TCB? Just because there is a dispute among scientists regarding the best way to represent time and distance does not mean that there is uncertainty in the value of an AU. It just means that scientists cannot agree.
D H
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#8
Dec7-10, 04:55 AM
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Quote Quote by twofish-quant View Post
It's intentionally not defined that way since we don't know exactly what the distance is between the earth and the sun in km.
This also is wrong. It is intentionally not defined that way (average distance between the Earth and Sun) precisely because we do know very well the distance between the Earth and Sun in km (or au) and we now know that it is not constant. (Technically, 1 AU is not constant either because the Sun is losing mass due to burning hydrogen in its core and solar wind at its surface, but that is a different matter.)
twofish-quant
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#9
Dec7-10, 10:04 AM
P: 6,863
Quote Quote by D H View Post
The sole source of uncertainty here is the heliocentric gravitational constant, and that is known to almost eleven significant digits, GM = 1.32712440041(10)1020 m3/s2 (TDB-compatible value).
I stand corrected about the length conversions, and I think I got confused with something else.

GM is known very accurately, but G and M are known only to 10^-4, which means that you can't do celestial mechanics in SI units, but that affects masses and not distances.
D H
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#10
Dec7-10, 10:42 AM
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Baloney. Here is a paper that discusses doing exactly that: http://iau-comm4.jpl.nasa.gov/EPM2004.pdf.

The solution is simple: You don't use G or M. You use μ=GM or the sun/planet mass ratio coupled with GM. There is no need to separate out the mass term. The standard gravitational parameters for each of the eight planets are known with greater precision than is G. See http://asa.usno.navy.mil/SecK/2011/A...tants_2011.pdf for an up-to-date set of astronomical constants.

What can't be done with any precision is to use natural units, G=1. Arbitrarily setting G to 1 makes the meaning of length and time rather uncertain.
twofish-quant
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#11
Dec7-10, 05:42 PM
P: 6,863
Quote Quote by D H View Post
Baloney.
I think we are agreeing here. I was wrong about km.

The solution is simple: You don't use G or M. You use μ=GM or the sun/planet mass ratio coupled with GM. There is no need to separate out the mass term. The standard gravitational parameters for each of the eight planets are known with greater precision than is G. See http://asa.usno.navy.mil/SecK/2011/A...tants_2011.pdf for an up-to-date set of astronomical constants.
But those aren't SI units.
D H
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#12
Dec7-10, 06:59 PM
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What? m3/s2 isn't SI?

The JPL Development Ephemerides internally expresses length in kilometers and time in Teph days (86400 ephemeris seconds), and hence the gravitational coefficients are in km3/day2. Not quite SI, but distance is obviously just a simple scale factor away from SI. TIme is a bit tougher, but still not all that tough. An ephemeris second differs from the TAI second because the distance between the Earth and Sun varies over the course of a year.
twofish-quant
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#13
Dec7-10, 07:58 PM
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Quote Quote by D H View Post
What? m3/s2 isn't SI?
SI consists of seven fundamental units, one of which is the kilogram, and I think that we both agree that you can't do celestial mechanics in kilograms (and you don't have to either).

There's also another problem in that both length and time are defined in terms of universal experiments whereas the kg is defined in terms of a physical object. This is a problem because you can do a thought experiment in which you imagine how a cesium atom vibrates near Mars, whereas because the kg is a physical object and it's non trivial to imagine how you bring that physical object on Mars.
D H
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#14
Dec7-10, 09:23 PM
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You don't need to know mass of the objects in the solar system to do solar system mechanics. You don't need G, either. Given the incredible lack of precision in G it is downright silly to use G.

You only need to know the mass of vehicles that move around in the solar system, and you only need to know that because they have things like thrusters that modify the state of the vehicles.

Do you really think the world's space agencies use normalized units to model how their spacecraft move about the solar system?
twofish-quant
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#15
Dec7-10, 10:07 PM
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Quote Quote by D H View Post
You don't need to know mass of the objects in the solar system to do solar system mechanics. You don't need G, either. Given the incredible lack of precision in G it is downright silly to use G.
Exactly. I don't see us disagreeing here.

Do you really think the world's space agencies use normalized units to model how their spacecraft move about the solar system?
No, but that only works because the gravitational pull of space craft is negligible. That doesn't work for things like binary stars or collapsing supernova.
Caramon
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#16
Dec7-10, 11:31 PM
P: 133
I'm sure we can all agree here that twofish-quant is right. Just give up DH.
twofish-quant
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#17
Dec8-10, 03:36 AM
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Quote Quote by Caramon View Post
I'm sure we can all agree here that twofish-quant is right. Just give up DH.
Since I agree with DH that my initial claim was wrong, I'm not sure what's there to give up.
D H
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#18
Dec8-10, 05:24 AM
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I also have a concession to make.

Quote Quote by D H View Post
What can't be done with any precision is to use natural units, G=1. Arbitrarily setting G to 1 makes the meaning of length and time rather uncertain.
This was wrong. While I look at what we do as using the "standard gravitational parameters" of the Sun and the planets in lieu of mass, another way to look at it is that we are using G=1. What setting G to 1 does is to change mass from being an essential unit to being a derived unit with dimensions length3*time-2, at least in terms of gravitational interactions.

A similar situation occurs in Newtonian mechanics. Newton's second law as expressed by Newton is not F=ma. It is F=kma, where k is some constant of proportionality. This constant of proportionality is 1/32.1740486 (lbfs2)/(lbmft) in English units where mass and force are expressed in pounds. What setting k=1 as is done in SI units is to make force a derived unit rather than an essential unit.

We do of course still use kilograms for spacecraft mass, kilograms*meters2 for its inertia tensor, and newtons for non-gravitational forces acting on the spacecraft.


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