
#1
Dec510, 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 



#2
Dec510, 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 EarthSun distance. But the technical definition is a little more complicated.
I quote wikipedia: 



#3
Dec510, 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! 



#4
Dec610, 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.




#5
Dec610, 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.




#6
Dec610, 07:47 PM

P: 6,863





#7
Dec710, 04:50 AM

Mentor
P: 14,440

[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)×10^{20} m^{3}/s^{2} (TDBcompatible 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. 



#8
Dec710, 04:55 AM

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P: 14,440





#9
Dec710, 10:04 AM

P: 6,863

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. 



#10
Dec710, 10:42 AM

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P: 14,440

Baloney. Here is a paper that discusses doing exactly that: http://iaucomm4.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 uptodate 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. 



#11
Dec710, 05:42 PM

P: 6,863





#12
Dec710, 06:59 PM

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P: 14,440

What? m^{3}/s^{2} isn't SI?
The JPL Development Ephemerides internally expresses length in kilometers and time in T_{eph} days (86400 ephemeris seconds), and hence the gravitational coefficients are in km^{3}/day^{2}. 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. 



#13
Dec710, 07:58 PM

P: 6,863

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. 



#14
Dec710, 09:23 PM

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P: 14,440

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? 



#15
Dec710, 10:07 PM

P: 6,863





#16
Dec710, 11:31 PM

P: 133

I'm sure we can all agree here that twofishquant is right. Just give up DH.




#17
Dec810, 03:36 AM

P: 6,863





#18
Dec810, 05:24 AM

Mentor
P: 14,440

I also have a concession to make.
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 (lbf·s^{2})/(lbm·ft) 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*meters^{2} for its inertia tensor, and newtons for nongravitational forces acting on the spacecraft. 


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