[QUOTE=xilman;285546]Bob: time for some Socratic questions.

Do you think that astronomers are unaware of the ellipticity of the Earth's orbit? If not, how might they frame the definition of the parsec in order to take ellipticity into account?
[/QUOTE]

Of [i]course[/i] they are aware! My questions were intended as rhetorical.
I studied orbital mechanics as part of freshman physics. (A long time ago....)

One would frame the definition in terms of measurement from either the
minor or major axis. But even these change slightly over time. Jupiter
tugs on us and adds/subtracts a miniscule amount of angular momentum....
Venus, although much smaller, tugs from another direction......

The N-body problem is well known to be chaotic.

[QUOTE]
If you wish, I could recommend some good books on the subject of astronomical measurement.
Paul[/QUOTE]

I would ask, but I do not have the time to read them. So much to learn.
So little time.

[QUOTE=R.D. Silverman;285561]One would frame the definition in terms of measurement from either the minor or major axis. But even these change slightly over time. Jupiter tugs on us and adds/subtracts a miniscule amount of angular momentum.... Venus, although much smaller, tugs from another direction...... [/QUOTE]
You might define it in such a way, but the astronomers wouldn't because of the variation in those quantities, some of which you have identified.

AFAIK (meaning it may have changed recently without my becoming aware of it), one first defines a year. The standard year is the time taken for the earth to orbit the sun once with respect to the "fixed" stars. That avoids problems with the proper motions of any particular star. Then you have to specify the epoch when the year is measured, to avoid problems with secular changes of the earth's orbital period due, in part, to planetary perturbations. By convention, this year is that measured in 1900.

Once you have a year defined, you can find out the radius of a circular orbit around a body of one solar mass assuming Newtonian gravity and no perturbations from other material in the solar system. That defines the AU and consequently the parsec.

Paul

From the same page as has been linked by three different people, the parsec is based on the AU, which is (sort of) derived from the Earth's orbit. It is independent of whatever your definition of a year is. The AU is defined as the distance that a point particle of Earth's mass would orbit at, if it's orbit was exactly 365.2568983 days long. Thus the AU is actually slightly shorter than the mean Earth-Sun distance (and the parsec is well defined, independent of the Earth's orbit). Also note that the parsec is generally used in favor of the light year by astronomers.

I suggest everyone actually go read that aforementioned link (reproduced [url=http://www.iau.org/public/measuring/]here[/url]) before more discussion takes place.

[QUOTE=Dubslow;285591]The AU is defined as the distance that a point particle of Earth's mass would orbit at, if it's orbit was exactly 365.2568983 days long. Thus the AU is actually slightly shorter than the mean Earth-Sun distance (and the parsec is well defined, independent of the Earth's orbit[/QUOTE]Ok, one Gaussian year. Thanks for the clarification. Only goes to show how complex this situation has become over the centuries. I was clearly thinking of some other quantity which is/was measured on the basis of the J1900.0 siderial year.

Regardless of this, the definition of the AU is given in terms of the orbital period of a point mass in an unperturbed circular orbit about one solar mass under Newtonian gravity. The latter qualification is important, in principle, because under GR such a system radiates gravitational energy leading to a secular change in the orbit irrespective of other influences.

Paul

[QUOTE=Dubslow;285591]From the same page as has been linked by three different people, the parsec is based on the AU, which is (sort of) derived from the Earth's orbit. It is independent of whatever your definition of a year is. The AU is defined as the distance that a point particle of Earth's mass would orbit at, if it's orbit was exactly 365.2568983 days long. Thus the AU is actually slightly shorter than the mean Earth-Sun distance (and the parsec is well defined, independent of the Earth's orbit). Also note that the parsec is generally used in favor of the light year by astronomers.

I suggest everyone actually go read that aforementioned link (reproduced [url=http://www.iau.org/public/measuring/]here[/url]) before more discussion takes place.[/QUOTE]

Urm, no. One AU is the heliocentric distance at which a massless particle in a circular, unperturbed orbit would have a mean motion of k ( = 0.01720209895) radians/day.

The value of k is that adopted by Gauss, who derived a number of tables needed for orbit computations that depended on k. Rather than having to recompute
these tables each time the value of k was updated (due to better determination of the values involved in its computation), it was decided to keep k fixed
(the so-called Gaussian constant) and dispense with the notion that one AU is exactly the semimajor axis of the earth's orbit.

Gareth

[QUOTE=Graff;285632]Urm, no. One AU is the heliocentric distance at which a massless particle in a circular, unperturbed orbit would have a mean motion of k ( = 0.01720209895) radians/day.

Gareth[/QUOTE]

The IAU either disagrees, or has the same result with a different definition:
[quote]One of the most important of these is the Astronomical Unit, abbreviated AU, which is defined by the IAU as equal to the distance from the centre of the Sun at which a particle of negligible mass, in an unperturbed circular orbit, would have an orbital period of 365.2568983 days.[/quote]Again, the link in that post is suggested reading, and the source for this.

[QUOTE=Dubslow;285636]The IAU either disagrees, or has the same result with a different definition:
Again, the link in that post is suggested reading, and the source for this.[/QUOTE]

if I've got it right from reading the content of the link:

circumference = (3600*360)AU =1296000AU=(2*Pi)Parsec so each parsec is 129600/(2*Pi)AU to be exact. though admittedly I did a Google search for factors of the number claimed by stargate38 that may be a way to come across the statement in the first post of the thread.

[QUOTE=xilman;285618]... under Newtonian gravity. The latter qualification is important, in principle, because under GR such a system radiates gravitational energy leading to a secular change in the orbit irrespective of other influences.

Paul[/QUOTE]Like I said, the IAU clings to the auld ways.

I find it interesting that so many have posted an incorrect value for the length of a year. Even more interesting is that our current calendar is inaccurate by about 26 seconds per year. The Mayan calendar was inaccurate by 10 seconds per year. The difference between us and them is that we know ours is off by 26 seconds/year, they didn't know about theirs. If we wanted to, we could add another leap day about every 3323 years which would correct for the 26 second discrepancy.

Sidereal year = 365.256363004 days
Tropical year = 365.242197 days
Anomalistic year = 365.259636 days

The conventional Julian year is 365.25 days.
With the accuracy of measurement available today, we can measure changes in length of an earth day because the earth is slowing down by about 1 second in 10 years. This causes major headaches for astronomers. It is also a problem with some high tech communications systems that are very time sensitive. I suspect Paul could contribute re some of the effects for astronomy.

DarJones

[QUOTE=Fusion_power;285681]I find it interesting that so many have posted an incorrect value for the length of a year.[/QUOTE]It's so much not that they are incorrect, more that they refer to different things. There are several sensible ways to define a day and a year from astronomical measurements and several other ways from other physical measurements. For the time being, the official definition of a day is 24 * 60 * 60 seconds, the second being defined in terms of the frequency a particular radiation. Under this definition, the astronomical days vary in length over quite short periods.

As for the year, what do you measure? The time between successive perihelia? Between successive northwards passages through the plane of the equator? Between successive passages through the invariant plane of the solar system (roughly speaking, the mass-weighted average plane of the orbits of the planets)? Between the times when the sun appears as close as possible to the same position on the sky? Or do you define the year as a specific number of seconds (as in the Julian and Gregorian calendars)? There are other options, including the Gaussian year which, I've now learned or re-learned, is the basis for the A.U.

Paul

[QUOTE=Dubslow;285636]The IAU either disagrees, or has the same result with a different definition:
Again, the link in that post is suggested reading, and the source for this.[/QUOTE]

It doesn't disagree substantially. The definition on the IAU page is
almost correct. My definition using k is exact. 2 * [TEX]\pi[/TEX] / k = 365.2568983... days.
I'm an IAU member and my primary work is orbits. I will alert the
maintainers of that page to their imprecise definition.

Gareth