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#23 | ||
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Nov 2003
22·5·373 Posts |
Quote:
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:
So little time. Last fiddled with by R.D. Silverman on 2012-01-09 at 15:51 Reason: typo |
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#24 | |
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
101010001000102 Posts |
Quote:
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 |
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#25 |
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Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
3·29·83 Posts |
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 here) before more discussion takes place. Last fiddled with by Dubslow on 2012-01-09 at 19:37 |
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#26 | |
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×5,393 Posts |
Quote:
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 |
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#27 | |
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Jul 2006
USA (UT-5) via UK (UT)
22×59 Posts |
Quote:
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 Last fiddled with by Graff on 2012-01-09 at 23:07 Reason: Change deg/day to rad/day, d'oh. And add a historical note. |
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#28 | ||
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Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
3·29·83 Posts |
Quote:
Quote:
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#29 | |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
Quote:
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. Last fiddled with by science_man_88 on 2012-01-10 at 00:09 |
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#30 |
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"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×641 Posts |
Like I said, the IAU clings to the auld ways.
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#31 |
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Aug 2003
Snicker, AL
7×137 Posts |
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 Last fiddled with by Fusion_power on 2012-01-10 at 08:24 |
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#32 | |
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2·5,393 Posts |
Quote:
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 Last fiddled with by xilman on 2012-01-10 at 17:56 |
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#33 | |
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Jul 2006
USA (UT-5) via UK (UT)
22×59 Posts |
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almost correct. My definition using k is exact. 2 * I'm an IAU member and my primary work is orbits. I will alert the maintainers of that page to their imprecise definition. Gareth |
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