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Spacecraft Doppler&Light Speed Extrapolation



 
 
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  #91  
Old August 8th 03, 04:01 PM
George Dishman
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Default Sidereal Time (was Spacecraft Doppler&Light Speed Extrapolation)


"Oriel36" wrote in message om...
"George Dishman" wrote in message ...

snip gibberish


The only gibberish you snipped was the NASA gibberish.


The paragraphs I snipped had no mention of NASA, just
your stuff about coordinates. They bear no relation to
the question you were supposed to be answering:

http://www.starsforfun.com/gaot01a.jpg

The stars appear to rotate about the celestial pole, turning
through 360 degrees in (to the nearest second):

[a] 24 hours exactly
[b] 23 hours, 56 minutes, 4 seconds
[c] 24 hours on average but varying during the year
[d] 365.2422 days of 86400 seconds each
[e] something else (state your value) _____________


Since the question says "to the nearest second", your
excuse [choice of stars] doesn't work.




snip more waffle

.. 360 degrees in that time is
a litle over a millionth of a degree per year or
5 billionths of a degree per day. It would affects
the answer to this question by about a millionth of
a second if you chose to observe Sag A* for example:


You are missing the really good stuff,


No, I am leaving it aside until you look at other more
fundamental aspects. We can come back to this later.
These paragraphs were quite well written Gerald and give
me a clear idea of your views so I have kept them to
answer later.

snip bit on distant supernovae

http://www.dishman.me.uk/George/SolarDay/index.htm

The only way for the rotation to be 360 degrees in 24 hours
would be if you stopped the Earth in its orbit. ...

snip
I suppose I should be suitably insulted


No, you shouldn't. You are simply supposed to look at
the web page and either agree or tell me what is wrong
with it. Writing lots of paragraphs that do not address
our differences doesn't move the conversation forward
and is as much a waste of your time as mine.

Since the errors you make regarding supernovae originate
with this orbital aspect, there is no point moving on
until you sort it out.

snip rest - again not relevant to the questions

George


  #92  
Old August 1st 13, 01:32 PM posted to sci.astro
oriel36[_2_]
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Default Sidereal Time (was Spacecraft Doppler&Light Speed Extrapolation)

On Thursday, July 24, 2003 11:08:44 AM UTC+1, Paul Schlyter wrote:
In article ,
Oriel36 wrote:
(Paul Schlyter) wrote in message ...
In article ,
George Dishman wrote:

The Equation of Time tells you the difference between the effect
of a circular and an elliptical orbit.

No it doesn't!

The Equation of Time contains two conponents:

1. The component due to the Earth's elliptical orbit:
1 cycle/year


That is correct,it is the difference betwen axial and elliptical
rotation via Kepler's second law,it is a property of the Earth's
rotation and not the Sun's observed motion.



2. The component due to the 23.4 degree tilt of the Earth's axis:
2 cycles/year


You did not read Maskelyne who correctly points out that Equatorial
orientation does not affect the apparent motion of the Sun for it is
neither accelerated or retarded against axial rotation,


Of course the apparent motion itself of the Sun (i.e. the orbital
motion of the Earth) is unaffected by the tilt of the Earth's axis.

But it's not quite that simple: the selected coordinate system
matters too. And if we use the RA/Decl coordinate system, that
system will be dependent on the axial tilt of the Earth.

To clarify matters, let's consider two simplified cases:

1. Earth's orbit is a perfect circle, and the axial tilt is exactly
zero degrees. In this case the Sun will always appear at the
celestial equator, moving uniformly at 360/365.2422 degrees per day.
The equation of time will always be zero.

2. Earth's orbit is a perfect circle, and the axial tilt is exactly
90 degrees. The north and south polar circles will reside at the
equator, and the Tropic of Cancer will reside at the North Pole and
the Tropic of Capricorn at the South Pole. The Sun will move along
RA=0h and RA=12h. Starting at the December solstice, the Sun's RA
will be 0h while it appears to move straight northward at
360/365.2422 degrees/day, while the Equation of Time increases by
4*360/365.2422 minutes per day. This continues until the June
solstice, when something dramatic happens: the Sun's RA suddenly
jumps from 0h to 12h, and the Equation of Time also suddenly jumps
12h (we can say it increases by 12h or decreases by 12h, it doesn't
matter which). Now the Sun's RA will he 12h while it appears to move
straight southward at 360/365.2422 degrees/day, while the Equation of
Time still increases by 4*360/365.2422 minutes per day. At the
December solstice the Sun's RA suddenly changes from 12h to 0h,
and there's another sudden change in the Equation of Time by 12h.

Now compare the two cases: the Sun's apparent motion (= the Earth's
orbital motion) is exactly the same in both cases. But we flipped
the coordinate system (RA/Decl) by 90 degrees, and this caused some
dramatic changes in the way the Equation of Time behaved.

Now, even if the tilt is larger than 0 degrees but less than 90
degrees, there will be some effect on the Equation of Time from
this. Not because the apparent motion of the Sun changes (it
doesn't, as you correctly pointed out), but because our
coordinate system changed.

--
----------------------------------------------------------------

Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN

e-mail: pausch at stockholm dot bostream dot se

WWW:
http://www.stjarnhimlen.se/

http://home.tiscali.se/pausch/


Ten years on from when you wrote this response you look like a dope and possibly why all the other academics disappeared from the forum.What made guys like Rod Molisse really silly was their insistence that it was finding Sir Isaac wrong when the endeavor is to demonstrate how astronomy has changed with the advent of the internet,imaging and graphic advances.

Were the Earth inclination 0 degrees and equatorial conditions as its orbited the Sun it would still display variations in the natural noon cycle as two surface rotations are involved with each pass of the central Sun.The orbital component is best appreciated by the polar day/night cycle as the North/South coordinates are carried around in an uneven way by the orbital motion of the planet and when combined with constant rotation at lower latitudes are responsible for the 1461 natural noon cycles covering 4 orbital circuits.

I do not like the new Google groups apart from this feature where it is possible to reply to old threads with better information and more images,time laopse footage and so on compared to 10 years ago.



 




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