Re: [Phys-l] celestial motion

["John SOHL" <JSOHL@weber.edu>]

You have a misconception here. The cause of sunset is Earth, not the
Sun. Tracking photons leaving the Sun while discussing Earth's rotation
is a bit like apples and oranges. 
 
True, the light that came from the sun was emitted 8 minutes ago and it
is also true that when we see those particular photons at sunset the
photons currently leaving the surface of the Sun will be arriving at
Earth when the Sun is below the local horizon. But, I don't think there
is much value in this line of thinking. (Your numbers and results ARE
correct though.) 
 
Perhaps more useful is to think about the diameter of the sun being
about 0.5 degrees and that at a rate of 0.25 deg/min the result is that
it takes about 2 minutes for the Sun to set. 
 
All the best,
 
John
 
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John E. Sohl, Ph.D.
Professor of Physics
Weber State University
2508 University Circle
Ogden, UT 84408-2508

voice: (801) 626-7907, fax: (801) 626-7445 
e-mail: jsohl@weber.edu 
web: http://physics.weber.edu/sohl/ 

> > > "Anthony Lapinski" <Anthony_Lapinski@pds.org> 8/22/2008 8:24 AM
What I want to know is the Sun's angular speed in the sky. The Sun is
about 0.5* across. So in a 12-hour day, it moves 180* from east to
west.
This is 15*/hr, or 0.25*/min. When we look at a sunset, it already
happened. Since light takes about 8 min to go from the Sun to Earth,
the
Sun has moved 2* in this time -- about 4 solar diameters.! This seems
a
bit large, or did I make an error?