(Just a few more seconds)
Unless I misplaced a decimal point, my calculations show that 8.3 minutes is (about) 1/63000 of a year. 1/63000 of 360 degrees is .00057. Thus, the Sun is actually .00057 degrees (or about 1/3 minute of arc) from where it appears to be. Insignificant.
Waitaminute...I'm not taking the rotation of the Earth into account.
The earth rotates through 360 degrees every 24 hours. 8.3 minutes is 1/173 of 24 hours. 1/173 x 360 degrees is...
2.1 degrees. Huh.
Am I missing something here? As far as I can tell, the math works.
No, you aren't missing anything at all. Except that some people choose not to see reality. They see without understanding.
According to this method of computing aberration, what would the aberration of the sun be for an observer on Neptune? And what would the aberration be for a star 10 light years away, computed by the same method? No doubt you will quickly see that, under this "LeGrandean" cosmology, many of the stars you see in front of you may actually be behind you, or anywhere at all.
Granted, it takes the light from the Sun 8.3 minutes to reach the Earth. Let me do a quick bit of math...
(Just a few more seconds)
Unless I misplaced a decimal point, my calculations show that 8.3 minutes is (about) 1/63000 of a year. 1/63000 of 360 degrees is .00057. Thus, the Sun is actually .00057 degrees (or about 1/3 minute of arc) from where it appears to be. Insignificant.
Waitaminute...I'm not taking the rotation of the Earth into account.
The earth rotates through 360 degrees every 24 hours. 8.3 minutes is 1/173 of 24 hours. 1/173 x 360 degrees is...
2.1 degrees. Huh.
Am I missing something here? As far as I can tell, the math works.
You and LaGrande are missing something. If the earth rotates 2.1 degrees in the time it takes light to travel from the sun, you calculate the angle offset of the light by finding the distance the earth moved in that time compared with the distance from the sun. Its about relative velocities, not angles.
Assuming the circumference of the earth is 24901 miles and rotates once every 24 hours, the surface speed at the equator is
24901 miles / 24 hours = 1,037.5 mph
To find the angle between the observed sun and the actual position due to the rotation of the earth, you use the inverse tangent of the speed of light and the speed of earth at the equator.
Assuming light travels at 670,616,629 mph
Tan(a) = (1,037.5 / 670,616,629) >>> a = tanh( 1.54708362 × 10-6)
a = 0.000001547 deg.
Thats a lot less than 2.1 deg. These calculations are assuming only Newtonian physics are in effect. I have no idea what part if any relativity plays it the observed angle offset. Im guessing that density differences in the atmosphere has a much larger effect.
Hope this helps clear up the confusion LeGrande.