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If you check the local time for solar noon is different every day. Why is it so? Is it because Earth doesn't make a complete rotation in exactly 24 hours?

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Can you provide a reference for these different times? There are some subtleties regarding different time keeping methods, like sidereal vs. solar, or variations due to the eccentricity of Earth's orbit. In order to answer with certainty, we need to know what kind of variation you're seeing. –  Chris White Sep 25 '12 at 7:58
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I say different by watching your watch. The Sun is highest on local meridian and if you look at your watch, that time is different from day to day. –  Florentin Sep 25 '12 at 10:56
    
You essentially have the answer in your question. The earth takes 23hrs56min to rotate a full 360 degrees. But since it's travelled a little along it's orbit around the sun (in that time), it will not be facing the sun. Since a year has ~360 days, it will in fact be off by ~1 degree. To turn that extra degree (and face the sun) it takes 4 minutes and so we need 24 hours for the sun to rise to max height again. –  Siva Apr 2 '13 at 2:31
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6 Answers

Here's how I explain this to my students. There are actually two suns in the sky. One is responsible for causing objects to cast shadows and is the daytime star we're all accustomed to observing. Let's call it the "true Sun" or the "apparent Sun." It, in effect, governs our lives. However, it has one main problem relevant to timekeeping, and that is that its speed along the ecliptic throughout the year varies. It's average (angular) speed is about $360/365$ deg/day or about $0.986$ deg/day. Now, this variation is important because if we want to use this Sun as a prime mover for timekeeping, we have to be aware of the variation. It means that an hour, which is defined as one twenty-fourth of the time it takes this Sun to go from the celestial meridian around to the celestial meridian again, varies in duration throughout the year when compared to the same interval measured with a timekeeping device that does NOT vary in rate. As for the cause of the variation, there are several factors at work here. One is the shape of Earth's orbit, which causes Earth to move at different speeds at different places in its orbit. Of course, we see this as differences in true Sun's speed along the ecliptic because we're on the moving Earth. Another is the fact that true Sun's path, the ecliptic, makes a $23.5^\circ$ angle with the celestial equator, which also causes seasonal variations in true Sun's speed along the celestial equator.

To get around these problems, astronomers invented the "mean Sun" for timekeeping, and it moves at a uniform rate (not a variable rate) along the celestial equator (not the ecliptic). Both it and the true Sun have the same average angular speed, taking one year to complete one trip around the sky (in different planes, however). Being fictitious, one can't actually observe mean Sun. Nevertheless, mean Sun is the prime mover that we use for all civil timekeeping. We can track its motion relative to the celestial meridian with a mechanical (or nowadays, digital) device called a "clock."

Now, the variations in the true Sun's motion causes it to sometimes be ahead of, sometimes lag behind, and sometimes be neck and neck with the mean Sun. This variation is called the equation of time and can be tabulated for any date.

When we say "noon" most people think of Sun being at its highest point in the sky, which need not be the zenith, but will ALWAYS be somewhere on the celestial meridian (yes, things get weird in the arctic and antarctic regions). However, "noon" really only applies to true Sun's motions. We keep time by mean Sun's motions, and a mean solar clock ALWAYS reads 12:00 when mean Sun is on the meridian. This moment, except for four days of the year, does not coincide with "noon" as embodied by true Sun's motions.

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A solar noon is defined when the Sun is at the zenith (directly above). Since the Earth revolves round the Sun, the point that is directly above would have changed because Earth is in a different point from the previous noon.

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True, but I don't think that's the answer sought. If we set our clocks by the Sun rather than the fixed stars (which is what most time systems do these days), then you've shown that then meridional crossings of the stars shift by ~ 4 minutes each day. However, solar noon would not change, since we define time of day based on it. –  Chris White Sep 25 '12 at 8:06
    
NO! Solar noon, more correctly 12:00 Local Apparent Solar Time (LAST), is NOT defined in terms of the zenith. If it were, it would never occur for the majority of the observers on the planet! Only observers within 23.5 degrees of the equator would experience noon. Instead, the concept of "noon" is related to meridian passage. Noon is when Sun (either true or apparent) is on the meridian. –  user11266 Sep 25 '12 at 10:52
    
I downvoted this answer because it give a patently incorrect definition of solar noon, but found my own rep downvoted by 1 point too. Why did this happen? –  user11266 Sep 25 '12 at 14:04
    
So that users will not anyhow downvote others. But I agree my error. I should have mention its the highest point of the observer at that latitude, not the zenith. –  Standstill Sep 25 '12 at 14:25
    
@Standstill I don't understand your response. I shouldn't have been downvoted because my response was correct. Maybe I don't fully understand how reputation points work yet. –  user11266 Sep 25 '12 at 15:11
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This is discussed in some detail in http://www.sundials.co.uk/equation.htm and the links therein.

The main cause is that the length of a day changes during the year. For example the Earth moves faster relative to the Sun at perihelion than it does at aphelion. This means the perion from noon to noon measured by a sundial is not the same as the period measured using one of the standard time schemes like universal time. The other significant cause of changes in tne day length is the the tilting of the Earth's orbit.

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the answer is much simpler - the earth's orbit is an ellipse, not a circle . . so it is traveling faster when closer to the sun (january) and slower when farthest from the sun (july). The time from solar noon to solar noon is a sidereal day (23h56m, fixed to the stars) PLUS the time it takes to "compensate" for how far it has moved in its orbit. This is about 4 minutes, but it is a bit MORE when the earth is moving fast (january) and a bit LESS when the earth is moving slowly - thus the time from solar noon is a bit longer than 24 hours in january, and a bit shorter in july. This is why the solar noon time "wobbles" throughout the year.

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It's not quite this simple. You must account for Earth's obliquity and not just Earth's orbital eccentricity. –  user11266 Jan 14 '13 at 16:54
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Solar noon becomes earlier and then back again more than once a year; it moves 18 minutes earlier from about Feb. 10th until May, then 12 minutes later by July , then back 23 minutes earlier by late October/early November, then forwards again by half an hour by Feb tenth. So the biggest change in Solar noon occurs between early November and mid February by 31 minutes. I hope that has answered questions, but it is also a question, because I don't know why direction of solar noon time changes 4 times a year; two per direction, and also why the largest difference is in between early November and mid February. If anybody can answer that I'd appreciate it. Also, what are the 4 days user 11266 is referring to the end of his answer?

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This does not appear to answer the question, it only gives more information to verify that the difference exists. If you have a new question, please click the Ask Question button and ask it there. It might be pertinent to include a link to this question for reference. –  Kyle Kanos Apr 12 at 20:39
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Is it because Earth doesn't make a complete rotation in exactly 24 hours?

No, it doesn't. It rotates in 23h 56m 4.098 903 691s.

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Careful! The numerical value of Earth's rotation period depends on the timescale used. It is indeed 24 hours of apparent solar time, but only 23h 56m of sidereal time. (I don't distinguish here between mean and apparent sidereal time.) –  user11266 Sep 25 '12 at 10:54
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protected by Qmechanic Apr 12 at 20:30

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