Is there a formula for how to calculate the angular rate at which the Sun apparently moves around the Earth? I know it takes 0.25 degrees/min, but as I understand it depends on other parameters as well (such as how close a certain place to the equator, months of the year etc.), but I'm looking for a formula that makes it easy to calculate based on my location proportional to the equator and different time of the year (unless if I'm mistaken and it's constant)

  • $\begingroup$ I mean simply to my location proprtional to the equator. $\endgroup$ Commented Feb 21, 2021 at 7:20
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    $\begingroup$ it moves over Earth from East to West once a day, so the rotation speed is constant as hence is the speed of the apparent movement of the Sun. What varies over the year is the maximum height of the Sun over the horizon. $\endgroup$ Commented Feb 21, 2021 at 7:25
  • $\begingroup$ @planetmaker if I stand 1 meter from the south pole and run around it you could say my angular speed was 1 rad/sec about Earth's axis, but my great circle angular speed drawn fro the center of the Earth would be only $1.6 \times 10^{-7}$ rad/sec. $\endgroup$
    – uhoh
    Commented Feb 21, 2021 at 8:12

2 Answers 2


If you don't have to have that precise value, then the angular speed of the sun is simply this:

$$\frac{360°}{24\text{ h}}$$

We can simplify this to SI units (we won't be using radians here, for simplicity):

$$\frac{360°}{24\text{ h}}=\frac{360°}{24\text{ h}}\times \frac{1\text{ h}}{3600\text{ s}}=\frac{1}{240}\frac{°}{s}$$

Let's suppose, that the angular speed of the sun is not constant on Earth. What would happen? The time between two same noons would be different. That would mean, that Earth would rotate with different angular rate on different points. We know that this isn't true, so the supposition is false. (But the Sun does rotate on different latitudes with different angular rates, so the angular speed of some star, looking from the Sun, depends on your location. Of course, this is theoretical and nobody has tried that yet.)

The angular speed doesn't depend on the latitude, but it depends on the current distance between the Sun and the Earth. For that you need true anomaly $\nu$, which is given by Kepler equation, which is transcedent (it can't be solved with finite steps = you have to use computer). Why does it depend on the current distance? If the Earth would not circle around the Sun and it would be stationary, the day would last for around 23 h 56 min. But it does circle around the Sun, thus the some point on the Earth has to wait a little more, to reach the culmination of the Sun, so it lasts for 24 h. The angular speed around the Sun is through the year changing, so the true day deviates from the mean day by a really tiny amount (i think that around 30 s).

  • $\begingroup$ Thank you. But this simple one I knew) I looked for the precise one, considering the location, proportional to the equator. Basically, I'm looking to calculate the astronomical twilight (which is different from place to place) $\endgroup$ Commented Feb 21, 2021 at 8:00
  • $\begingroup$ Note that the angular speed of the Sun doesn't change with the latitude, but it changes with the position of the Earth around the Sun (the date). More correct formula would differ from mine for about 1%, I believe. $\endgroup$
    – User123
    Commented Feb 21, 2021 at 8:03
  • $\begingroup$ You can also give us your source of this precise formula, but unless you are doing some programming project with display of the Sun or length of the day, the more precise formula isn't really that useful. $\endgroup$
    – User123
    Commented Feb 21, 2021 at 8:04
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    $\begingroup$ Right now the problem is "what angle" is it exactly whose rate of change is being discussed. As long as everybody talks about it but it's not carefully defined, it's going to be hard to resolve this. $\endgroup$
    – uhoh
    Commented Feb 21, 2021 at 8:14
  • $\begingroup$ @uhoh I believe that the angular speed of the Sun on the celestial sphere. It could be hardly anything else, because he said: I know it takes 0.25 degrees/min ... $\endgroup$
    – User123
    Commented Feb 21, 2021 at 8:16

Since you said in a comment that you want to calculate astronomical twilight, here some links that may help.

Some general info


Online Tool


Answer on Physics- Stackexchange


Some more involved calculations with programming instructions



(the second one does not deal with astronomical twilight (only civil and nautical) but could probably be easily modified).

No guarantee that either of these are useful. I just googled them and never used them.


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