<p>Let <span class="math-container">$\vec{r}$</span> be the position of the Sun in the sky, from Earth. How could I estimate the magnitude of <span class="math-container">$d\vec{r}/dt$</span>, to measure the rate of change in time of the arc it traces? I suppose it would be proportional to Earth's rotation. I guess this has been already calculated before.</p>
<p>I'd like to estimate this quantity to compare with ephemeris and, then, answer to: how fast should I be moving (e.g., travelling on a plane) to always observe the Sun in the same relative position in sky?</p>

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2022-10-08T21:22:04.207Z
Related: https://en.wikipedia.org/wiki/Subsolar_point
PM 2Ring
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2022-10-08T22:14:16.693Z
It is roughly one degree every 4 minutes.  But all degrees are not created equal.  Since degrees of declination get closer as you move away from the equator, the declination of the sun affects the angular distance traveled.  As well as the longitude of the observer trying to keep pace.
Greg Miller
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2022-10-09T01:35:10.333Z
@GregMiller what about degrees per minute measured in the Sun's apparent direction of motion, are those constant? I'm imagining myself at the South Pole at "sunrise" (southern hemisphere's spring) and "noon" (southern hemisphere's summer) where the sun completes 360° in 23h 56m, but at different elevations above the horizon, so I guess the answer is "no, not the same degrees per minute, even in the Sun's apparent direction of motion".
uhoh
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2022-10-09T04:45:12.853Z
@uhoh, If you use the average solar day, rather than sidereal day and look at the Equation of Time.  The maximum variation is less than 30 seconds spread out over a day.
Greg Miller
CC BY-SA 4.0
2022-10-09T05:01:29.833Z
@GregMiller in the spiring or fall at the south pole the Sun moves in a great circle as seen from the observer (i.e. the horizon) covering 360° in 3.99 minutes. In the summer at the south pole the Sun circles at an elevation of 23.5° above the horizon. The instantaneous angular speed along the direction of motion is reduced by cos(23.5°) such that the rate is now one degree in 4.35 minutes along its path. Extreme example is Polaris, which circles the celestial pole at 1 degree every 3.99 minutes but moves at an instantaneous rate corresponding to one degree every 310.7 minutes.
uhoh
CC BY-SA 4.0
2022-10-09T13:54:28.087Z
@uhoh, agreed.  That is essentially what I said in my first comment.
Greg Miller
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2022-10-13T07:24:11.723Z
Maybe these kind of consideration have been made for sofia
planetmaker
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2022-10-13T07:56:42.063Z
@uhoh *"covering 360° in 3.99 minutes"* I suspect that's a typo. ;)
PM 2Ring
CC BY-SA 4.0
2022-10-13T14:23:51.140Z
@PM2Ring yes, alas it was supposed to be "covering 1° in 3.99 minutes". I could petition a moderator to make the correction, but I don't think it rises to the appropriate level of urgency. I once saw a moderator edit someone's joke in a comment to make the punch line less offensive, but that was in a different universe.
uhoh
CC BY-SA 4.0
2022-10-15T03:11:19.780Z
A current question on flying fast enough to keep up with the Sun: https://aviation.stackexchange.com/q/95404
PM 2Ring
<observational-astronomy><positional-astronomy><celestial-mechanics><astrometry>
Positional astronomy - Speed of the sun in sky
nuwe
# Positional astronomy - Speed of the sun in sky

(Asked by: username_0 on 2022-10-08)

Let $\vec{r}$ be the position of the Sun in the sky, from Earth. How could I estimate the magnitude of $d\vec{r}/dt$, to measure the rate of change in time of the arc it traces? I suppose it would be proportional to Earth's rotation. I guess this has been already calculated before.

I'd like to estimate this quantity to compare with ephemeris and, then, answer to: how fast should I be moving (e.g., travelling on a plane) to always observe the Sun in the same relative position in sky?

**Tags:** observational-astronomy, positional-astronomy, celestial-mechanics, astrometry
> **Comments:**
> **username_1**: Related: https://en.wikipedia.org/wiki/Subsolar_point  
> **username_2**: It is roughly one degree every 4 minutes.  But all degrees are not created equal.  Since degrees of declination get closer as you move away from the equator, the declination of the sun affects the angular distance traveled.  As well as the longitude of the observer trying to keep pace.  
> **username_3**: @GregMiller what about degrees per minute measured in the Sun's apparent direction of motion, are those constant? I'm imagining myself at the South Pole at "sunrise" (southern hemisphere's spring) and "noon" (southern hemisphere's summer) where the sun completes 360° in 23h 56m, but at different elevations above the horizon, so I guess the answer is "no, not the same degrees per minute, even in the Sun's apparent direction of motion".  
> **username_2**: @uhoh, If you use the average solar day, rather than sidereal day and look at the Equation of Time.  The maximum variation is less than 30 seconds spread out over a day.  
> **username_3**: @GregMiller in the spiring or fall at the south pole the Sun moves in a great circle as seen from the observer (i.e. the horizon) covering 360° in 3.99 minutes. In the summer at the south pole the Sun circles at an elevation of 23.5° above the horizon. The instantaneous angular speed along the direction of motion is reduced by cos(23.5°) such that the rate is now one degree in 4.35 minutes along its path. Extreme example is Polaris, which circles the celestial pole at 1 degree every 3.99 minutes but moves at an instantaneous rate corresponding to one degree every 310.7 minutes.  
> **username_2**: @uhoh, agreed.  That is essentially what I said in my first comment.  
> **username_4**: Maybe these kind of consideration have been made for sofia  
> **username_1**: @uhoh *"covering 360° in 3.99 minutes"* I suspect that's a typo. ;)  
> **username_3**: @PM2Ring yes, alas it was supposed to be "covering 1° in 3.99 minutes". I could petition a moderator to make the correction, but I don't think it rises to the appropriate level of urgency. I once saw a moderator edit someone's joke in a comment to make the punch line less offensive, but that was in a different universe.  
> **username_1**: A current question on flying fast enough to keep up with the Sun: https://aviation.stackexchange.com/q/95404  

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