<p>All points on the celestial sphere execute a full circle every (sidereal) day, but the speed of a point with nonzero declination is slower than a point on the celestial equator because it's traveling on a small circle, not a great circle. This is exactly the same as how a point on the terrestrial equator travels at a higher speed than a point not on the equator. Eg, a point with a latitude or declination of 60° travels at half the speed of a point on the equator, because cos(60°) = 0.5</p>

<p>At the solstices, the Sun is on either the tropic of Cancer or Capricorn, so it has its maximum or minimum declination, approximately ±23.4°. So its speed is cos(23.4) <span class="math-container">$\approx$</span> 0.9178 relative to a point on the celestial equator, or about 8% slower, as Mike G mentioned in that comment.</p>

<p>Here's a diagram from an <a href="https://space.stackexchange.com/a/21069">answer about latitude speed</a> by SF. on the Space Exploration Stack Exchange:</p>

<p><img src="https://i.sstatic.net/XiyIp.png" alt="latitude speed diagram"></p>

CC BY-SA 4.0
2020-03-02T12:47:49.203Z
I guess I didn't phrase my question right. I wanted to know about the angular speed, because that's what is mentioned in the linked answer.
usernumber
CC BY-SA 4.0
2020-03-02T13:05:55.910Z
@usernumber Well, "angular speed" can be a bit ambiguous. ;) But my answer is talking about the same thing that Mike G is. Eg, in 1 hour a star on the celestial equator traces out an arc across the sky of 15°, but a star at 60° declination traces out an arc of half that length, and a star close to the celestial pole only moves along a very small arc, although its arc will have a lot more curvature.
PM 2Ring
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2020-03-06T14:05:00.060Z
@usernumber Is there some reason why you are still not happy with this answer?
PM 2Ring
CC BY-SA 4.0
2020-03-06T14:23:36.200Z
@PM2Ring I still haven't been able to wrap my head around the fact that the Sun can move through the sky 8% slower than 15° per hour.
usernumber
CC BY-SA 4.0
2020-03-06T14:40:04.417Z
@usernumber Firstly, I guess I should make it clear that in my answer I'm ignoring the fact that the Sun's RA & declination change, since that's only about 1° per day. I'm just talking about the motion of a fixed point on the celestial sphere, i.e., with fixed RA & declination.
PM 2Ring
CC BY-SA 4.0
2020-03-06T14:43:30.627Z
@usernumber (cont) A point on the celestial equator & a point on the small circle with declination 60° both travel 360° per sidereal day around the axis. But the circle at dec 60° has half the circumference of the equator, hence in one hour a point on the dec 60° small circle traces an arc of half the length that a point on the celestial equator (which is a great circle) traces.
PM 2Ring
CC BY-SA 4.0
2020-03-06T14:53:41.613Z
Maybe it's helpful to consider the equivalent terrestrial situation. A nautical mile is defined as the length of an arc of 1 minute on any meridian of an ideal spherical Earth, so 60 NM (nautical miles) equals 1 degree on the meridian. If you travel exactly north or south along a meridian by 60 NM your latitude changes by 1°. Like the meridians, the equator is a great circle, so if you go 60 NM along the equator your longitude changes by 1°. But if you travel along the *small* circle of latitude 60° for 60 NM your longitude changes by 2°.
PM 2Ring
CC BY-SA 4.0
2020-03-07T02:33:44.403Z
+1 Useful. @usernumber, the last paragraph of [this answer](/a/35377) may or may not help.
Mike G
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2020-03-02T12:30:19.583Z
2020-03-02T12:40:03.553Z
2020-03-02T12:40:03.553Z
PM 2Ring
<p>A comment under <a href="https://astronomy.stackexchange.com/a/35235/9527">this answer</a> states that the apparent angular speed of the Sun is 8% slower during the solstices. This is rather counter-intuitive, since the rotation speed of the Earth is constant (or close enough for the timescales considered).</p>

<p>Why does the Sun appear to move slower in the sky at the solstices?</p>

<the-sun>
Why does the Sun move slower during the solstices?
usernumber
# Why does the Sun move slower during the solstices?

(Asked by: username_0 on 2020-03-02)

A comment under [this answer](https://astronomy.stackexchange.com/a/35235/9527) states that the apparent angular speed of the Sun is 8% slower during the solstices. This is rather counter-intuitive, since the rotation speed of the Earth is constant (or close enough for the timescales considered).

Why does the Sun appear to move slower in the sky at the solstices?

**Tags:** the-sun


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## 1 answers

### Answer by username_1 on 2020-03-02. Score: 5

All points on the celestial sphere execute a full circle every (sidereal) day, but the speed of a point with nonzero declination is slower than a point on the celestial equator because it's traveling on a small circle, not a great circle. This is exactly the same as how a point on the terrestrial equator travels at a higher speed than a point not on the equator. Eg, a point with a latitude or declination of 60° travels at half the speed of a point on the equator, because cos(60°) = 0.5

At the solstices, the Sun is on either the tropic of Cancer or Capricorn, so it has its maximum or minimum declination, approximately ±23.4°. So its speed is cos(23.4) $\approx$ 0.9178 relative to a point on the celestial equator, or about 8% slower, as username_2 mentioned in that comment.

Here's a diagram from an [answer about latitude speed](https://space.stackexchange.com/a/21069) by SF. on the Space Exploration Stack Exchange:

![latitude speed diagram](https://i.sstatic.net/XiyIp.png)

> **Comments:**
> **username_0**: I guess I didn't phrase my question right. I wanted to know about the angular speed, because that's what is mentioned in the linked answer.  
> **username_1**: @username_0 Well, "angular speed" can be a bit ambiguous. ;) But my answer is talking about the same thing that username_2 is. Eg, in 1 hour a star on the celestial equator traces out an arc across the sky of 15°, but a star at 60° declination traces out an arc of half that length, and a star close to the celestial pole only moves along a very small arc, although its arc will have a lot more curvature.  
> **username_1**: @username_0 Is there some reason why you are still not happy with this answer?  
> **username_0**: @PM2Ring I still haven't been able to wrap my head around the fact that the Sun can move through the sky 8% slower than 15° per hour.  
> **username_1**: @username_0 Firstly, I guess I should make it clear that in my answer I'm ignoring the fact that the Sun's RA & declination change, since that's only about 1° per day. I'm just talking about the motion of a fixed point on the celestial sphere, i.e., with fixed RA & declination.  
> **username_1**: @username_0 (cont) A point on the celestial equator & a point on the small circle with declination 60° both travel 360° per sidereal day around the axis. But the circle at dec 60° has half the circumference of the equator, hence in one hour a point on the dec 60° small circle traces an arc of half the length that a point on the celestial equator (which is a great circle) traces.  
> **username_1**: Maybe it's helpful to consider the equivalent terrestrial situation. A nautical mile is defined as the length of an arc of 1 minute on any meridian of an ideal spherical Earth, so 60 NM (nautical miles) equals 1 degree on the meridian. If you travel exactly north or south along a meridian by 60 NM your latitude changes by 1°. Like the meridians, the equator is a great circle, so if you go 60 NM along the equator your longitude changes by 1°. But if you travel along the *small* circle of latitude 60° for 60 NM your longitude changes by 2°.  
> **username_2**: +1 Useful. @username_0, the last paragraph of [this answer](/a/35377) may or may not help.  

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