<blockquote>
<p>All stars have to cover an angle of 360° in one sidereal day.</p>
</blockquote>
<p>is close but incomplete; we should really say that <em>right ascension</em> covers 360° in one <em>sidereal</em> day.</p>
<p>Let me state that analogy: Imagine that you are near the north pole. You start walking with the velocity of 5 km/h to the east. In a short amount of time, you are at the beginning. You went through all the time zones and geographical longitudes. Now you fly to the equator and do the same. Now you need a lot of time to go around the Earth. Change of geographical longitude is same in both cases. But the angle, measured from the center of the Earth, is different - at the north pole 0° and at the equator 360°.</p>
<p>With that analogy, you can understand the stars movement.</p>
<p>Let me give the last proof: has the north pole star same angular speed as Rigel?</p>

CC BY-SA 4.0
2021-01-15T20:24:27.773Z
What confuses me is the following-
The celestial sphere rotates around the earth at an angular velocity of $\frac{360}{24}$degrees/hr. So, all stars rotate at the same angular velocity(except the north pole star which doesn't rotate at all assuming it's at the exact celestial pole). So, why should the angular velocity depend on declination. I understand the case of north pole star but I don't get why other stars should have different angular velocities depending on their declinations.
James
CC BY-SA 4.0
2021-01-16T09:23:55.910Z
Don't look at the change of right ascension. Here we are interested in the angle for the observer on Earth. Look at some images of star trails. Celestial sphere is just some astronomical construct. Rotation of the sphere is given on the equator. But just some stars are near equator.
User123
CC BY-SA 4.0
2022-09-22T08:36:43.997Z
Hmm... if $\delta$ is $90^\circ$, how can a point on the North Celestial Pole have infinite angular velocity?
Cheng
CC BY-SA 4.0
2021-01-15T19:23:38.863Z
2021-01-16T00:47:12.960Z
2021-01-16T00:47:12.960Z
User123
<p>The angular drift velocity of stars is not the same for all stars and goes as <span class="math-container">$\frac{1}{\cos(\delta)}$</span>. I don't understand why it should depend on the declination. All stars have to cover an angle of 360° in one sidereal day. Why should the angular velocity depend on the declination at all?</p>

CC BY-SA 4.0
2021-01-15T12:38:30.380Z
Compare the apparent motion of a star near a celestial pole vs a star near the celestial equator.
PM 2Ring
CC BY-SA 4.0
2022-09-22T08:34:11.367Z
For better context, https://astronomy.stackexchange.com/questions/48895/deriving-the-apparent-angular-speed-of-a-star
Cheng
<observational-astronomy>
A factor of cos(declination) in the formula of apparent of drift velocity of stars
James
# A factor of cos(declination) in the formula of apparent of drift velocity of stars

(Asked by: username_0 on 2021-01-15)

The angular drift velocity of stars is not the same for all stars and goes as $\frac{1}{\cos(\delta)}$. I don't understand why it should depend on the declination. All stars have to cover an angle of 360° in one sidereal day. Why should the angular velocity depend on the declination at all?

**Tags:** observational-astronomy

> **Comments:**
> **username_1**: Compare the apparent motion of a star near a celestial pole vs a star near the celestial equator.  
> **username_2**: For better context, https://astronomy.stackexchange.com/questions/48895/deriving-the-apparent-angular-speed-of-a-star  

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## 1 answers

### Answer by username_3 on 2021-01-15. Score: 1

> All stars have to cover an angle of 360° in one sidereal day.

is close but incomplete; we should really say that *right ascension* covers 360° in one *sidereal* day.

Let me state that analogy: Imagine that you are near the north pole. You start walking with the velocity of 5 km/h to the east. In a short amount of time, you are at the beginning. You went through all the time zones and geographical longitudes. Now you fly to the equator and do the same. Now you need a lot of time to go around the Earth. Change of geographical longitude is same in both cases. But the angle, measured from the center of the Earth, is different - at the north pole 0° and at the equator 360°.

With that analogy, you can understand the stars movement.

Let me give the last proof: has the north pole star same angular speed as Rigel?

> **Comments:**
> **username_0**: What confuses me is the following- The celestial sphere rotates around the earth at an angular velocity of $\frac{360}{24}$degrees/hr. So, all stars rotate at the same angular velocity(except the north pole star which doesn't rotate at all assuming it's at the exact celestial pole). So, why should the angular velocity depend on declination. I understand the case of north pole star but I don't get why other stars should have different angular velocities depending on their declinations.  
> **username_3**: Don't look at the change of right ascension. Here we are interested in the angle for the observer on Earth. Look at some images of star trails. Celestial sphere is just some astronomical construct. Rotation of the sphere is given on the equator. But just some stars are near equator.  
> **username_2**: Hmm... if $\delta$ is $90^\circ$, how can a point on the North Celestial Pole have infinite angular velocity?  

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