<p>The Earth rotates approximately <span class="math-container">$ 15^\circ$</span> per hour.  Or <span class="math-container">$ .25^\circ $</span> per minute.  So, multiplying <span class="math-container">$\frac{.25^\circ}{min}$</span>by the cosine of the declination would give you the resulting angular speed in deg/min.</p>
<p>I would like to point out that this equation is off a bit.  The <span class="math-container">$15^\circ$</span> deg/hour is based on a 24 hour rotational period of the Earth.  However, the actual rotation is called a Sidereal day, and is 23 hours, 56 minutes, and ~ 4 seconds.  It won't add up to much over the course of a minute or two, but it becomes obvious over the course of several hours, and does trip a lot of people up.</p>

CC BY-SA 4.0
2022-03-23T04:49:58.470Z
2022-03-23T04:49:58.470Z
Greg Miller
<p>As you mention, the angular speed of the celestial sphere on the equator is 0.25° per minute. The speed on a declination circle is proportional to the circumference of that circle, relative to the circumference of the celestial equator.</p>
<p>Of course, the circumference of a circle is proportional to its radius. It's convenient to use a unit sphere for the celestial sphere, so the radius of a declination circle is simply the cosine of the declination angle. This leads us to the speed formula given in the question, <span class="math-container">$\cos(\delta) × 0.25°$</span> per minute.</p>
<p>These diagrams may be helpful.</p>
<p><a href="https://i.sstatic.net/TKujH.png" rel="noreferrer"><img src="https://i.sstatic.net/TKujH.png" alt="Unit circle with sin &amp; cos" /></a></p>
<p>Here's a 3D version:</p>
<p><a href="https://i.sstatic.net/2iSEI.png" rel="noreferrer"><img src="https://i.sstatic.net/2iSEI.png" alt="3D declination diagram" /></a></p>
<p>It's easier to see what's going on in the <a href="https://sagecell.sagemath.org/?z=eJyFU8GK2zAQvfsrBl1iZ7VZx6FQCoZCoT0Vkl56KEvQ2hNbrWMJWfZu9us7Izsbky1pSMRoZp7e09NECAFfsMHOa9VAZ2t0KOFZ-xoUlFg0ulVemxYK7YoGI6DPT6e9xxaeTrD9DtkP3VaQpVm2SjerbBMJIaJoUC5e9HJYJJG1kINVTh3RO13sbWP8poxKrDiv4QHWH9OoJMYc1h9gSbRVROUDvJxe40Z5CY1pk0-B21FTYTpOJ2MCfe9acMuQpT5JcafbKQ4R90bGMV8qwzfqS3OUMNBKybiXcG_1QyaBV4LFQ8jwPom-sU4bsxpqHKh8QUtS0xiXi8ohtkKCsarQ_pSnqzRj6F0OXX-ML_hRV38buwF94E7ApkNY81UPxoWMbqFzqq0wTlnekkSTd8l7ruAcq_2PzpFL-Wsuyly4iGLm0Jly5LRGt_SiMTl8RTOJoiFCqv8ixam8X7O_HK2Tx9v94cB4Ta3pP1qjz0SLThU-DMueL5zHi9nQLiRsUmKz6DqLhdcD5l8V3ZL0K_fnHKvem31vS-Wn8jRr7MAyD-PI2y2NAcmDm_Y-NT2KcTJfJJwkvBJqR783SDpWre8oO11xnMpXOmxHuxFIV55oZxYT6o3KYTkxbWem8bl307m7x6vm0O1rPCJxC_ZA8ONzML6-aHRVezEqdOY3m0YzR80zDwNmtp-gxvnaVE7ZWhfjCdtVV5vn-MB__bPZgT0Pq5xR5Jcw-QsfV12r&amp;lang=sage" rel="noreferrer">interactive version</a>.</p>

CC BY-SA 4.0
2022-03-23T10:26:59.200Z
2022-03-23T10:26:59.200Z
PM 2Ring
<p>The angular speed of stars in the sky is constant but the apparent angular speed is not, due to the star moving faster on the celestial equator and slower near the celestial pole. The apparent angular speed, modified with <span class="math-container">$\cos(\delta)$</span>, where <span class="math-container">$\delta$</span> is declination. The full equation:
<span class="math-container">$$\frac{0.25^{\circ}}{min}\cos(\delta)$$</span>
How is this equation derived? Resolving the speed on the celestial equator yields the same equation, but the method doesn't seem to be very convincing to me.</p>
<p><a href="https://i.sstatic.net/3gWoF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3gWoF.png" alt="enter image description here" /></a></p>

<star><orbital-mechanics><declination><apparent-motion>
Deriving the apparent angular speed of a star
radastro
# Deriving the apparent angular speed of a star

(Asked by: username_0 on 2022-03-23)

The angular speed of stars in the sky is constant but the apparent angular speed is not, due to the star moving faster on the celestial equator and slower near the celestial pole. The apparent angular speed, modified with $\cos(\delta)$, where $\delta$ is declination. The full equation:
$$\frac{0.25^{\circ}}{min}\cos(\delta)$$
How is this equation derived? Resolving the speed on the celestial equator yields the same equation, but the method doesn't seem to be very convincing to me.

[![enter image description here](https://i.sstatic.net/3gWoF.png)](https://i.sstatic.net/3gWoF.png)

**Tags:** star, orbital-mechanics, declination, apparent-motion

---

## 2 answers

### Answer by username_2 on 2022-03-23. Score: 6 | Accepted answer

As you mention, the angular speed of the celestial sphere on the equator is 0.25° per minute. The speed on a declination circle is proportional to the circumference of that circle, relative to the circumference of the celestial equator.

Of course, the circumference of a circle is proportional to its radius. It's convenient to use a unit sphere for the celestial sphere, so the radius of a declination circle is simply the cosine of the declination angle. This leads us to the speed formula given in the question, $\cos(\delta) × 0.25°$ per minute.

These diagrams may be helpful.

[![Unit circle with sin & cos](https://i.sstatic.net/TKujH.png)](https://i.sstatic.net/TKujH.png)

Here's a 3D version:

[![3D declination diagram](https://i.sstatic.net/2iSEI.png)](https://i.sstatic.net/2iSEI.png)

It's easier to see what's going on in the [interactive version](https://sagecell.sagemath.org/?z=eJyFU8GK2zAQvfsrBl1iZ7VZx6FQCoZCoT0Vkl56KEvQ2hNbrWMJWfZu9us7Izsbky1pSMRoZp7e09NECAFfsMHOa9VAZ2t0KOFZ-xoUlFg0ulVemxYK7YoGI6DPT6e9xxaeTrD9DtkP3VaQpVm2SjerbBMJIaJoUC5e9HJYJJG1kINVTh3RO13sbWP8poxKrDiv4QHWH9OoJMYc1h9gSbRVROUDvJxe40Z5CY1pk0-B21FTYTpOJ2MCfe9acMuQpT5JcafbKQ4R90bGMV8qwzfqS3OUMNBKybiXcG_1QyaBV4LFQ8jwPom-sU4bsxpqHKh8QUtS0xiXi8ohtkKCsarQ_pSnqzRj6F0OXX-ML_hRV38buwF94E7ApkNY81UPxoWMbqFzqq0wTlnekkSTd8l7ruAcq_2PzpFL-Wsuyly4iGLm0Jly5LRGt_SiMTl8RTOJoiFCqv8ixam8X7O_HK2Tx9v94cB4Ta3pP1qjz0SLThU-DMueL5zHi9nQLiRsUmKz6DqLhdcD5l8V3ZL0K_fnHKvem31vS-Wn8jRr7MAyD-PI2y2NAcmDm_Y-NT2KcTJfJJwkvBJqR783SDpWre8oO11xnMpXOmxHuxFIV55oZxYT6o3KYTkxbWem8bl307m7x6vm0O1rPCJxC_ZA8ONzML6-aHRVezEqdOY3m0YzR80zDwNmtp-gxvnaVE7ZWhfjCdtVV5vn-MB__bPZgT0Pq5xR5Jcw-QsfV12r&lang=sage).


---

### Answer by username_1 on 2022-03-23. Score: 2

The Earth rotates approximately $ 15^\circ$ per hour. Or $ .25^\circ $ per minute. So, multiplying $\frac{.25^\circ}{min}$by the cosine of the declination would give you the resulting angular speed in deg/min.

I would like to point out that this equation is off a bit. The $15^\circ$ deg/hour is based on a 24 hour rotational period of the Earth. However, the actual rotation is called a Sidereal day, and is 23 hours, 56 minutes, and ~ 4 seconds. It won't add up to much over the course of a minute or two, but it becomes obvious over the course of several hours, and does trip a lot of people up.


---

