<p>You can think of the position given in your title as the coordinates of the star that is directly over the Sun's north pole. (Similar to how Polaris is <em>almost</em> directly over the Earth's north pole.) Likewise, the ecliptic plane has a north pole which can be located by right ascension and declination.</p>

<p>7.25° to 7.26° appears to be the correct calculation, based on the following:</p>

<ul>
<li>North pole of sun: 286.13° RA, 63.87° declination</li>
<li>North pole of ecliptic: 270° RA, 66.5738° declination (per <a href="https://en.wikipedia.org/wiki/Orbital_pole" rel="noreferrer">Wikipedia</a>)</li>
</ul>

<p><span class="math-container">$$cos(angle)=sin(dec1)sin(dec2)+cos(dec1)cos(dec2)cos(RA1-RA2)$$</span>
<span class="math-container">$$cos(angle)=sin(63.87)sin(66.5738)+cos(63.87)cos(66.5738)cos(286.13-270)$$</span>
<span class="math-container">$$angle=7.26$$</span></p>

CC BY-SA 4.0
2018-10-15T22:54:30.863Z
Ohh I used: North pole of ecliptic 282° RA, 66.57° declination for calculating, dont ask me why. So they use the Aequatorial System centered in the Sun, I didnt know that works, but it makes sense since Equinox is the only "invariable" direction in the ecliptic plane. Thanks a lot John, you helped me a lot.
Andres Huster
CC BY-SA 4.0
2023-05-20T22:56:44.313Z
I have a hard time making sense of this. Isn't 0° of RA defined to be at the March equinox? If so, how can the rotational axis of Sol possibly be towards ~286° RA when the axis of rotation points roughly towards or away from Earth at the equinoctes? I would have expected the pole to have an RA of closer to 0° or 180°.
Outis Nemo
CC BY-SA 4.0
2023-05-21T01:03:51.883Z
Hi @OutisNemo. I think you are confusing the Earth's axis (which points toward the Sun at summer) and the Sun's axis which is the topic of this question. The Sun's axis has no relationship to the Earth (at least none for this question.) In theory, the Sun's axis could be pointing anywhere in space. Also, a line of right ascension stretches from the celestial south pole (-90 declination) to the north pole (+90 declination). You need both right ascension and declination to locate a point on the sky.
JohnHoltz
CC BY-SA 4.0
2023-05-21T10:57:27.133Z
@JohnHoltz: Not really, I'm talking about the Solar rotational axis. However, since I posted this yesterday I've figured it out, and as expected it points to roughly ~350° in terms of ecliptic longitude. It was just somewhat unintuitive that ~286° RA could correspond to that.
Outis Nemo
CC BY-SA 4.0
2018-10-15T16:53:01.070Z
2018-10-15T16:53:01.070Z
JohnHoltz
<p>first post here.</p>

<p>I found 1 related Question, which has not solved my Problem.</p>

<p>SHORT QUESTIONs VERSION:</p>

<p>Suns axis is 7,25° tilted from the ecliptic. But in which direction?</p>

<p>LONG QUESTION:</p>

<p>Nasa defines the Suns Rotation axis as described in the title:
<a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html" rel="nofollow noreferrer">https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html</a></p>

<p><a href="https://i.sstatic.net/0mkXW.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0mkXW.gif" alt="enter image description here"></a></p>

<p>Why is it described with (Ra, Declination)? Is this a (Earth) Aequatorial System placed in the middle of the Sun?</p>

<p>If true, then this should work (<a href="https://astronomy.stackexchange.com/questions/11849/what-is-the-angle-between-the-equator-of-the-sun-and-the-plane-of-the-earths-o">What is the angle between the equator of the sun, and the plane of the Earth&#39;s orbit</a>). As the best anwer says, the angle between ecliptical pole and sun Rotation axis should be 7.25°. I get 4°, what is going on here? Have I messed up the angle calculation?</p>

<p>Thanks a lot, this is making my Brain burn : D</p>

<the-sun><earth><coordinate>
What does it mean that the Sun`s rotation axis is: RA 286.13° , Dekl: 63.87°?
Andres Huster
# What does it mean that the Sun`s rotation axis is: RA 286.13° , Dekl: 63.87°?

(Asked by: username_0 on 2018-10-15)

first post here.

I found 1 related Question, which has not solved my Problem.

SHORT QUESTIONs VERSION:

Suns axis is 7,25° tilted from the ecliptic. But in which direction?

LONG QUESTION:

Nasa defines the Suns Rotation axis as described in the title:
<https://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html>

[![enter image description here](https://i.sstatic.net/0mkXW.gif)](https://i.sstatic.net/0mkXW.gif)

Why is it described with (Ra, Declination)? Is this a (Earth) Aequatorial System placed in the middle of the Sun?

If true, then this should work ([What is the angle between the equator of the sun, and the plane of the Earth's orbit](https://astronomy.stackexchange.com/questions/11849/what-is-the-angle-between-the-equator-of-the-sun-and-the-plane-of-the-earths-o)). As the best anwer says, the angle between ecliptical pole and sun Rotation axis should be 7.25°. I get 4°, what is going on here? Have I messed up the angle calculation?

Thanks a lot, this is making my Brain burn : D

**Tags:** the-sun, earth, coordinate

---

## 1 answers

### Answer by username_1 on 2018-10-15. Score: 5

You can think of the position given in your title as the coordinates of the star that is directly over the Sun's north pole. (Similar to how Polaris is *almost* directly over the Earth's north pole.) Likewise, the ecliptic plane has a north pole which can be located by right ascension and declination.

7.25° to 7.26° appears to be the correct calculation, based on the following:

* North pole of sun: 286.13° RA, 63.87° declination
* North pole of ecliptic: 270° RA, 66.5738° declination (per [Wikipedia](https://en.wikipedia.org/wiki/Orbital_pole))

$$cos(angle)=sin(dec1)sin(dec2)+cos(dec1)cos(dec2)cos(RA1-RA2)$$
$$cos(angle)=sin(63.87)sin(66.5738)+cos(63.87)cos(66.5738)cos(286.13-270)$$
$$angle=7.26$$

> **Comments:**
> **username_0**: Ohh I used: North pole of ecliptic 282° RA, 66.57° declination for calculating, dont ask me why. So they use the Aequatorial System centered in the Sun, I didnt know that works, but it makes sense since Equinox is the only "invariable" direction in the ecliptic plane. Thanks a lot John, you helped me a lot.  
> **username_2**: I have a hard time making sense of this. Isn't 0° of RA defined to be at the March equinox? If so, how can the rotational axis of Sol possibly be towards ~286° RA when the axis of rotation points roughly towards or away from Earth at the equinoctes? I would have expected the pole to have an RA of closer to 0° or 180°.  
> **username_1**: Hi @OutisNemo. I think you are confusing the Earth's axis (which points toward the Sun at summer) and the Sun's axis which is the topic of this question. The Sun's axis has no relationship to the Earth (at least none for this question.) In theory, the Sun's axis could be pointing anywhere in space. Also, a line of right ascension stretches from the celestial south pole (-90 declination) to the north pole (+90 declination). You need both right ascension and declination to locate a point on the sky.  
> **username_2**: @JohnHoltz: Not really, I'm talking about the Solar rotational axis. However, since I posted this yesterday I've figured it out, and as expected it points to roughly ~350° in terms of ecliptic longitude. It was just somewhat unintuitive that ~286° RA could correspond to that.  

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