<p>@MikeG's <a href="https://astronomy.stackexchange.com/a/27973/7982">answer</a> is excellent. This is a supplementary answer only.</p>

<p>Part of the problem finding the coordinates of the "center" is defining the center. One example might be the center of mass, but finding all the mass is quite a challenge as we can't even see all of the galaxy's mass due to dust clouds, as discussed in <a href="https://astronomy.stackexchange.com/q/23927/7982">How was the galactic plane established?</a>, links therein and @RobJeffries <a href="https://astronomy.stackexchange.com/a/27981/7982">excellent answer</a>, and of course dark matter being invisible and all. </p>

<p>But even though you can't see and identify all of the mass, you can <em>try to have a look</em> at the way that <em>all of the stars that you can see</em> seem to be moving via their radial Doppler shift, and at least try to estimate it. </p>

<p>However, for the purposes of your question, I'm not sure if that's the only definition of "center" that could work for you.</p>

<p>If instead you'd like to consider something more catchy and visual, how about the <a href="https://astronomy.stackexchange.com/q/23927/7982">supermassive black hole</a> that's almost for sure at the "center" of our galaxy, affectionately known as <a href="https://en.wikipedia.org/wiki/Sagittarius_A*" rel="nofollow noreferrer">Sagittarius A*</a>? That's just Sgr A* to her friends (see also <a href="https://astronomy.stackexchange.com/q/23692/7982">Why is Sagittarius A* called so?</a>). According to that article, the distance from our Sun to Sgr A* is 7,860 ±140 ±40 pc, which is a remarkable thing in itself because it has <em>error bars on the error bars!</em></p>

<p>That number is linked to the ArXiv preprint <a href="https://arxiv.org/abs/1607.05726" rel="nofollow noreferrer">An Improved Distance and Mass Estimate for Sgr A* from a Multistar Orbit Analysis</a>, which is also an analysis of how some very special stars move. </p>

<p>Here's a GIF, found in <a href="https://physics.stackexchange.com/a/406609/83380">this excellent answer</a> to <em>What is the evidence for a supermassive black hole at the center of Milky Way?</em></p>

<p><a href="https://i.sstatic.net/HLjZZ.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HLjZZ.gif" alt="stars orbiting Sgr A*"></a></p>

<p>The reason that I call this a supplementary answer is that you end up with very nearly the same answer as @MikeG. Here I've used the same procedure:</p>

<pre><code>               X                     Y                    Z
parsecs:     -430±8,              -6861±122,           -3812±68
lightyears:  -1401±30,            -22,379±399,         -12433±221
AU:          -8.86(0.16)E+07,     -1.415(0.025)E+09,   -7.86(0.14)E+08
km:          -1.326(0.024)E+16,   -2.117(0.038)E+17,   -1.176(0.021)E+17
</code></pre>

CC BY-SA 4.0
2018-10-12T09:11:22.147Z
The second error bar is an *additional* systematic error. i.e. they have just separated out an additional source of error.
ProfRob
CC BY-SA 4.0
2018-10-12T09:17:00.147Z
@RobJeffries thanks! Decades ago I did hear of a determination of uncertainty in the error in a different context. If I wanted to recombine them and recalculate, can I just do it in quadrature (e.g. ±100±10 → 100.5)?
uhoh
CC BY-SA 4.0
2018-10-12T05:52:14.347Z
2018-10-12T08:05:34.943Z
2018-10-12T08:05:34.943Z
uhoh
<p>Distance estimates to the <a href="https://en.wikipedia.org/wiki/Galactic_Center" rel="noreferrer">center of our galaxy</a> vary around <span class="math-container">$r \approx$</span> 8000 <span class="math-container">$\pm$</span> 200 parsecs.
Its spherical <a href="https://en.wikipedia.org/wiki/Equatorial_coordinate_system" rel="noreferrer">equatorial coordinates</a> are <span class="math-container">$\alpha$</span> = 17h45m40s, <span class="math-container">$\delta$</span> = -29<span class="math-container">$^\circ$</span>00'28" (J2000).
Converting those to rectangular form, we have:
<span class="math-container">$$\begin{align}
X &amp;= r \cos \delta \cos \alpha \approx -437 \pm 11 \ \mathrm{pc} \\
Y &amp;= r \cos \delta \sin \alpha \approx -6983 \pm 175 \ \mathrm{pc} \\
Z &amp;= r \sin \delta \approx -3879 \pm 97 \ \mathrm{pc}
\end{align}$$</span>
As with most celestial objects, its direction is known much more precisely than its distance.
We can't pinpoint its 3D position, but we can narrow it down to a long, fuzzy ellipsoid pointing away from the Sun.</p>

CC BY-SA 4.0
2018-10-12T04:52:03.290Z
Nice answer! Slightly related and currently unanswered: [How was the galactic plane established?](https://astronomy.stackexchange.com/q/23927/7982)
uhoh
CC BY-SA 4.0
2018-10-11T17:49:09.220Z
2018-10-11T17:49:09.220Z
Mike G
<p>I have downloaded a star database from this page (The HYG Database):<br>
<a href="http://astronexus.com/node/34" rel="nofollow noreferrer">http://astronexus.com/node/34</a>  </p>

<p>And all the stars have (among other parameters) their XYZ coordinates in parsecs according to this definition:  </p>

<blockquote>
  <p><strong>X,Y,Z</strong>: The Cartesian coordinates of the star, in a system based on the
  equatorial coordinates as seen from Earth. +X is in the direction of
  the vernal equinox (at epoch 2000), +Z towards the north celestial
  pole, and +Y in the direction of R.A. 6 hours, declination 0 degrees</p>
</blockquote>

<p>The problem is: I cannot figure out how to calculate the XYZ coordinates of the center of the galaxy (the Milky Way). So the question is: What are the coordinates of the Milky Way center? (Or is it not possible to calculate it just with that information?). This is an example of the data:<br>
<a href="https://i.sstatic.net/q4PJB.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/q4PJB.jpg" alt="enter image description here"></a></p>

CC BY-SA 4.0
2018-10-11T21:27:09.430Z
Note that your catalog only contains bright and/or nearby stars and the distances are from Hipparcos and are not likely to be trustworthy beyond a few hundred parsecs. So your stars are going to be in a pretty local bubble around the Sun and the furthest star will be less than ~10 % of the way to the Galactic center. A catalog based on Gaia-DR2 parallaxes will give a better idea of the structure of the Milky Way and the location of the galactic center.
astrosnapper
CC BY-SA 4.0
2018-10-12T05:53:39.440Z
`+1` for  a well explained and well-sourced first question by a new user!
uhoh
<distances><milky-way>
Where is the center of the Milky Way located?
Carlos Zamora
# Where is the center of the Milky Way located?

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

I have downloaded a star database from this page (The HYG Database):  
<http://astronexus.com/node/34>

And all the stars have (among other parameters) their XYZ coordinates in parsecs according to this definition:

> **X,Y,Z**: The Cartesian coordinates of the star, in a system based on the
> equatorial coordinates as seen from Earth. +X is in the direction of
> the vernal equinox (at epoch 2000), +Z towards the north celestial
> pole, and +Y in the direction of R.A. 6 hours, declination 0 degrees

The problem is: I cannot figure out how to calculate the XYZ coordinates of the center of the galaxy (the Milky Way). So the question is: What are the coordinates of the Milky Way center? (Or is it not possible to calculate it just with that information?). This is an example of the data:  
[![enter image description here](https://i.sstatic.net/q4PJB.jpg)](https://i.sstatic.net/q4PJB.jpg)

**Tags:** distances, milky-way
> **Comments:**
> **username_1**: Note that your catalog only contains bright and/or nearby stars and the distances are from Hipparcos and are not likely to be trustworthy beyond a few hundred parsecs. So your stars are going to be in a pretty local bubble around the Sun and the furthest star will be less than ~10 % of the way to the Galactic center. A catalog based on Gaia-DR2 parallaxes will give a better idea of the structure of the Milky Way and the location of the galactic center.  
> **username_2**: `+1` for  a well explained and well-sourced first question by a new user!  

---

## 2 answers

### Answer by username_4 on 2018-10-11. Score: 6 | Accepted answer

Distance estimates to the [center of our galaxy](https://en.wikipedia.org/wiki/Galactic_Center) vary around $r \approx$ 8000 $\pm$ 200 parsecs.
Its spherical [equatorial coordinates](https://en.wikipedia.org/wiki/Equatorial_coordinate_system) are $\alpha$ = 17h45m40s, $\delta$ = -29$^\circ$00'28" (J2000).
Converting those to rectangular form, we have:
$$\begin{align}
X &= r \cos \delta \cos \alpha \approx -437 \pm 11 \ \mathrm{pc} \\
Y &= r \cos \delta \sin \alpha \approx -6983 \pm 175 \ \mathrm{pc} \\
Z &= r \sin \delta \approx -3879 \pm 97 \ \mathrm{pc}
\end{align}$$
As with most celestial objects, its direction is known much more precisely than its distance.
We can't pinpoint its 3D position, but we can narrow it down to a long, fuzzy ellipsoid pointing away from the Sun.

> **Comments:**
> **username_2**: Nice answer! Slightly related and currently unanswered: [How was the galactic plane established?](https://astronomy.stackexchange.com/q/23927/7982)  

---

### Answer by username_2 on 2018-10-12. Score: 3

@MikeG's answer is excellent. This is a supplementary answer only.
Part of the problem finding the coordinates of the "center" is defining the center. One example might be the center of mass, but finding all the mass is quite a challenge as we can't even see all of the galaxy's mass due to dust clouds, as discussed in How was the galactic plane established?, links therein and @RobJeffries excellent answer, and of course dark matter being invisible and all. 
But even though you can't see and identify all of the mass, you can try to have a look at the way that all of the stars that you can see seem to be moving via their radial Doppler shift, and at least try to estimate it. 
However, for the purposes of your question, I'm not sure if that's the only definition of "center" that could work for you.
If instead you'd like to consider something more catchy and visual, how about the supermassive black hole that's almost for sure at the "center" of our galaxy, affectionately known as Sagittarius A*? That's just Sgr A* to her friends (see also Why is Sagittarius A* called so?). According to that article, the distance from our Sun to Sgr A* is 7,860 ±140 ±40 pc, which is a remarkable thing in itself because it has error bars on the error bars!
That number is linked to the ArXiv preprint An Improved Distance and Mass Estimate for Sgr A* from a Multistar Orbit Analysis, which is also an analysis of how some very special stars move. 
Here's a GIF, found in this excellent answer to What is the evidence for a supermassive black hole at the center of Milky Way?

The reason that I call this a supplementary answer is that you end up with very nearly the same answer as @MikeG. Here I've used the same procedure:
               X                     Y                    Z
parsecs:     -430±8,              -6861±122,           -3812±68
lightyears:  -1401±30,            -22,379±399,         -12433±221
AU:          -8.86(0.16)E+07,     -1.415(0.025)E+09,   -7.86(0.14)E+08
km:          -1.326(0.024)E+16,   -2.117(0.038)E+17,   -1.176(0.021)E+17

> **Comments:**
> **username_3**: The second error bar is an *additional* systematic error. i.e. they have just separated out an additional source of error.  
> **username_2**: @RobJeffries thanks! Decades ago I did hear of a determination of uncertainty in the error in a different context. If I wanted to recombine them and recalculate, can I just do it in quadrature (e.g. ±100±10 → 100.5)?  

---

