<p>Obviously the answer depends on what the density of such a sphere was!</p>
<p>The only sensible way I can interpret this question is to ask what is the combined volume occupied by all of the galaxy's stars and then get a spherical radius from this.</p>
<p>A modal star in the galaxy has a mass of around 0.3 times the Sun and a radius of 0.3 times the Sun. Since the stellar mass distribution is quite peaked at that value, we could do worse than simply assume that we have about <span class="math-container">$10^{11}$</span> stars like this, since stars of higher and lower masses are less common.</p>
<p>In which case, the radius containing a similar total volume is just <span class="math-container">$0.3\times 10^{11/3}\ R_\odot= 1400R\odot$</span>.</p>
<p>The problem with this calculation is that although the masses of stars are quite peaked, their radii are not. Some evolved stars individually have radii that are a significant fraction of the combined radius figure. Thus although the figure above might be good for the combined radius of all main sequence stars it certainly isn't a good number if you wanted to include all giant stars, with the densities they have now.</p>

CC BY-SA 4.0
2021-04-17T16:06:50.650Z
I suppose the problem is always that the underlying point of curiosity involves all sorts of things that defy reality. What my 9 year old was really asking was - if the empty space between the stars was removed BUT if the stars maintained their structural integrity as balls of spheres of whatever diameter they currently are, pressed against one another - how large would this "galaxy" be. It's very hard to ask such a question because in the real world such a construct is not possible but I was hoping that an abstraction of this type could still be calculable.
Yevgeny Simkin
CC BY-SA 4.0
2021-03-12T09:09:24.560Z
2021-03-12T09:09:24.560Z
ProfRob
<p>There is some simple back-of-the envelope reverse-engineering which might be worth sharing:</p>
<ol>
<li><p>The following <a href="https://en.wikipedia.org/wiki/Milky_Way" rel="nofollow noreferrer">Wikipedia</a> quote gives an approximate of the overall mass of the Milky way:</p>
<blockquote>
<p>In March 2019, astronomers reported that the mass of the Milky Way galaxy is 1.5 trillion solar masses within a radius of about 129,000 light-years, over twice as much as was determined in earlier studies, and suggesting that about 90% of the mass of the galaxy is dark matter.</p>
</blockquote>
<p>This would give around <span class="math-container">$1.5 \cdot 10^{11} M_\odot$</span> for the overall visible mass of the Milky way.</p>
</li>
<li><p>Assuming around 100 billion stars for the Milky way, we would end up with <span class="math-container">$1.5 M_\odot$</span> for an average star, if we assumed that all visible matter are stars. As <a href="https://astronomy.stackexchange.com/users/16685/pm-2ring">PM 2Ring</a> pointed out in a comment, this assumption is far from being correct.</p>
</li>
</ol>
<p>In a nutshell, in terms of mass (not volume), approximating all stars with a single solar mass per star is at least roughly the same order of magnitude, which I found rather suprising.</p>
<p>Converting mass to volume via density, is not too straight forward though, because we would have to assume a size and a density (distribution) for all of the <span class="math-container">$n \approx  10^{11}$</span> stars.</p>

CC BY-SA 4.0
2021-02-09T17:44:47.040Z
Thanks - I definitely want to exclude the dark matter (since it doesn't contribute to the volume in any way). I'm not looking for anything too accurate, and I figured I'd probably need to do the math myself - just thought there may be an agreed-upon diameter of such a construct.
Yevgeny Simkin
CC BY-SA 4.0
2021-02-09T19:55:29.127Z
The mass *in stars* is fairly well constrained to be about $5 \times 10^{10}$ solar masses (e.g., [Gerhard & Bland-Hawthorn 2016](https://arxiv.org/abs/1602.07702)).
Peter Erwin
CC BY-SA 4.0
2021-03-12T08:55:05.607Z
The average stellar mass is certainly not 1.5 solar masses. We know the average mass of a star far better than we know the number of stars in the galaxy.
ProfRob
CC BY-SA 4.0
2021-02-09T08:41:24.390Z
2021-02-09T22:17:27.873Z
2021-02-09T22:17:27.873Z
B--rian
<p>Neglecting the fact that this ball of gas would just collapse on itself - I'm curious if there's an agreed-upon measurement that takes into account the volume of all the ~100 billion stars to predict the diameter of a milky way where all the gas is in one orb. If not, is it safe to just use the volume of our Sun (which is an &quot;average&quot; star) and multiply by 100 billion or is there a more accurate data set to include in the calculation?</p>
<p>Asking for a 6-year-old :)</p>

CC BY-SA 4.0
2021-02-09T03:57:34.623Z
It's a bit tricky. :) I *think* that just multiplying the Sun's volume by the number of stars would give a value that's too big, because there are a *lot* of red dwarfs. You may enjoy a related (vut much simpler) calculation I did recently in [this answer](https://astronomy.stackexchange.com/a/41005/16685): how thick would the Milky Way be if we squashed it down into a flat disk, with the same density as water? That calculation includes dark matter, as well as all the stray gas and dust that isn't inside star systems (there's quite a lot of that).
PM 2Ring
CC BY-SA 4.0
2021-03-12T07:31:16.600Z
@PM2Ring Our sun is well above the _median_ size for stars, but it’s probably below the _mean_. Which is what matters here. No star is more than one solar mass smaller than the sun, but big stars can be dozens (or in extreme cases, hundreds)  of solar masses bigger.
Mike Scott
CC BY-SA 4.0
2021-03-12T08:37:52.163Z
@Mike Sure, the mean is what we want for this calculation, but I think the Sun's mass is *above* the mean. Using the value of 50 billion solar masses from Peter Erwin's comment, and 100 to 400 billion stars, that gives a mean stellar mass of $\frac18$ to $\frac12$ $M_\odot$.
PM 2Ring
CC BY-SA 4.0
2021-03-12T08:52:30.697Z
It obviously depends on the density of the sphere. This isn't a sensible question.
ProfRob
<milky-way>
How big would the Milky Way be if all the stars were emptied into one sphere?
Yevgeny Simkin
# How big would the Milky Way be if all the stars were emptied into one sphere?

(Asked by: username_0 on 2021-02-09)

Neglecting the fact that this ball of gas would just collapse on itself - I'm curious if there's an agreed-upon measurement that takes into account the volume of all the ~100 billion stars to predict the diameter of a milky way where all the gas is in one orb. If not, is it safe to just use the volume of our Sun (which is an "average" star) and multiply by 100 billion or is there a more accurate data set to include in the calculation?

Asking for a 6-year-old :)

**Tags:** milky-way
> **Comments:**
> **username_1**: It's a bit tricky. :) I *think* that just multiplying the Sun's volume by the number of stars would give a value that's too big, because there are a *lot* of red dwarfs. You may enjoy a related (vut much simpler) calculation I did recently in [this answer](https://astronomy.stackexchange.com/a/41005/16685): how thick would the Milky Way be if we squashed it down into a flat disk, with the same density as water? That calculation includes dark matter, as well as all the stray gas and dust that isn't inside star systems (there's quite a lot of that).  
> **username_2**: @PM2Ring Our sun is well above the _median_ size for stars, but it’s probably below the _mean_. Which is what matters here. No star is more than one solar mass smaller than the sun, but big stars can be dozens (or in extreme cases, hundreds)  of solar masses bigger.  
> **username_1**: @Mike Sure, the mean is what we want for this calculation, but I think the Sun's mass is *above* the mean. Using the value of 50 billion solar masses from Peter Erwin's comment, and 100 to 400 billion stars, that gives a mean stellar mass of $\frac18$ to $\frac12$ $M_\odot$.  
> **username_3**: It obviously depends on the density of the sphere. This isn't a sensible question.  

---

## 2 answers

### Answer by username_3 on 2021-03-12. Score: 2

Obviously the answer depends on what the density of such a sphere was!

The only sensible way I can interpret this question is to ask what is the combined volume occupied by all of the galaxy's stars and then get a spherical radius from this.

A modal star in the galaxy has a mass of around 0.3 times the Sun and a radius of 0.3 times the Sun. Since the stellar mass distribution is quite peaked at that value, we could do worse than simply assume that we have about $10^{11}$ stars like this, since stars of higher and lower masses are less common.

In which case, the radius containing a similar total volume is just $0.3\times 10^{11/3}\ R\_\odot= 1400R\odot$.

The problem with this calculation is that although the masses of stars are quite peaked, their radii are not. Some evolved stars individually have radii that are a significant fraction of the combined radius figure. Thus although the figure above might be good for the combined radius of all main sequence stars it certainly isn't a good number if you wanted to include all giant stars, with the densities they have now.

> **Comments:**
> **username_0**: I suppose the problem is always that the underlying point of curiosity involves all sorts of things that defy reality. What my 9 year old was really asking was - if the empty space between the stars was removed BUT if the stars maintained their structural integrity as balls of spheres of whatever diameter they currently are, pressed against one another - how large would this "galaxy" be. It's very hard to ask such a question because in the real world such a construct is not possible but I was hoping that an abstraction of this type could still be calculable.  

---

### Answer by username_4 on 2021-02-09. Score: 0

There is some simple back-of-the envelope reverse-engineering which might be worth sharing:

1. The following [Wikipedia](https://en.wikipedia.org/wiki/Milky_Way) quote gives an approximate of the overall mass of the Milky way:

   > In March 2019, astronomers reported that the mass of the Milky Way galaxy is 1.5 trillion solar masses within a radius of about 129,000 light-years, over twice as much as was determined in earlier studies, and suggesting that about 90% of the mass of the galaxy is dark matter.

   This would give around $1.5 \cdot 10^{11} M\_\odot$ for the overall visible mass of the Milky way.
2. Assuming around 100 billion stars for the Milky way, we would end up with $1.5 M\_\odot$ for an average star, if we assumed that all visible matter are stars. As [PM 2Ring](https://astronomy.stackexchange.com/users/16685/pm-2ring) pointed out in a comment, this assumption is far from being correct.

In a nutshell, in terms of mass (not volume), approximating all stars with a single solar mass per star is at least roughly the same order of magnitude, which I found rather suprising.

Converting mass to volume via density, is not too straight forward though, because we would have to assume a size and a density (distribution) for all of the $n \approx 10^{11}$ stars.

> **Comments:**
> **username_0**: Thanks - I definitely want to exclude the dark matter (since it doesn't contribute to the volume in any way). I'm not looking for anything too accurate, and I figured I'd probably need to do the math myself - just thought there may be an agreed-upon diameter of such a construct.  
> **username_5**: The mass *in stars* is fairly well constrained to be about $5 \times 10^{10}$ solar masses (e.g., [Gerhard & Bland-Hawthorn 2016](https://arxiv.org/abs/1602.07702)).  
> **username_3**: The average stellar mass is certainly not 1.5 solar masses. We know the average mass of a star far better than we know the number of stars in the galaxy.  

---

