# How big would the Milky Way be if all the stars were emptied into one sphere?

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?

• 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: 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 Feb 9 at 3:57
• @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 Mar 12 at 7:31
• @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 Mar 12 at 8:37
• It obviously depends on the density of the sphere. This isn't a sensible question. – ProfRob Mar 12 at 8:52

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.

• 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 Apr 17 at 16:06

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

1. The following Wikipedia 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 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.

• 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 Feb 9 at 17:44
• The mass in stars is fairly well constrained to be about $5 \times 10^{10}$ solar masses (e.g., Gerhard & Bland-Hawthorn 2016). – Peter Erwin Feb 9 at 19:55
• 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 Mar 12 at 8:55