I want to know the volumetric ratio of empty space to planets, stars, and all other solid matter combined. Not considering any dark matter, dark energy, or black holes.

For example, you can say X percent of the observable space is just empty space. The remainings are comets, stars, planets, and so on.


  1. I meant solid as in all the things that are not gas.
  2. I know that space is not exactly empty, let's consider that empty means all other things except solid.
  3. I'm not looking for exact numbers but just an estimation.
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    $\begingroup$ Are you including gases in the "solid" matter? Most of the normal matter (i.e , not dark matter) in the universe is hydrogen or helium, and most of that is in a gaseous or plasma state. Admittedly, the density of star plasma gets rather high as you approach the core of the star. $\endgroup$
    – PM 2Ring
    Dec 30, 2020 at 16:03
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    $\begingroup$ BTW, that "empty space" isn't exactly empty. The matter in the space between stars & even between galaxies has a very low density, but there's a lot of it, and it makes up a significant percentage of the normal matter in the universe. $\endgroup$
    – PM 2Ring
    Dec 30, 2020 at 18:00
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    $\begingroup$ Don't have the time for full answer, but see ned.ipac.caltech.edu/level5/March04/Fukugita/Fukugita1.html $\endgroup$ Dec 30, 2020 at 18:24
  • $\begingroup$ @AndersSandberg That page deals with mass, the OP asked about "volume" for which I think the answer is $100\% (-\epsilon)$ is empty (but some clarification on what "solid" and "empty") mean before anyone can do better) $\endgroup$
    – James K
    Dec 30, 2020 at 21:49
  • $\begingroup$ I good approximation would be to multiple the average size ( -> volume) of the stars with the count of the stars of the visible Universe. Then calculate the volume of the visible universe, and divide it. The result will be very close to 100% (I think about the 20th or 30th digit will you see the first not-9). $\endgroup$
    – peterh
    Dec 30, 2020 at 22:53

2 Answers 2


Any non-star is negligible, their volume is practically zero.

Most matter in the Universe is gravitationally bound plasma, resulting that they do not have a well-specifiable volume, because their density continuously changes from the interstellar vacuum to many times of lead. But choosing a surface at any arbitrary point (for example, at 1 bar pressure) does not modify the result significantly (this is why we can say a radius for the Sun).

This source says that the estimated count of the stars in the visible Universe is $10^{21}$. This is obviously not very exact, there are probably better estimations which you substitute into this answer.

The majority of the stars in the Universe is a red dwarf. Their radius varies between 9% and 50% of the Sun, we get their geometric average, this is $\approx$ 21% of the Sun. Their volume is $\frac{4 \pi}{3}(0.21 \cdot 700000km)^3 \approx 1.37 \cdot 10^{16} km^3$.

Thus, the volume of all stars in the Universe is $\approx 1.37 \cdot 10^{37} km^3$.

The observable Universe is a sphere with the radius of 46billion light years.1 A light year is $\approx 9.5 \cdot 10^{12} km$. Thus, the volume of the observable Universe is $\frac{4 \pi}{3} (23 \cdot 10^{12} \cdot 9.5 \cdot 10^{12})^3 \approx 4.37 \cdot 10^{79} km^3$.

After a division, we get that

$$\frac{1.37 \cdot 10^{37}}{4.37 \cdot 10^{79}} \approx \underline{\underline{\frac{1}{3.2 \cdot 10^{42}}}}$$

part of the volume of the visible Universe is not empty space.

About 1 digit difference from the reality is possible due to the lack of not enough exact approximations.

1This is not the well-known and false 13.7 billion light years, which is the way what light could do since the Big Bang. The difference is coming from that meanwhile the Universe expanded, thus the stars we can see (as if they would be 13.7 billion ly far away) are now 46billion ly away. More details are on the wiki link.

  • 4
    $\begingroup$ (I think you have your fraction upside down) To interpret this, this means that about 99.9999999999999999999999999999999999999999% of the universe is "empty space" (not stars, etc) $\endgroup$
    – James K
    Dec 31, 2020 at 0:09
  • $\begingroup$ @JamesK Thanks, fixed. $\endgroup$
    – peterh
    Dec 31, 2020 at 0:19
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    $\begingroup$ This was my first question and I think I got my best answer. Brilliant. $\endgroup$ Dec 31, 2020 at 17:41

One way of estimating this is to consider the fraction of matter that is solid, and then make use of estimates of its density. The cosmic energy inventory includes a fraction 0.00036 ± 0.00008 in the form of white dwarf stars, 0.00005 ± 0.00002 as neutron stars, $10^{-6}$ as planets, $10^{-5.6}$ as other condensed matter. The rest is dark matter, gas, plasma and similar non-solid matter.

It turns out that solid molecular matter has a density on the order of 1000 kg per cubic meter, varying within an order of magnitude. Degenerate matter in white dwarf stars range over $10^7$ to $10^{10}$ kg per cubic meter, while the neutron stars lie around $10^{17}$. The latter case is a borderline case, since they may actually be largely liquid rather than solid.

The universe has a mass density of $9.9\cdot 10^{-27}$ kg per cubic meter (WMAP). If we invert it, we get that one kg of generic mass-energy takes up $10^{26}$ cubic meters. Of this a fraction 0.00036 kg is white dwarf stars, taking up a volume $1.1\cdot 10^{-12}$ cubic meters (assuming density $10^{8.5}$). Neutron stars make up $5\cdot 10^{-22}$ , and solid matter $4.1\cdot 10^{-9}$ cubic meters. So, summing up, I get a volume fraction of solid matter of $4.2\cdot 10^{-35}$. Most of the volume is likely interstellar dust.

This calculation is independent of the size of the visible universe. But the estimates in Peebles and Fukugita are also likely just within an order of magnitude for the planets and dust.


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