I was just reading "Rare Earth" by Peter Ward and Donald Brownlee and they claim that $95\%$ of stars are less massive than the Sun. But, I was always told the Sun is an average star. Which is it?

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    $\begingroup$ physics.stackexchange.com/q/153150 $\endgroup$
    – BowlOfRed
    Nov 21, 2018 at 0:26
  • $\begingroup$ An average planet? Stars and planets are different things. $\endgroup$
    – G. Smith
    Nov 21, 2018 at 0:58
  • $\begingroup$ Being in the top 5% isn't very unusual. Still pretty average, imo $\endgroup$
    – JollyJoker
    Nov 21, 2018 at 8:44
  • $\begingroup$ An average star can still have 95% of the stars less massive. Average doesn't mean typical - it can be skewed by the extremes in a non-normal distribution. The median provides a better measure of "typical". $\endgroup$
    – J...
    Nov 21, 2018 at 11:57

3 Answers 3


The notion that the sun is an 'average' star, is a popular notion, not a scientific one. It is to my knowledge based on the spectral colour-classification (OBAFGKM..) of stars, where the sun (G2-class) hits about the center of this classification, depending how many classes one adds at the end.

However by number, the sun is not average, it is on the rarer end of the mass distribution function. See also this graphic, provided by the University of Colorado:

enter image description here

This has to do with two effects:

1. Heavier stars burn through their fuel much faster than low-mass stars.

The lifetime $t_{\rm life}$ of a star on the main sequence goes like $$t_{\rm life} \sim 10^{10}yrs \left(\frac{M}{M_{\odot}}\right)^{-2.5}$$ so you can do the math yourself how short more massive stars live.

2. Lower mass stars are formed more frequently.

In star-forming environments it is the 3D-turbulence that clumps gas together at smaller and smaller scales (as opposed to larger and larger scales). It is because of this, that there are more small clumps that become gravitationally unstable than large clumps, which then form stars of a given clump mass.

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    $\begingroup$ Where does the sun sit in that graphic? $\endgroup$ Nov 21, 2018 at 12:46
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    $\begingroup$ @DavidTheWin: At stellar mass = one solar mass on the x-axis, right where the yellow ones sit. The numbers on the y-axis are normalized relative to the abundances of OB-stars (the ones between 10 and 150 solar masses.) $\endgroup$ Nov 21, 2018 at 12:50
  • $\begingroup$ How accurate is that equation? It seems to be too simple. How accurate is it? Like, margin of error? Also, It's frustrating this graph goes from 0.5 to 2. lol. Like, really? I know we can obviously imagine the one solar mass line, but still. lol. $\endgroup$
    – Jimmy G.
    Nov 21, 2018 at 22:04
  • $\begingroup$ @JimmyG.: The equation is a scaling relation that comes from a simplified theoretical view of stellar evolution. Therefore no margin of error can be assigned. One could go through all the quantities that this timescale is based on and assign individual measurment uncertainties, but this goes way beyond the scope of a se.com answer, you're free to do your own research, now that you know the keywords. $\endgroup$ Nov 21, 2018 at 23:46
  • $\begingroup$ The graph doesn't go from 0.5 to 2, but very clearly from 0.08 (deuterium burning limit) until 150 (theoretical high-mass limit). It is a simplified graph that I've plugged in there in order to not confront you with the actual data, as you didn't provide any clues as to your prior science education. Again, you're free to do your own research now. $\endgroup$ Nov 21, 2018 at 23:49

"I was always told the sun is an average star" is not really an established scientific fact. It doesn't hold true for any reasonable definition of "average", unless you're ready to accept a definition specifically crafted to support the statement above.

Simply put, it's not true.

  • $\begingroup$ There is a historical context that you seem to be ignoring. The sun sits roughly in the middle of the color, mass, and luminosity ranges of stars for which each property could be reasonably easily measured. Certainly that is driven to a significant extent by the difficulty in counting the very numerous low-mass (i.e. very dim) stars, but it was still a valid observation for much of the twentieth century. $\endgroup$ Nov 21, 2018 at 21:52
  • $\begingroup$ @dmckee Well it's not true anymore. To put this into perspective, Sun was once a single known object of its kind, so it was certainly "average" back then. $\endgroup$ Nov 22, 2018 at 7:54

So I think other people have given great answers, but I think the question still remains, what do people mean when people say that the sun is an "average" star?

I think what they mean that the sun is "average" in the sense that the in terms of the biggest and smallest stars that we've ever found, with radii $R_\text{max}$ and $R_\text{min}$ respectively, the solar radius $R_\odot$ is approximately the following:

$$R_\odot \approx \frac{R_\text{max} + R_\text{min}}{2}$$

So this has little to do with how many stars there actually are with radii around $R_\text{min}$ or radii around $R_\text{max}$ (it turns out there are a lot lot more stars with radii smaller than the sun than radii bigger than the sun), it's just saying that the sun is not the biggest or the smallest star, its size is average. It's just this average is not a weighted average, it's just averaging the extreme radii that we've seen in the universe.

Hope that helps!

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    $\begingroup$ The initial equation you give is certainly not true. The largest stars we know of may have radii of $\sim2000R_{\odot}$. It is clear that $R_{\text{max}}\gg R_{\text{min}}$, so your equation would have $R_{\odot}\approx R_{\text{max}}/2$, which is not the case. The same goes for mass and luminosity. $\endgroup$
    – HDE 226868
    Nov 21, 2018 at 6:20
  • $\begingroup$ @HDE226868 This was my immediate thought too. But could it perhaps be true on a log scale? $\endgroup$
    – jkej
    Nov 21, 2018 at 9:58
  • $\begingroup$ @jkej There certainly is factor for which this equation holds. In fact, any non-trivial linear equation can be turned into an equity with a suitable factor. $\endgroup$ Nov 21, 2018 at 12:35

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