I know there are two criteria to meet in order for nuclear fusion to occurs.

  • High temperature (many times temperature at Sun's core)
  • High pressure (protons are very close to each other)


However I want to know what equations allows me to calculate the amount of hydrogen gas needed so that the internal pressure can produce the right amount of heat for nuclear fusion to occur.

[My own understandings]

I also understand that the temperature of our Sun's core is not sufficient to overcome the coulomb barrier, it make use of quantum tunneling to achieve fusion instead.


What is the minimum amount of hydrogen gas required to form a star? (also please include the name of the equation used)


  • shape is perfectly spherical
  • no angular velocity, not rotating
  • homogenous, consist of 95% hydrogen and 5% helium
  • sustain fusion for at least 1 million years
  • 2
    $\begingroup$ what-if.xkcd.com/99 and en.wikipedia.org/wiki/Jeans_instability $\endgroup$ – Mithoron Mar 31 '15 at 14:45
  • $\begingroup$ I love the starling example above. This isn't an equation, obviously but you're talking about a red dwarf star. A brown dwarf doesn't burn hydrogen, though clearly there's no exact cut off point between the two and it would be hard to tell from the outside the difference between slow hydrogen burn and heat from coalescing. The smallest type of Red Dwarf is an M9V and it's listed as 7.5% the mass of the sun, or about 70 Jupiters. en.wikipedia.org/wiki/Red_dwarf. $\endgroup$ – userLTK Apr 1 '15 at 15:36
  • $\begingroup$ @user6760 Would you mind unaccepting my answer? It needs some heavy revisions, and I'd rather remove while those are in progress, but I can't do so if the answer is still accepted. $\endgroup$ – HDE 226868 Apr 28 '17 at 0:11

A proposal for the minimum mass of a gas cloud before collapsing to become a star was made by James Jeans; because of this, it was termed the Jeans mass. It is calculated as $$M=\frac{4 \pi}{3} \rho R^3$$ where $\rho$ is density and $R$ is the radius of the cloud - one half of the Jeans length, which is dependent on the speed of sound and the density of the cloud. It was originally thought that any mass above the Jeans mass would not be in hydrostatic equilibrium, and would collapse.

But Jeans didn't realize that any regions outside this radius would also collapse. So his arguments were flawed, though they're still sound in many applications.

  • $\begingroup$ But brown dwarfs probably form in the same way as stars do. So how does this answer the question? $\endgroup$ – Rob Jeffries Apr 27 '17 at 23:41

There is no equation, you need a detailed model of the interior physics of very low mass stars. Very roughly, you can say that hydrogen fusion occurs when the central temperature exceeds about $10^{7}\ K$ (the density dependence is secondary) and that from the virial theorem, the central temperature is given approximately by $T \simeq 1.6\times 10^{7} M/R$, where the mass and radius are in solar units. $M/R$ decreases slowly down the main sequence towards lower mass stars, but then what happens is that electron degeneracy pressure becomes important and less massive objects do not become much smaller than Jupiter-sized and thus $M/R$ decreases and fusion stops - or rather it never starts.

In detail the minimum mass for hydrogen fusion in a manner that is capable of sustaining a star in equilibrium against gravitational contraction is about $0.075M_{\odot}$. With an uncertainty of about $0.002M_{\odot}$.

It is slightly more complicated than this, since at lower internal temperatures the deuterium in a star can fuse. This will happen during the early life of any ball of gas more massive than about 13 Jupiter masses.

Your assumptions about the relative H and He abundances are completely wrong. Even in the early universe the gas is $\sim 25$% He by mass.

  • $\begingroup$ 25% He by mass is about 6%-7% He by content. I assume that's what he meant, meaning, no heavy elements in the star. I'm also curious, are they really certain to within 0.002 solar masses? That's sounds awfully precise. I thought there was more uncertainty in there. $\endgroup$ – userLTK Apr 1 '15 at 15:40
  • 1
    $\begingroup$ @userLTK yes, 8% by number of nuclei would be about right. The uncertainty on the threshold is largely how you define it. Different models do agree to these sorts of precision. But you are also correct that there is no observational test other than to somehow estimate an age and then an old star will be hotter than an old brown dwarf. $\endgroup$ – Rob Jeffries Apr 1 '15 at 15:59

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