What’s the lightest you could make a "star" if you made it out of different materials? How large would the "star" be?

For example according to here https://astronomy.stackexchange.com/q/20448/ a star made of water only need 13 Jupiter masses.

I'll count objects that don't burn protonium as stars for the purposes of this question. I intially briefly considered Tritium and Deutrium but that doesn't work due to Tritium being radioactive. So how low can you get this figure?

  • $\begingroup$ How do you want to define a "star"? The normal definition is protium fusion - which means you need protons. $\endgroup$
    – ProfRob
    Commented Jul 17, 2023 at 8:23
  • $\begingroup$ You'll get nearly as much energy per kilo of fuel as with hydrogen fusion, so despite the lank of protium you'll be able to treat it as an ordinary star. $\endgroup$ Commented Jul 17, 2023 at 8:37
  • $\begingroup$ But as I answered in that question - deuterium burning has a lower threshold temperature. $\endgroup$
    – ProfRob
    Commented Jul 17, 2023 at 8:45
  • $\begingroup$ I want to see how low we can go. $\endgroup$ Commented Jul 17, 2023 at 8:46
  • $\begingroup$ But deuterium burning happen in normal composition matter down to 13 Jupiter masses and we don't call those stars. What's your definition of a star? Sorry to be awkward. $\endgroup$
    – ProfRob
    Commented Jul 17, 2023 at 8:48

2 Answers 2


Let's assume by a "star" you mean something that undergoes hydrogen (protium) fusion. In which case, the simplest analysis is to assume it is made of multiples of: a proton plus electron, and something else that has $A$ nucleons and $Z$ electrons (which could consist of more than 1 nucleus).

The analysis in the link you refer to shows that if we assume the temperature dependence of the fusion is high enough that we can assume the fusion is initiated as the same temperature as a "normal star" and the minimum mass is determined by reaching this temperature at densities where the electrons in the star remain non-degenerate, then the threshold mass scales as $$ M_{\rm min} \simeq 0.0365\ \mu^{-3/2} \mu_e^{-1/2}\ M_{\odot}\ , $$ where $\mu$ is the number of atomic mass units per particle in the gas and $\mu_e$ is the number of atomic mass units per free electron. In "normal" composition stellar matter, $\mu \simeq 16/27$ and $\mu_e \simeq 8/7$, giving $M_{\rm min} \simeq 0.075 M_\odot$. i.e. A protostar above this minimum mass will (eventually) contract until it reaches the fusion ignition temperature. An object with lower mass will become supported by electron degeneracy pressure and then cool, before it reaches this temperature.

So the trick is to minimise $\mu^{-3/2} \mu_e^{-1/2}$ with respect to $A$ and $Z$, where $$ \mu = \left(\frac{A+1}{Z+3}\right)\ ,\ \ \ \mu_e = \left(\frac{A+1}{Z+1}\right) $$

This function has no minimum - you want $A$ to be as large as possible and $Z$ to be as small as possible. If $A$ amd $Z$ are large, then you want $A/Z$ to be as large as possible. Thus mixing hydrogen with something like iron would lead to a lower $M_{\rm min}$ than mixing with oxygen.

The maths also seems to indicate that you could mix a little bit of hydrogen in with a large amount of heavier material. However, I caution that the maths assumes that the density of hydrogen plays no role in the ignition temperature because of the steep dependence on temperature of the reaction rate. That is only true up to a point and if you move to a mixture that is dominated by the other element then the ignition temperature and minimum mass will rise again.

If your definition of a "star" is one that undergoes any kind of nuclear fusion that can sustain its radiative losses for any brief period then you must also consider deuterium burning. This occurs at lower temperatures than protium burning and hence $M_{\rm min} \simeq 0.013 M_\odot$ in material with a "normal" composition. If you consider a mixture of deuterium plus some other heavier element then $M_{\rm min}$ is also reduced by a similar factor given by the equation above but the leading constant in the equation would be $\simeq 0.0063 M_\odot$ (assuming that deuterium burning with deuterium occurs at a similar temperature to deuterium and protium).

  • $\begingroup$ How is the number mass units per particle<1 ? are they dividing the mean atomic mass by 2 or somehting? $\endgroup$ Commented Jul 17, 2023 at 11:30
  • $\begingroup$ The Equation for deutrium Burning assumes p+d fusion I presume? And that assumes the the deutrium proportion is small, but If the Deutrium fraction is large enough you could end up with large quantities of d+d fusion which I do believe is easier, is that correct or am I making a mistake here? $\endgroup$ Commented Jul 17, 2023 at 11:35
  • $\begingroup$ "where 𝜇 is the number of atomic mass units per particle in the gas and 𝜇𝑒 is the number of atomic mass units." Is this a typo, it seems like a typo? $\endgroup$ Commented Jul 17, 2023 at 11:37
  • $\begingroup$ @blademan9999 Why fo you believe D+D fusion is easier (I presume you mean occurs at a lower temperature) than p+D fusion? It has the same Coulomb repulsion and a slightly lower Q value. I have no information on the reaction rate. I think it would require an almost identical ignition temperature. If it were easier, then there would be no p+D fusion in a star, since all the D would have been previously burned in D+D reactions and I feel that this detail would have been mentioned in the papers that discuss early stellar evolution. The typo is corrected, $\endgroup$
    – ProfRob
    Commented Jul 17, 2023 at 11:50
  • $\begingroup$ Well there's the fact that due to Deutrium's relative rarity, d-d interactions owuld be much rarer then d-p interactions. $\endgroup$ Commented Jul 17, 2023 at 12:09

This question has come up on occasion. I have a developed answer in one of my older books, but I can no longer find it. I would swear it is in Amasa Bishop's Project Sherwood, but I bought another copy and it's not there (versions perhaps?). It's from that era though, late 1950s.

In any event, if you consider the star to lose energy primary as a black body, then the star is "working" when the energy loss through radiation is equal to the rate of fusion. That rate has a temperature dependency, as does the radiation, so you solve for the root.

For a body consisting of pure deuterium, for instance, the resulting "star" is about the size of the moon. You can develop similar answers for all of the common fuels.

If I ever find it I'll post it.

  • $\begingroup$ Except the fusion rate depends on the central temperature but the blackbody radiation depends on the surface temperature. How are you connecting the two? $\endgroup$
    – ProfRob
    Commented Aug 14, 2023 at 20:05
  • $\begingroup$ I am not, the book in question did. As I noted, if I can find the reference in question I will update all of these posts with a link. $\endgroup$ Commented Aug 15, 2023 at 13:47
  • $\begingroup$ Well can you put the details of the calculation/method. As it stands this is not an answer. Can you also clarify what you mean by "the size" of the Moon. $\endgroup$
    – ProfRob
    Commented Aug 15, 2023 at 15:25
  • $\begingroup$ The moon? That can't be right. Deuterium fusion normally requires 13 Jupiter masses, and fusion is HIGHLY deutrium dependant. $\endgroup$ Commented Aug 16, 2023 at 11:49

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