I have a generic zero-age main sequence star. The only thing I know about it is its temperature. How can I estimate time time it spends on the main sequence (in millions of years)? I've seen equations that estimate it based on mass, like $L=M^{-2.5}\cdot10^{10}$, but I do not have the mass, only the temperature. Is there any definite formula (accurate to a reasonable degree) for this?

  • $\begingroup$ @slowerthanstopped I don't exactly understand the previous comment, but if you are using two different user accounts on the same site to interact with each other this is called a "sock puppet" and it's against the rules. If you are a new user you may not be aware of this but it's important to ask for moderator help if indeed you might have done this accidentally. $\endgroup$
    – uhoh
    Commented Feb 21, 2021 at 3:16
  • $\begingroup$ Main sequence stars don't have a fixed temperature (or luminosity) during their main sequence lifetimes, so which temperature would you be using? $\endgroup$
    – ProfRob
    Commented Feb 22, 2021 at 8:42
  • $\begingroup$ @uhoh Actually, having multiple accounts in and of itself is not against the rules, per se. It just isn't encouraged. There's a nice list over at MSE about which behaviors are inappropriate for a user of multiple accounts. It doesn't appear slowerthanstopped has performed any of these problematic actions, so whatever their reasons for having multiple accounts they can keep them for now if they wish. $\endgroup$
    – called2voyage
    Commented Feb 22, 2021 at 13:08
  • $\begingroup$ @called2voyage yep, I said "if... interact with each other" and that says "Supporting your own arguments", so as long as the sock is either silent towards the "hand's" point or actively argues against it it's okay :-) $\endgroup$
    – uhoh
    Commented Feb 22, 2021 at 14:10
  • 2
    $\begingroup$ @uhoh Well "interact with each other" is a bit unclear. It could be interpreted as any interactions, like editing posts. Editing posts is allowed. $\endgroup$
    – called2voyage
    Commented Feb 22, 2021 at 14:18

1 Answer 1


The website on Main sequence stars fromr the Austalian national telescope facility lists star mass, temperature and life span:

Mass/MSun Luminosity/LSun T=Effective Temperature/K Radius/RSun t=Main sequence lifespan/yrs
0.10 $3×10^{-3}$ 2,900 0.16 $2×10^{12}$
0.50 0.03 3,800 0.6 $2×10^{11}$
0.75 0.3 5,000 0.8 $3×10^{10}$
1.0 1 6,000 1.0 $1×10^{10}$
1.5 5 7,000 1.4 $2×10^9$
3 60 11,000 2.5 $2×10^8$
5 600 17,000 3.8 $7×10^7$
10 10,000 22,000 5.6 $2×10^7$
15 17,000 28,000 6.8 $1×10^7$
25 80,000 35,000 8.7 $7×10^6$
60 790,000 44,500 15 $3.4×10^6$

If we suspect a lifespan, $t$ is related to the temperature $T$ by a power relationship, $t=aT^k$, we can plot log(t) against log(T) and look for a linear relationship enter image description here

That looks plausible, and gives a relationship of $t = 10^{28}\times T^{-4.7}$ (rounding to 2sf)

looking more closely, this tends to underestimate the life span of mid-range stars, but overestimate large and small stars. You can adapt for this by considering smaller and larger stars separately: For stars less than 10000K, a model $t = 10^{37}\times T^{-7.6}$ and for hotter stars $t = 10^{20}\times T^{-3.0}$ gives reasonable fit to the data.

This is, of course, a purely empirical fitting of a curve, and isn't based on an astrophysical model. (though I speculate that the change in gradient is related to the change from the proton-proton to the CNO cycle fusion of hydrogen.)

  • $\begingroup$ I tested your two formulas for $T=5778$ and it didn't work very well. Am I doing something wrong? $\endgroup$ Commented Feb 21, 2021 at 15:39
  • $\begingroup$ Also, is there any place where I can find a detailed table of temperature-lifespan relations (of step 100 K or something up to a reasonable degree of accuracy)? $\endgroup$ Commented Feb 21, 2021 at 16:15

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