So, I saw this question (which is quite old) and that's all well and good, but what if I need to (very) roughly calculate the luminosity of a star from the end of the main sequence? Let's assume that this is a non-extreme mass star (K-A) and can use $10M^{-2.5}$ To calculate (roughly) when the star will exit the main sequence. Would there be a way to include the expansion into a red giant? What about simulating the planetary nebula? Simplify as much as is needed.

  • $\begingroup$ Now that there's an answer posted to address all aspects of this question, I don't think closing and preventing answer posts is necessary or even productive. voting to leave open $\endgroup$
    – uhoh
    Commented Feb 18, 2023 at 22:28

1 Answer 1


Rough reaction for stars between 0.5 and 2.5 solar masses: We first need to know the initial(ZAMS) luminosity, the luminosity at the end of the main sequence(TAMS) and the MS lifetime: $$L_\text{ZAMS}=0.7×M^4$$


$$T=9×M^{-2.6}\text{ (in Gyr)}$$

Reg giant stage: luminosity exponentially increases from the TAMS luminosity to about 3500×solar luminosity(at the tip of RGB) lasts about 10% of main sequence

Helium burning:luminosity exponentially increases from 50 to 150×solar luminosity, lasts about 100 Myr

AGB: luminosity increases exponentially from 150×solar luminosity to 10000×mass of star(in solar masses)×solar luminosity, lasts 20 Myr

Planetary nebula: roughly constant luminosity at 10000×mass of star(in solar masses)×solar luminosity, lasts roughly 30000yr

For other mass ranges it gets much more complicated.

Calculations: https://www.desmos.com/calculator/l2nelshpnx

  • 2
    $\begingroup$ Ive updated the formatting. In English don't use a dot to separate thousands, because the dot is used as a decimal point (ie 10.001 = 10001/1000, not ten thousand and one) What is the meaning of "L" and "M" in 150L to 10000M×solar luminosity? $\endgroup$
    – James K
    Commented Feb 14, 2023 at 8:27
  • 1
    $\begingroup$ I edited the my answer for clarity $\endgroup$
    – HugoF
    Commented Feb 14, 2023 at 11:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .