Main Sequence Star Lifetime Calculator
Estimate how long a star spends on the main sequence based on its mass.
Uses the mass-luminosity relation to calculate stellar lifetime.
How long does a star live?
A star’s lifetime on the main sequence (the period of hydrogen fusion in its core) depends on two things: how much fuel it has (proportional to mass) and how fast it burns it (luminosity).
Basic lifetime formula:
t ≈ (M/L) × t☉
Where t☉ ≈ 10 billion years is the Sun’s estimated main sequence lifetime.
Using the mass-luminosity relation (L ∝ M^3.5 to M^4 for main sequence stars):
t ≈ (M/M☉)^(-2.5) × 10 billion years
More precisely for different mass ranges:
- M < 0.43 M☉: L = 0.23 × M^2.3 → longer-lived, dimmer red dwarfs
- 0.43–2 M☉: L ≈ M^4 → t ∝ M^(-3)
- 2–55 M☉: L ≈ 1.4 × M^3.5 → massive stars burn fast
- M > 55 M☉: L ≈ 32,000 × M → near the Eddington luminosity limit
Key examples:
- The Sun (1 M☉): ~10 billion years (about 4.6 Gyr elapsed, ~5.4 Gyr remaining)
- Sirius (2.1 M☉): ~1 billion years
- A 10 M☉ star: only ~23 million years
- A 0.1 M☉ red dwarf: nearly 900 billion years (over 60 times the current age of the universe)
Important note: This calculation is an approximation for main sequence stars. Very massive stars (O-class) and very low-mass stars (late M-class) deviate significantly. Post-main-sequence evolution (red giant, supernova, white dwarf) is not included here.
Why heavier stars die so much faster
The intuition people arrive with is backwards. A star ten times the Sun’s mass has ten times the fuel, so surely it should last ten times as long? It lasts about four hundred times less. The reason is that luminosity climbs far faster than mass does. Roughly speaking L goes as M to the power 3.5, so ten times the mass burns something like three thousand times faster. Divide the fuel by the burn rate and the lifetime falls as M to the power minus 2.5.
That single relation shapes most of what we can see in the sky. Every O and B star visible tonight formed within the last few tens of millions of years, because nothing that heavy survives longer. The red dwarfs, meanwhile, will still be fusing hydrogen long after the last Sun-like star has gone. Not one of them has ever died. The universe is not old enough.
A caveat about the fuel fraction
The formula assumes every star burns the same fraction of its hydrogen, around 10%, because only the core gets hot enough. That is a decent approximation in the middle of the range and a poor one at the ends. Red dwarfs below about 0.35 M☉ are fully convective, which drags the whole star’s hydrogen through the core and stretches their lives well past what this calculation gives. At the top end, radiation-driven mass loss strips an O star as it burns, which cuts its life short in the other direction.
How we build and check this calculator
This calculator runs entirely in your browser, so the numbers you enter stay on your device. The math behind it is written by hand and tested against worked examples and standard references before the page goes live.
SuperGlobalCalculator is independently built and maintained. See how we build and verify our calculators.
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