Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 2·0 uses

Main-Sequence Star Lifetime Calculator

Estimate how long a star of given mass stays on the main sequence, or which mass lasts a given time, with an editable mass–luminosity exponent and fuel share.

Star Lifetime Calculator

What do you want to find?
1 M☉ = 1.98841 × 10³⁰ kg (PDG 2024). Empirically calibrated mass–luminosity data covers 0.179–31 M☉ (Eker et al. 2018); below about 0.08 M☉ hydrogen fusion never ignites and there is no main sequence at all.
Used in the inverse route. In Julian years of exactly 31 557 600 s. For reference, the Hipparcos age of the Hyades cluster is 0.625 ± 0.050 Gyr (Perryman et al. 1998) and the universe is 13.787 Gyr old.
The single most important assumption on this page, and an empirical fit rather than a law. 3.5 is the conventional value for roughly 2–20 M☉. Eker et al. (2018) measure it from 509 eclipsing-binary components as 2.028 (0.179–0.45 M☉), 4.572 (0.45–0.72), 5.743 (0.72–1.05), 4.329 (1.05–2.40), 3.967 (2.40–7) and 2.865 (7–31). Cannot be exactly 1 in the inverse route.
The share of the star's total mass actually consumed in the core during the main sequence. 0.1 is the standard textbook figure; the true value depends on the size of the convective core and therefore on mass and metallicity. The lifetime is exactly proportional to it.
Fraction of rest mass released as energy by 4 ¹H → ⁴He. AME2020 atomic masses give (4 × 1.00782503190 − 4.00260325413)/(4 × 1.00782503190) = 0.0071185; about 2% escapes as neutrinos, leaving ≈ 0.006976 thermally available. The conventional 0.007 sits between them.
Main-sequence lifetime (Gyr)
10.3555
The estimated main-sequence lifetime in billions of Julian years, assuming constant luminosity and a single power-law mass–luminosity relation.
Main-sequence lifetime (years)
10,355,462,213.3
Lifetime ÷ the Sun's
1
Luminosity (L☉)
1
Luminosity (W)
3.828 × 10²⁶
Mass (M☉)
1
Mass (kg)
1.988 × 10³⁰
Hydrogen consumed (kg)
1.988 × 10²⁹
Nuclear energy reservoir (J)
1.251 × 10⁴⁴
Model assumptions and limits for this result
Nuclear-timescale estimate, not a stellar-evolution model. It assumes the star shines at a CONSTANT luminosity for its whole main sequence (real stars brighten — the Sun is about 30% brighter now than at the zero-age main sequence), a single power-law mass–luminosity relation, no mass loss, no rotation and no convective-core overshoot. All three model parameters are yours to change: the mass–luminosity exponent, the core fuel fraction and the hydrogen-burning efficiency. Read the answer as an order-of-magnitude scaling law. For this mass, Eker et al. (2018) measure an exponent of 5.743 (0.72–1.05 M☉) from 509 detached eclipsing-binary components — compare that with the value you entered. NOTE: the conventional single exponent of 3.5 is quoted for roughly 2–20 M☉. This mass is outside that band, so a single power law is a poorer approximation here than usual.

Background.

A star ten times as massive as the Sun carries ten times as much hydrogen, yet it runs out about 300 times sooner. The reason is luminosity. Along the main sequence, brightness climbs far faster than mass: with the conventional mass–luminosity exponent of 3.5, ten solar masses shine 10^3.5, about 3,162, times as bright as the Sun. Ten times the fuel spent 3,162 times as fast leaves roughly 1/316 of the time, some 33 million years against the Sun's ten billion.

That argument is the nuclear timescale, and it runs here in either direction: from a mass to a lifetime, or from a lifetime back to the mass that lasts exactly that long. The energy on hand is a small fraction of the star's rest mass, ε f M c², and the lifetime is that energy divided by the luminosity.

Three ingredients are modelling choices rather than constants of nature, so all three are inputs. The exponent α is the sensitive one: measurements of eclipsing binaries by Eker and colleagues (2018) put it anywhere from 2.0 to 5.7 depending on the mass range, rather than a steady 3.5. The core fuel fraction f, conventionally 0.1, and the burning efficiency ε, about 0.007, each scale the answer in direct proportion.

The result is a scaling law, not a stellar-evolution model. It assumes the star shines at constant brightness, although the Sun is about 30 percent brighter today than when it began fusing hydrogen, and it leaves out mass loss, rotation and mixing beyond the core. What it captures well is the shape of the dependence, which is enough to show why the most massive stars are always young and why no red dwarf has yet had time to burn out.

What is star lifetime calculator?

The main-sequence lifetime is the time a star spends fusing hydrogen into helium in its core, the long, stable stage that takes up most of a star's life and the stage the Sun is in now. On the nuclear timescale it is the energy that core hydrogen burning can release divided by the rate at which the star radiates: t = ε f M c² / L. Because L rises roughly as a power of mass, M^α, the lifetime scales as M^(1−α), which is M^−2.5 for α = 3.5. Below about 0.08 solar masses hydrogen never ignites, so a brown dwarf has no main sequence and no lifetime in this sense.

How to use this calculator.

  1. Under 'What do you want to find?', choose mass to lifetime for a star of known mass, or lifetime to mass to ask which star would just be leaving the main sequence after a given time, such as the age of a cluster.
  2. Enter the mass in solar masses, or the lifetime in billions of years (Julian years of 31,557,600 s).
  3. Set the mass–luminosity exponent. Keep 3.5 for a rough figure between about 2 and 20 solar masses; for a closer match use the Eker et al. value for your mass, which the results note quotes: 2.028 up to 0.45 M☉, 4.572 to 0.72, 5.743 to 1.05, 4.329 to 2.40, 3.967 to 7 and 2.865 to 31.
  4. Leave the core fuel fraction at 0.1 unless you have a reason to change it. Red dwarfs below about a third of a solar mass are convective throughout and stir nearly all their hydrogen through the core, so a far larger fraction suits them.
  5. Pick the efficiency: 0.007 by convention, 0.0071185 for the full mass defect of hydrogen-to-helium fusion, or about 0.006976 if you discount the roughly 2 percent carried off by neutrinos.
  6. Read the lifetime relative to the Sun alongside the absolute figure. It depends only on mass and exponent, so it is unaffected by whatever you chose for f and ε.

The formula.

L ⁄ L☉ = (M ⁄ M☉)^α · t = ε f M c² ⁄ L · t(M) ⁄ t(☉) = (M ⁄ M☉)^(1 − α) · inverse: M = (t ⁄ K)^(1 ⁄ (1 − α)), K = ε f M☉ c² ⁄ L☉

Fusion turns four hydrogen atoms into one helium-4 atom, and the helium is lighter. With AME2020 atomic masses the loss is (4 × 1.00782503190 − 4.00260325413) ÷ (4 × 1.00782503190) = 0.0071185 of the rest mass; that fraction is ε. If a share f of the star's mass M is burned before the star leaves the main sequence, the energy available is E = ε f M c², and dividing by the luminosity gives the time: t = ε f M c² / L.

The luminosity comes from the mass–luminosity relation L = L☉ (M/M☉)^α, with L☉ = 3.828 × 10²⁶ W, the IAU 2015 nominal value. Substituting gives t = K (M/M☉)^(1−α), where K = ε f M☉ c² / L☉ is the lifetime of a one-solar-mass star under the same assumptions. With M☉ = 1.98841 × 10³⁰ kg, ε = 0.007 and f = 0.1, K = 3.268 × 10¹⁷ s, or 10.355 billion years. The inverse route solves M = (t/K)^(1/(1−α)).

Since α sits in an exponent, small changes to it move the answer a long way, while f and ε only rescale it in proportion. At α = 1 the lifetime would be the same for every mass, so the inverse has no solution and is refused. Near α = 1 the inverse becomes wildly unstable: a 5 Gyr lifetime implies 0.000689 M☉ at α = 0.9 and 1,452 M☉ at α = 1.1.

A worked example.

Example

Take a ten-solar-mass star, an early B-type star of the kind that ends in a supernova. Its luminosity is 10^3.5 = 3,162.28 L☉, and 3,162.28 × 3.828 × 10²⁶ W = 1.2105 × 10³⁰ W. The fuel is a tenth of its 1.98841 × 10³¹ kg, so 1.98841 × 10³⁰ kg of hydrogen, and the energy it can release is 0.007 × 1.98841 × 10³⁰ × (299,792,458)² = 1.2510 × 10⁴⁵ J. Energy over power gives 1.2510 × 10⁴⁵ ÷ 1.2105 × 10³⁰ = 1.0334 × 10¹⁵ s. Dividing by 31,557,600 s per year gives 32.75 million years, shown as 0.0327 Gyr, which is 0.0032 of the Sun's 10.355 Gyr. Swap in Eker's measured exponent for this mass range, 2.865, and the same star lasts 0.1413 Gyr, about 141 million years: more than four times longer from a single change of exponent.

lifetime Gyr10.355
mass Luminosity Exponent3.5
hydrogen Burning Efficiency0.007
mass Solar10
core Fuel Fraction0.1
solve ForlifetimeFromMass

Frequently asked questions.

How long will the Sun stay on the main sequence?
With the defaults, 10.355 billion years; with the full mass defect of 0.0071185 instead of 0.007, 10.531 billion. At one solar mass the exponent drops out entirely, because 1 raised to any power is 1. The Sun is about 4.6 billion years old, so it is a little under halfway through, though its gradual brightening makes the constant-luminosity figure approximate.
Why are the most massive stars always young?
Because they cannot be anything else. A 30-solar-mass star at α = 3.5 lasts about 0.0021 Gyr, roughly two million years, which is brief beside the hundreds of millions of years the Sun takes to orbit the galaxy. Any such star you see must have formed recently, which is why the hottest, bluest stars mark regions where stars are still forming.
Has any red dwarf ever run out of hydrogen?
Not according to this scaling. With the defaults, every star below about 0.89 solar masses has a lifetime longer than the universe's 13.787 billion years. A 0.2 M☉ red dwarf with Eker's exponent of 2.028 and a fuel fraction of 1, appropriate for a fully convective star, comes out at 541.6 billion years, nearly forty times the present age of the universe.
Can the inverse route tell me a star cluster's age?
Only roughly. Hipparcos dates the Hyades at 625 ± 50 million years. Asking which mass lives exactly 0.625 Gyr gives 3.074 M☉ at α = 3.5, 2.549 M☉ at α = 4.0 and 2.324 M☉ at α = 4.329, Eker's value for that mass range. Real cluster ages come from fitting the whole colour–magnitude diagram to stellar-evolution isochrones.
Why does a star burn only a tenth of its mass?
Only the core is hot and dense enough to fuse hydrogen, and in most stars the outer layers never mix down into it. Once the core hydrogen is gone the star leaves the main sequence, even with most of its hydrogen untouched. The true share depends on the size of the convective core, which grows with mass and varies with metallicity.
What if my star is below 0.179 or above 31 solar masses?
The results note warns that no measured mass–luminosity relation covers it, since Eker et al. calibrated 0.179 to 31 M☉ from 509 eclipsing-binary components. Below about 0.08 M☉ there is no hydrogen fusion at all, and above roughly 100 M☉ radiation pressure and heavy mass loss govern the star's evolution, neither of which a single power law describes.

References& sources.

  1. [1]Eker, Z., et al. (2018), "Interrelated main-sequence mass–luminosity, mass–radius, and mass–effective temperature relations", Monthly Notices of the Royal Astronomical Society 479, 5491; arXiv:1807.02568, full text retrieved 2026-07-29. Uses absolute parameters of 509 main-sequence stars from detached eclipsing spectroscopic binaries. Table 4 gives the six-piece classical mass–luminosity relation L ∝ M^α with α = 2.028 (0.179–0.45 M☉), 4.572 (0.45–0.72), 5.743 (0.72–1.05), 4.329 (1.05–2.40), 3.967 (2.40–7) and 2.865 (7–31), and states that the exponent "is smallest (α=2.028) in the ultra low mass domain … then it increases and reaches its maximum (α=5.743) in the low mass domain". This is the second, independent authority: it contradicts the single conventional exponent 3.5 everywhere, which is why α is an editable input. Peer-reviewed; open-access preprint.
  2. [2]Perryman, M. A. C., et al. (1998), "The Hyades: distance, structure, dynamics, and age", Astronomy & Astrophysics 331, 81; arXiv:astro-ph/9707253, abstract retrieved 2026-07-29. States a cluster age of 625 ± 50 Myr from Hipparcos astrometry. Used here only as an external calibration point for the inverse route; the turnoff MASS is deliberately not quoted, because the abstract does not give one. Peer-reviewed; open-access preprint.
  3. [3]Wang, M., Huang, W. J., Kondev, F. G., Audi, G. and Naimi, S. (2021), "The AME 2020 atomic mass evaluation (II)", Chinese Physics C 45, 030003. Source of the atomic masses m(¹H) = 1.00782503190 u and m(⁴He) = 4.00260325413 u from which the hydrogen-burning efficiency ε = 0.0071185 is computed. Peer-reviewed; the evaluation and its data files are freely distributed by the AMDC. Bibliographic citation — the two mass values are standard and appear in every AME2020 mirror.
  4. [4]Prša, A., et al. (2016), "Nominal Values for Selected Solar and Planetary Quantities: IAU 2015 Resolution B3", Astronomical Journal 152, 41; arXiv:1510.07674, retrieved 2026-07-29. Table 1 gives the nominal solar luminosity 3.828 × 10²⁶ W used to normalise every luminosity here, and states that these are standard conversion factors rather than current best estimates of the real Sun. Peer-reviewed; open-access preprint.
  5. [5]Particle Data Group (Workman et al.), Review of Particle Physics, "Astrophysical Constants and Parameters", Table 2.1, revised August 2023; 2024 PDF retrieved 2026-07-29. Source of the solar mass M☉ = 1.98841(4) × 10³⁰ kg. Open access.
  6. [6]NIST/CODATA 2022, "Speed of light in vacuum": c = 299 792 458 m s⁻¹, exact. Retrieved 2026-07-29. Standards body; open access.

How this page was produced

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Quanta Calculator
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Method
L ⁄ L☉ = (M ⁄ M☉)^α · t = ε f M c² ⁄ L · t(M) ⁄ t(☉) = (M ⁄ M☉)^(1 − α) · inverse: M = (t ⁄ K)^(1 ⁄ (1 − α)), K = ε f M☉ c² ⁄ L☉
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