The big idea: A star shines by fusing hydrogen. It only has a limited tank of fusible hydrogen, and it spends that fuel at a steady rate — its luminosity.
So its lifetime is just: how much energy the fuel is worth, divided by how fast the star uses it.
And because that energy comes from converting mass (E = mc²), the star is slowly getting lighter as it shines.
What runs the star
- Its fuel = the hydrogen in the core it can actually fuse
- Its burn rate = the luminosity L (energy radiated each second)
- Each kilogram of fuel is worth a fixed amount of energy (E = mc²)
How long it lasts
- Total energy available = fuel turned into energy
- Lifetime = energy available ÷ how fast it is used
- So a bright, hungry star burns out faster than a dim one
New words, plainly: Main sequence = the long, stable middle of a star's life, when it is steadily fusing hydrogen into helium.
Luminosity (L) = the total energy a star radiates every second — its power output, in watts (W = J s⁻¹).
Fusible hydrogen = only the hydrogen in the hot, dense core can fuse, and only a small fraction of its mass is released as energy. So the usable fuel is far less than the star's whole mass.
| You need | What it is | Symbol / unit |
|---|---|---|
| The fuel's energy | energy the fusible hydrogen releases when it fuses | E (J) |
| The burn rate | luminosity — energy the star radiates per second | L (W = J s⁻¹) |
| The lifetime | how long the fuel lasts at that burn rate | t (s, then years) |
Why a brighter star dies younger: A bright star has a huge luminosity — it burns through its fuel fast.
A dim star sips its fuel slowly, so it lasts far longer.
That is why the most luminous stars live only millions of years, while faint ones can outlast the Universe so far.
Two steps. First find the energy the fusible hydrogen can release, using E = mc². Then divide by the luminosity (the energy used each second) to get the lifetime in seconds — and convert to years.
- energy released as the fusible hydrogen fuses (J)
- mass that is actually converted to energy (kg)
- speed of light, 3.00 × 10⁸ m s⁻¹ (given constant)
- main-sequence lifetime — how long the star keeps fusing hydrogen (s)
- total energy the fusible hydrogen can release (J)
- luminosity — the energy the star radiates each second (W = J s⁻¹)
Only a sliver of the mass is fuel: Two cuts shrink the usable fuel right down:
- Only the core's hydrogen fuses — typically about 10–12% of the star's mass. - Of that mass, only about 0.7% is actually released as energy when hydrogen turns into helium.
So multiply the star's mass by both fractions before using E = mc².
Worked example — show that the star lives for about 11 billion years
A Sun-like star, Helios-B, has mass M = 2.4 × 10³⁰ kg. About 12% of its mass is hydrogen in the core that can fuse, and 0.70% of that fusible mass is released as energy. Its luminosity is L = 5.0 × 10²⁶ W. Show that it stays on the main sequence for about 1.1 × 10¹⁰ years. (c = 3.00 × 10⁸ m s⁻¹; 1 year ≈ 3.16 × 10⁷ s.)
Solution
- Find the fusible-hydrogen mass. Take 12% of the star's mass:
- Find the energy released. Write the given first; only 0.70% of that fuel mass becomes energy, so multiply by 0.0070:
- Put in c = 3.00 × 10⁸ m s⁻¹:
- Find the lifetime. Write the relation , then divide by the luminosity:
- Convert seconds to years (÷ 3.16 × 10⁷):
Final answer
The fuel is worth E ≈ 1.8 × 10⁴⁴ J, and at L = 5.0 × 10²⁶ W that lasts t = E/L ≈ 3.6 × 10¹⁷ s ≈ 1.1 × 10¹⁰ years — about 11 billion years.
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How this is tested: Stellar lifetime and mass loss is a Paper 2 topic, and it comes in three linked parts:
- Paper 2 — show that: estimate the main-sequence lifetime from the fusible-hydrogen mass, the energy released (E = mc²) and the luminosity (t = E ÷ L), then convert to years. - Paper 2 — state: give one assumption behind that estimate (e.g. the luminosity stays constant, only the core's hydrogen fuses, a fixed fraction of mass is converted). - Paper 2 — estimate: find the mass the star loses by radiating energy, using Δm = E ÷ c².
Classic trap: using the star's whole mass as fuel. Only the core's hydrogen fuses, and only ~0.7% of that mass becomes energy — multiply by both fractions first.
| The question type | What you do |
|---|---|
| Show the lifetime ≈ N years | energy of fusible H ÷ luminosity, then ÷ (3.16 × 10⁷) to get years |
| State an assumption | e.g. luminosity stays constant; only core hydrogen fuses; fixed fusible fraction |
| Estimate the mass lost | Δm = E ÷ c², where E is the total energy radiated |
IB-style question — show that the star lives for about 2.7 billion years
A hot, bright main-sequence star, Lyra-A, has mass M = 4.0 × 10³⁰ kg. About 10% of its mass is fusible core hydrogen, and 0.70% of that fusible mass is released as energy. Its luminosity is L = 3.0 × 10²⁷ W. Show that its main-sequence lifetime is about 2.7 × 10⁹ years. (c = 3.00 × 10⁸ m s⁻¹; 1 year ≈ 3.16 × 10⁷ s.)
Solution
- Fusible-hydrogen mass — take 10% of the star's mass:
- Energy released — write the given , with only 0.70% of the fuel mass converted:
- Lifetime — use :
- Convert to years (÷ 3.16 × 10⁷):
Final answer
E ≈ 2.5 × 10⁴⁴ J and L = 3.0 × 10²⁷ W give t ≈ 8.4 × 10¹⁶ s ≈ 2.7 × 10⁹ years. (It is far brighter than Helios-B, so it burns out much sooner.)
IB-style question — state an assumption
State one assumption made when estimating a star's main-sequence lifetime this way. [1]
Solution
- Pick any one valid assumption the estimate relies on:
The star's luminosity stays constant over its whole main-sequence life (it actually changes a little). - Other answers that would score:
Only the core's hydrogen is available to fuse · a fixed fraction of the mass (≈ 0.7%) is converted to energy · the star fuses hydrogen at a steady rate the whole time.
Final answer
Any one of: the luminosity stays constant; only the core hydrogen fuses; a fixed fraction (~0.7%) of the fuel mass is converted to energy; the fusion rate is steady throughout.
IB-style question — estimate the mass the star loses
Over its whole main-sequence life, Helios-B radiates a total energy of about E = 1.8 × 10⁴⁴ J. Estimate the total mass it loses by radiating this energy. (c = 3.00 × 10⁸ m s⁻¹.)
Solution
- The radiated energy comes from lost mass. Write the given and rearrange for the mass:
- Substitute E = 1.8 × 10⁴⁴ J and c = 3.00 × 10⁸ m s⁻¹:
- Work it out — keep the unit:
Final answer
Δm = E ÷ c² ≈ 2.0 × 10²⁷ kg — roughly 0.08% of the star's mass, lost as radiated energy over billions of years.