Stellar Luminosity Calculator
Find a star's luminosity from its size and temperature.
Result
How it is calculated
Luminosity follows the Stefan–Boltzmann law. Measured against the Sun the formula reduces to L / L☉ = R² × (T / 5772)⁴, where R is the star's radius in solar radii and T its surface temperature in kelvin; 5772 K is the Sun's effective temperature. Multiplying by L☉ = 3.828 × 10²⁶ W gives the power in watts. Because temperature is raised to the fourth power, a modest temperature change has a large effect, while radius counts squared. Both inputs must be greater than zero. The result assumes a spherical black-body star and ignores the real spectrum.
Worked example
A star with a radius of 2 solar radii and a surface temperature of 6000 K. R² = 4. T / 5772 = 1.0395 and its fourth power is about 1.1676. Relative luminosity = 4 × 1.1676 = 4.6705 L☉. In watts, 4.6705 × 3.828 × 10²⁶ = 1.7879 × 10²⁷ W.
Common mistakes
- Entering the radius in kilometres or metres instead of solar radii; 1 means the Sun's radius.
- Using temperature in Celsius: the formula needs kelvin.
- Forgetting the fourth power and assuming doubling the temperature doubles the output; it multiplies it by 16.
- Mixing up luminosity (total power emitted) with apparent brightness seen from Earth, which also depends on distance.
- Expecting precise values for real stars from rounded temperature and radius inputs.
About this calculator
A star's luminosity — its total power output — follows the Stefan-Boltzmann law: L = 4πR²σT⁴, where R is the star's radius, T is its surface (effective) temperature, and σ is the Stefan-Boltzmann constant. Since luminosity scales with the square of the radius but the fourth power of temperature, a modest increase in temperature has a much bigger effect on brightness than the same relative increase in size — this calculator typically expresses the result relative to the Sun's luminosity (L☉) for easy comparison.
Astronomers use this relationship to place stars on the Hertzsprung-Russell diagram, estimate distances to stars via the inverse-square law once luminosity is known, and classify stars by evolutionary stage — a red giant is luminous mainly because of its huge radius, while a blue main-sequence star is luminous mainly because of its high temperature.
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Comments (1)
Luminosity grows with the fourth power of temperature, so a small error there matters a lot: if you enter a surface temperature 10 percent too high, the luminosity comes out about 46 percent too high (1.1 to the fourth power). Radius is squared, so it matters less. Use the effective temperature, not a colour guess. A quick check is to enter 1 and 5772 and see 1 L☉. Which star did you try, and did the result match the value you expected?