Stellar Luminosity
Calculator
Results
- Luminosity (solar luminosities)
- 0.999997
- Luminosity exponent (log₁₀ W)
- 26.58297
- Absolute bolometric magnitude
- 4.740002
Astronomy results
| Luminosity (solar luminosities) | 0.999997 |
| Luminosity exponent (log₁₀ W) | 26.58297 |
| Absolute bolometric magnitude | 4.740002 |
formula-map diagram
- Luminosity (solar luminosities)
- 0.999997
- Luminosity exponent (log₁₀ W)
- 26.58297
- Absolute bolometric magnitude
- 4.740002
Astronomical relationship
Formula
L = 4π R² σ T⁴= 0.99999762360308
Note
This result is a simplified model: it applies the displayed textbook formula to the values you entered, assuming ideal spherical bodies, circular orbits, blackbody radiation and perfect optics, and ignoring atmospheric seeing, relativistic corrections beyond those stated, cosmological models and measurement uncertainty. Use published ephemerides and catalogue data for real observations.
More in Astronomy and space
See all →Frequently asked questions
What does the Stefan-Boltzmann law calculate for a star?+
It calculates total luminosity (power output) from a star's radius and surface temperature, using L = 4πR²σT⁴, where σ is the Stefan-Boltzmann constant. The 4πR² term is the star's surface area, and σT⁴ is the power radiated per unit area.
Why does temperature affect luminosity so much more than size does?+
Luminosity scales with the fourth power of temperature but only the square of radius, so a modest temperature increase has an outsized effect. Doubling a star's temperature increases its luminosity 16-fold, while doubling its radius only increases luminosity 4-fold.
Why can a cool star still be more luminous than a hot one?+
Because luminosity depends on both radius and temperature together, a huge but cool star, like a red giant, can outshine a small but hot star, like a white dwarf. This is exactly why the Hertzsprung-Russell diagram needs both temperature and luminosity axes to classify stars usefully.
What units does the calculator use for a star's luminosity result?+
Results are typically given in watts or, more commonly in astronomy, in solar luminosities (L☉), where the Sun's output is the reference value of 1. Expressing luminosity relative to the Sun makes it far easier to compare stars, since raw watt values are astronomically large.
Does this formula assume the star radiates as a perfect blackbody?+
Yes, the Stefan-Boltzmann law assumes ideal blackbody radiation, which is a very good approximation for most stars including the Sun. Real stellar spectra have absorption lines and deviations, but for total luminosity estimates the blackbody assumption is accurate enough for standard use.