Stellar Luminosity & Radius Calculator

Apply the Stefan-Boltzmann law to find a star's luminosity or radius using solar units.

⭐ Stellar Luminosity & Radius Calculator
Stellar Radius1.00 R☉
R☉
0.1 R☉50 R☉
Surface Temperature5,778 K
K
2,000 K50,000 K
Stellar Luminosity100.000 L☉
L☉
Surface Temperature3,800 K
K
2,000 K50,000 K
Luminosity
Absolute Luminosity
Spectral Class
Stellar Radius
Radius in km
Spectral Class

⭐ What is the Stellar Luminosity & Radius Calculator?

Stellar luminosity is the total power output of a star across all wavelengths, measured in watts or in solar luminosities (L☉ = 3.828 × 1026 W). This calculator applies the Stefan-Boltzmann law to compute a star's luminosity from its radius and surface temperature, or to work backward and find the stellar radius when luminosity and temperature are known.

The Stefan-Boltzmann law states that the power radiated per unit area of a blackbody is σ × T4, where σ = 5.67 × 10-8 W m-2 K-4 is the Stefan-Boltzmann constant. For a spherical star of radius R and effective surface temperature T, integrating over the entire surface gives L = 4πR2 × σ × T4. Dividing this expression by the equivalent formula for the Sun yields the compact solar-unit form used here: L/L☉ = (R/R☉)2 × (T/T☉)4.

The IAU 2015 nominal solar constants used as reference are: T☉ = 5,778 K (solar effective temperature), L☉ = 3.828 × 1026 W (solar luminosity), and R☉ = 695,700 km (solar radius). Working in solar units keeps numbers manageable. Sirius A, for example, is expressed as 25.6 L☉ rather than 9.81 × 1027 W.

The calculator supports two modes. Find Luminosity accepts stellar radius in solar radii and surface temperature in Kelvin, then outputs luminosity in solar units, absolute luminosity in watts, and the Morgan-Keenan spectral class. Find Radius accepts luminosity in solar units and temperature, then outputs stellar radius in solar radii and in kilometres. Both modes are used in introductory astrophysics courses, stellar physics problem sets, and amateur astronomy to quickly compare stars to the Sun using well-established reference values.

📐 Formula

L / L☉  =  (R / R☉)2  ×  (T / T☉)4
L = stellar luminosity (W or L☉)
L☉ = solar luminosity = 3.828 × 1026 W (IAU 2015)
R = stellar radius (m or R☉)
R☉ = solar radius = 695,700 km
T = stellar effective surface temperature (K)
T☉ = solar effective temperature = 5,778 K
Find Radius (rearranged): R / R☉ = √(L / L☉) × (T☉ / T)2
Example: R = 2 R☉, T = 8500 K → L = 4 × (8500/5778)4 = 18.73 L☉

📘 How to Use This Calculator

Steps

1
Select a calculation mode - Choose Find Luminosity to compute L from radius and temperature, or Find Radius to compute stellar radius from luminosity and temperature.
2
Enter stellar radius or luminosity - In Find Luminosity mode, type the radius in solar radii (R☉) and drag the slider. In Find Radius mode, type the luminosity in solar units (L☉) directly into the input field.
3
Enter the surface temperature - Enter the star's effective surface temperature in Kelvin. The Sun is 5,778 K, O-type stars exceed 30,000 K, and M-type red dwarfs are below 3,700 K.
4
Read the results - The calculator shows luminosity or radius in solar units, the absolute value in watts or kilometres, and the Morgan-Keenan spectral class based on the temperature you entered.

💡 Example Calculations

Example 1 - The Sun

R = 1 R☉, T = 5,778 K (our Sun)

1
Ratio: (R/R☉)2 = (1/1)2 = 1.0
2
Ratio: (T/T☉)4 = (5778/5778)4 = 1.0
3
L = 1.0 × 1.0 = 1.0000 L☉ = 3.828 × 1026 W. The Sun, by definition, has a luminosity of exactly 1 solar luminosity.
Result: 1.0000 L☉ (3.828e+26 W) | Spectral class: G-type
Try this example →

Example 2 - Hot Main-Sequence Star

R = 2.0 R☉, T = 8,500 K (A-type main-sequence star)

1
Radius factor: (2.0/1)2 = 4.0
2
Temperature factor: (8500/5778)4 = (1.471)4 = 4.6835
3
L = 4.0 × 4.6835 = 18.7338 L☉. This star is nearly 19 times more luminous than the Sun despite being only twice its radius, because the higher temperature contributes a factor of 4.68.
Result: 18.7338 L☉ (7.171e+27 W) | Spectral class: A-type
Try this example →

Example 3 - Red Giant (Find Radius)

L = 100 L☉, T = 3,800 K (K-type red giant)

1
Rearranged formula: R/R☉ = √(L/L☉) × (T☉/T)2 = √100 × (5778/3800)2
2
√100 = 10; (5778/3800)2 = (1.5205)2 = 2.3120
3
R = 10 × 2.3120 = 23.1200 R☉ = 16,084,586 km. Despite being 100 times more luminous than the Sun, its low temperature forces it to be over 23 times larger to radiate that much energy.
Result: 23.1200 R☉ (16,084,586 km) | Spectral class: K-type
Try this example →

❓ Frequently Asked Questions

What is the Stefan-Boltzmann law for stellar luminosity?+
The Stefan-Boltzmann law states that a blackbody radiates power proportional to the fourth power of its temperature. For a spherical star of radius R and surface temperature T, total luminosity is L = 4πR² × σ × T4. Dividing by the solar equivalents gives the dimensionless form L/L☉ = (R/R☉)2 × (T/T☉)4, where T☉ = 5,778 K and σ = 5.67 × 10-8 W m-2 K-4.
What is solar luminosity L☉ in watts?+
The IAU 2015 nominal solar luminosity is L☉ = 3.828 × 1026 W. This is the total electromagnetic power radiated by the Sun across all wavelengths. Expressing stellar luminosities in solar units keeps the numbers compact. Sirius A is about 25.6 L☉ rather than 9.81 × 1027 W, making comparisons far more intuitive.
How do I find the luminosity of a star from its radius and temperature?+
Use L/L☉ = (R/R☉)2 × (T/T☉)4. Enter radius in solar radii and temperature in Kelvin into Find Luminosity mode. For a star with R = 2 R☉ and T = 8,500 K, the luminosity is 4 × (8500/5778)4 = 18.73 L☉, showing how both radius and temperature amplify the total output.
Why is temperature raised to the fourth power in the luminosity formula?+
The Stefan-Boltzmann law gives radiated power per unit area as σT4. When integrated over the full spherical surface (area = 4πR2), both factors combine to give L = 4πR2σT4. Temperature therefore dominates: doubling T increases luminosity 16-fold at constant radius, while doubling R increases it only 4-fold. This is why hot blue stars are so luminous despite sometimes being only moderately larger than the Sun.
What is the radius of the Sun in kilometres?+
The IAU 2015 nominal solar radius is R☉ = 695,700 km, approximately 109 Earth radii. Red supergiants like Betelgeuse reach about 700 R☉ (485 million km), large enough to engulf Jupiter's orbit. White dwarfs are typically 0.01 R☉ (roughly 7,000 km), comparable to Earth. Neutron stars have radii of only about 10 km, far below what the Stefan-Boltzmann formula alone can predict.
How does a hotter star compare in luminosity to a cooler star of the same size?+
Because luminosity scales as T4, a 20 percent increase in temperature raises luminosity by (1.2)4 = 2.07 times. An A-type star at 8,500 K radiates (8500/5778)4 = 4.68 times more energy per unit area than the Sun. A star twice the solar radius at that temperature is 4 × 4.68 = 18.7 times more luminous in total, illustrating how temperature and radius compound each other.
Which spectral class is the hottest and most luminous?+
O-type stars are the hottest (above 30,000 K) and most luminous main-sequence stars, often exceeding 100,000 L☉. The Morgan-Keenan sequence from hot to cool is O, B, A, F, G, K, M. The Sun is G-type at 5,778 K. M-type red dwarfs are the coolest at below 3,700 K. Hot blue O and B stars burn through their hydrogen fuel thousands of times faster than the Sun, giving them very short main-sequence lifetimes.
Can this Stefan-Boltzmann formula be applied to white dwarfs?+
Yes. White dwarfs radiate approximately as blackbodies. A typical white dwarf at T = 25,000 K and R = 0.01 R☉ gives L = (0.01)2 × (25000/5778)4 = 0.0001 × 70.1 = 0.007 L☉. Despite the high temperature, the tiny surface area makes it far less luminous than the Sun. Neutron stars are more complex because their radiation includes non-thermal components and general relativistic corrections become important.
What is the difference between stellar luminosity and apparent brightness?+
Luminosity is the intrinsic total power output of a star, independent of the observer's distance. Apparent brightness (flux) is how bright the star appears from Earth and follows the inverse-square law: F = L / (4πd2). Two stars of equal luminosity at different distances will have very different apparent brightnesses. Astronomers measure apparent magnitude, then use the distance modulus to derive absolute magnitude, which encodes luminosity.
How do astronomers measure the actual radius of a distant star?+
Direct methods include interferometry (resolving the stellar disk for nearby giants like Betelgeuse) and eclipse timing during binary transits. Indirect methods use the Stefan-Boltzmann law in reverse: measure luminosity from apparent magnitude and known distance, measure effective temperature from spectral fitting, then solve R = R☉ × √(L/L☉) × (T☉/T)2. This is exactly what Find Radius mode computes.
What assumptions does the Stefan-Boltzmann stellar formula make?+
The formula assumes the star radiates as a perfect blackbody with a single uniform effective temperature Teff over its photosphere. Real stars have limb darkening, starspots, and atmospheric absorption, so Teff is a luminosity-weighted average over the disk. For most main-sequence stars the blackbody approximation is accurate to within a few percent. It becomes less accurate for highly distorted stars like luminous blue variables, close binaries, or rapidly rotating stars with significant equatorial bulges.
Why is the solar effective temperature defined as 5778 K?+
The solar effective temperature is defined by rearranging the Stefan-Boltzmann law: Teff = (L☉ / (4πR☉2σ))0.25. Inserting the measured values L☉ = 3.828 × 1026 W and R☉ = 695,700 km gives Teff = 5,778 K. This is the temperature of the visible photosphere, not the core (about 15 million K) or corona (over 1 million K), and it determines the total radiated flux that sets the Sun's luminosity.

What is the Stefan-Boltzmann law for stellar luminosity?

The Stefan-Boltzmann law states that a blackbody radiates power proportional to the fourth power of its temperature. For a spherical star of radius R and surface temperature T, the total luminosity is L = 4 pi R squared times sigma times T to the fourth, where sigma = 5.67 x 10^-8 W per square metre per K^4. Dividing by the solar expression gives L/L_sun = (R/R_sun)^2 times (T/T_sun)^4.

What is solar luminosity L☉ in watts?

The IAU 2015 nominal solar luminosity is L☉ = 3.828 x 10^26 watts. This is the total electromagnetic power radiated by the Sun across all wavelengths. It is used as the standard unit for stellar luminosity because it keeps the numbers compact. Sirius A, for example, is about 25 L☉ rather than 9.57 x 10^27 W.

How do I find the luminosity of a star from its radius and temperature?

Use the formula L/L☉ = (R/R☉)^2 x (T/T☉)^4 where T☉ = 5778 K and L☉ = 3.828 x 10^26 W. Enter the radius in solar radii and the temperature in Kelvin in Find Luminosity mode. For Sirius A with R = 1.711 R☉ and T = 9940 K, the luminosity works out to about 25.6 L☉, consistent with the measured value.

Why is temperature raised to the fourth power in the luminosity formula?

Because the Stefan-Boltzmann law gives radiated power per unit area as sigma times T^4. When you integrate over the entire spherical surface of a star (area = 4 pi R^2), both the area factor (R^2) and the flux factor (T^4) multiply together. Temperature therefore has a far greater influence on luminosity than radius: doubling T increases luminosity 16-fold at constant radius, while doubling R increases it only 4-fold.

What is the radius of the Sun in kilometres?

The IAU 2015 nominal solar radius is R☉ = 695,700 km, approximately 109 times the radius of Earth. This value is used as the reference unit in stellar astronomy. Red supergiants like Betelgeuse can reach 700 R☉ (about 485 million km), while white dwarfs are typically 0.01 R☉ (roughly 7000 km, similar to Earth).

How does a hotter star compare in luminosity to a cooler star of the same size?

Because luminosity scales as T^4, even a moderate temperature increase produces a large luminosity boost. An A-type star at 8500 K is about (8500/5778)^4 = 4.68 times more luminous per unit area than the Sun. A star twice the solar radius at 8500 K would therefore be 4 x 4.68 = 18.7 times more luminous than the Sun in total.

Which spectral class is the hottest?

O-type stars are the hottest, with surface temperatures above 30,000 K. They appear blue-white and are extremely luminous, often 100,000 times more luminous than the Sun. The Morgan-Keenan sequence from hot to cool is O, B, A, F, G, K, M. The Sun is a G-type star at 5778 K. M-type red dwarfs are the coolest main-sequence stars, typically below 3700 K.

Can this Stefan-Boltzmann formula be applied to white dwarfs and neutron stars?

The formula applies to any object that radiates approximately as a blackbody, including white dwarfs. A typical white dwarf at T = 25,000 K and R = 0.01 R☉ has L = (0.01)^2 x (25000/5778)^4 = 0.0001 x 70.1 = 0.007 L☉. Neutron stars are trickier because their radiation is often non-thermal and the radius definition is complicated by general relativistic effects.

What is the difference between stellar luminosity and apparent brightness?

Luminosity is the total power output of a star, independent of distance. Apparent brightness (flux) is how bright the star appears from Earth and depends on both luminosity and distance via the inverse-square law: F = L / (4 pi d^2). Two stars with equal luminosity at different distances will have very different apparent brightnesses. Astronomers measure apparent magnitude and use the distance modulus to convert to absolute magnitude, which is proportional to luminosity.

How do astronomers measure the actual radius of a distant star?

Direct methods include interferometry (for nearby giants like Betelgeuse) and transit timing (when a planet or companion eclipses the star). Indirect methods use the Stefan-Boltzmann law in reverse: measure the luminosity from the distance and apparent magnitude, measure the temperature from spectral fitting, then solve for R = R☉ times sqrt(L/L☉) times (T☉/T)^2. This calculator uses that same relationship in Find Radius mode.

What assumptions does the Stefan-Boltzmann stellar formula make?

The formula assumes the star radiates as a perfect blackbody with a uniform effective temperature over its photosphere. Real stars have limb darkening, star spots, and atmospheric absorption, so the effective temperature T_eff is a luminosity-weighted average. For most main-sequence stars the blackbody approximation is accurate to within a few percent. It becomes less accurate for very extended or irregular objects like luminous blue variables.

Why is the solar effective temperature defined as 5778 K?

The solar effective temperature is defined by rearranging the Stefan-Boltzmann law: T_eff = (L☉ / (4 pi R☉^2 sigma))^0.25. Inserting the measured values L☉ = 3.828 x 10^26 W and R☉ = 695,700 km gives T_eff = 5778 K. This is not the temperature at the Sun's core (about 15 million K) or corona (over 1 million K) but the temperature of the visible photosphere that determines the total radiated flux.