Calculate luminosity, radius or temperature from the other two values using the Stefan-Boltzmann formula and unit conversions for W, m and K as needed.

Luminosity Radius Temperature Calculator

Enter any 2 values to calculate the missing variable





Related Calculators

Luminosity Radius Temperature Formula

The calculator uses the Stefan-Boltzmann law for a spherical blackbody radiator. Enter any two values, and the missing value is found by rearranging the same relationship.

L = 4*pi*R^2*sigma*T^4

To calculate radius:

R = sqrt(L/(4*pi*sigma*T^4))

To calculate temperature:

T = (L/(4*pi*R^2*sigma))^(1/4)
  • L = luminosity, in watts (W)
  • R = radius, in meters (m)
  • T = absolute temperature, in kelvin (K)
  • sigma = Stefan-Boltzmann constant, approximately 5.67 × 10-8 W/(m2 K4)
  • pi = 3.14159

The luminosity function calculates total radiated power from radius and temperature. The radius function solves for the size needed to produce a given luminosity at a given temperature. The temperature function solves for the blackbody temperature needed for a given luminosity and radius.

Unit selections are converted to base units before the formula is applied: luminosity to watts, radius to meters, and temperature to kelvin. The result is then converted back to your selected output unit.

Common Units and Base Conversions

Quantity Unit Conversion to base unit
Luminosity kW 1 kW = 1,000 W
Luminosity MW 1 MW = 1,000,000 W
Radius cm 1 cm = 0.01 m
Radius km 1 km = 1,000 m
Temperature °C K = °C + 273.15
Temperature °F K = (°F – 32) × 5/9 + 273.15

Typical Reference Values

Object or surface Approximate temperature Notes
Room temperature object 293 K About 20 °C
Boiling water temperature 373 K About 100 °C at standard pressure
Incandescent filament 2,500 K to 3,000 K Approximate glowing tungsten range
Sun’s photosphere About 5,778 K Often used as a stellar reference temperature

Examples

Example 1: Calculate luminosity

Suppose an object has a radius of 1 m and a temperature of 300 K.

L = 4*pi*1^2*(5.67*10^-8)*300^4

The result is approximately 5,770 W, or about 5.77 kW.

Example 2: Calculate radius

Suppose an object has a luminosity of 1,000 W and a temperature of 500 K.

R = sqrt(1000/(4*pi*(5.67*10^-8)*500^4))

The result is approximately 0.15 m.

FAQ

Why does temperature have such a large effect on luminosity?

Temperature is raised to the fourth power in the Stefan-Boltzmann law. If radius stays the same and temperature doubles, luminosity increases by 24, or 16 times. This is why small temperature changes can produce large changes in emitted power.

Why does the calculator use kelvin for temperature?

The formula requires absolute temperature, so kelvin is used internally. Celsius and Fahrenheit can be entered, but they must be converted to kelvin before the fourth-power calculation is applied.

Does this give the true luminosity of any real object?

It gives the blackbody luminosity for a spherical object. Real objects may emit less or more at certain wavelengths depending on emissivity, surface properties, atmosphere, and geometry. For many physics and astronomy problems, this formula is the standard starting point.