Stellar Radius from Luminosity | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Temperature controls power per square metre

The Stefan–Boltzmann law says that a black-body surface emits power per unit area proportional to the fourth power of temperature. A spherical star has a surface area of $4\pi r^2$ .

L=4\pi r^2\sigma T^4

r=\sqrt{\dfrac{L}{4\pi\sigma T^4}}

Symbols

$L$ = luminosity of the star (W)
$r$ = radius of the star (m)
$\sigma$ = Stefan–Boltzmann constant (W m⁻² K⁻⁴)
$T$ = surface temperature (K)

Radius fixed: doubling temperature makes luminosity 16 times larger.
Temperature fixed: doubling radius makes luminosity 4 times larger.
Luminosity fixed: a hotter star must have a smaller radius.

Use the spectrum first, then find the radius

Read the peak wavelength from the spectrum.
Use Wien's law to calculate surface temperature.
Use luminosity and temperature in Stefan–Boltzmann.
Rearrange for the stellar radius.

Worked example

Estimate a Sun-like star's radius

A star has a 501 nm peak and a luminosity of 3.85 × 10²⁶ W.

T=\dfrac{2.90\times10^{-3}}{501\times10^{-9}}

T=5.79\times10^3\ \mathrm{K}

4\pi\sigma T^4=8.01\times10^8\ \mathrm{W\,m^{-2}}

r=\sqrt{\dfrac{3.85\times10^{26}}{8.01\times10^8}}

Answer

r\approx6.9\times10^8\ \mathrm{m}

Your turnrelationship check

Two stars have the same surface temperature. Star B has four times the luminosity of star A. Compare their radii.

Show worked answer

At fixed temperature, luminosity is proportional to radius squared. Star B therefore has twice the radius of star A.