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A LevelPhysicsCIERevision Notes25. Astronomy & Cosmology25.2 Stellar RadiiStefan-Boltzmann Law & Stellar Radii

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Stefan-Boltzmann Law & Stellar Radii (CIE A Level Physics)

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Ashika

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Ashika

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Stefan-Boltzmann law

  • A star’s luminosity depends on its:
    • surface temperature
    • radius
  • The relationship between these is known as the Stefan-Boltzmann law, which states:

The total energy emitted by a black body per unit area per second is proportional to the fourth power of the absolute temperature of the body

  • It is equal to:

L space equals space 4 straight pi r squared sigma T to the power of 4

  • Where:
    • L = luminosity of the star (W)
    • r = radius of the star (m)
    • σ = the Stefan-Boltzmann constant
    • T = surface temperature of the star (K)

Estimating the radius of stars

  • The radius of a star can be estimated by combining Wien’s displacement law and the Stefan–Boltzmann law
  • The procedure for this is as follows:
    • Use Wien’s displacement law to find the surface temperature of the star
    • Use the inverse square law of flux equation to find the luminosity of the star (if given the radiant flux and stellar distance)
    • Then, use the Stefan-Boltzmann law to calculate the radius of the star

Summary of Stellar Equations (1), downloadable AS & A Level Physics revision notesSummary of Stellar Equations (2), downloadable AS & A Level Physics revision notes

Worked example

Betelgeuse is our nearest red giant star.

It has a luminosity of 4.49 × 1031 W and emits radiation with a peak wavelength of 850 nm.

Calculate the ratio of the radius of Betelgeuse rB to the radius of the Sun rs.

Radius of the sun rs = 6.95 × 108 m.

Answer:

Step 1: Write down Wien’s displacement law

lambda subscript m a x end subscript T space equals space 2.9 space cross times space 10 to the power of negative 3 end exponent space straight m space straight K

Step 2: Rearrange Wien’s displacement law to find the surface temperature of Betelgeuse

T space equals space fraction numerator 2.9 space cross times 10 to the power of negative 3 end exponent space over denominator lambda subscript m a x end subscript end fraction space equals space fraction numerator 2.9 space cross times space 10 to the power of negative 3 end exponent space over denominator 850 space cross times 10 to the power of negative 9 end exponent end fraction space equals space 3410 space straight K

Step 3: Write down the Stefan-Boltzmann law

L space equals space 4 straight pi r squared sigma T to the power of 4

Step 4: Rearrange for r and calculate the stellar radius of Betelgeuse

r subscript B space equals space square root of fraction numerator L over denominator 4 straight pi sigma T to the power of 4 end fraction end root space equals space square root of fraction numerator 4.49 space cross times space 10 to the power of 31 over denominator 4 straight pi space cross times space open parentheses 5.67 space cross times space 10 to the power of negative 8 end exponent close parentheses space cross times space open parentheses 3410 close parentheses to the power of 4 end fraction end root space equals space 6.83 space cross times space 10 to the power of 11 space straight m space

Step 5: Calculate the ratio r subscript B over r subscript s

r subscript B over r subscript s space equals space fraction numerator 6.83 space cross times space 10 to the power of 11 space over denominator 6.95 space cross times space 10 to the power of 8 end fraction space equals space 983

  • Therefore, the radius of Betelgeuse is about 1000 times larger than the Sun’s radius

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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.