A LevelPhysicsOCRRevision Notes5. Newtonian World & AstrophysicsEM Radiation From StarsEstimating the Radius of Stars

Estimating the Radius of Stars (OCR A Level Physics): Revision Note

Exam code: H556

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Katie M

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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 T of the star

    • Use the inverse square law of intensity equation to find the luminosity L of the star (if given the intensity I and stellar distance d)

    • Then, use the Stefan-Boltzmann law to determine the radius r of the star

Summary of equations

Inverse square law of intensity

I space equals space P over A

  • For a star:

    • the power output is its luminosity, so P space equals space L

    • the area over which the light spreads is A space equals space 4 straight pi d squared

  • Therefore, the inverse square law of intensity for a star is:

I space equals space fraction numerator L over denominator 4 straight pi d squared end fraction

  • Where:

    • I = intensity of light received on Earth (W m-2)

    • L = luminosity of the star (W)

    • d = distance between the star and the Earth (m)

Wien's displacement law

  • Wien's law for a star is given by:

lambda subscript m a x end subscript T space equals space constant

  • Where:

    • lambda subscript m a x end subscript = wavelength emitted by the star at maximum intensity (m) 

    • T = surface temperature of the star (K) 

Stefan-Boltzmann law

  • Stefan's law for a star is given by:

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

    • sigma = the Stefan-Boltzmann constant

    • T = surface temperature of the star (K) 

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.

The Sun has a surface temperature of 5800 K and emits radiation with a peak wavelength of 500 nm.

Calculate the ratio of the radius of Betelgeuse r subscript B to the radius of the Sun r subscript S.

Radius of the Sun, r subscript S = 6.96 × 108 m

Answer:

Step 1: List the known quantities

  • Luminosity of Betelgeuse, L = 4.49 × 1031 W

  • Peak wavelength emitted by Betelgeuse, lambda subscript B = 850 nm

  • Peak wavelength emitted by the Sun, lambda subscript S = 500 nm

  • Surface temperature of the Sun, T subscript S = 5800 K

  • Radius of the Sun, r subscript S = 6.96 × 108 m

Step 2: Write down Wien’s displacement law

lambda subscript m a x end subscript T space equals space constant

Step 3: Use Wien’s law to find the surface temperature of Betelgeuse

lambda subscript S over lambda subscript B space equals space T subscript B over T subscript S

T subscript B space equals space fraction numerator T subscript S lambda subscript S over denominator lambda subscript B end fraction space equals space fraction numerator 5800 cross times 500 over denominator 850 end fraction space equals space 3410 space straight K (3 s.f.)

Step 4: Write down the Stefan-Boltzmann law

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

Step 5: 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 subscript B superscript 4 end fraction end root

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

Step 6: Calculate the ratio rB / rs

r subscript B over r subscript S space equals space fraction numerator 6.83 cross times 10 to the power of 11 over denominator 6.96 cross times 10 to the power of 8 end fraction space equals space 981

  • Therefore, Betelgeuse is approximately 1000 times larger than the Sun

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    Katie M

    Author: Katie M

    Expertise: Physics Content Creator

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.