Theory of Intelligence
12TH•Physics•Space And Environment•Luminosity of a Star

21.1 Luminosity of a Star

Stefan-Boltzmann law $L = 4\\pi r^{2}\\sigma T^{4}$ and Example 21.1.

21.1 Luminosity of a Star

Luminosity is the total amount of energy emitted by a star per unit time in the form of electromagnetic radiations. It is a fundamental property that defines the intrinsic brightness of a star. Luminosity is represented by the symbol LL and is measured in watts, indicating the total power of radiation emitted by a star. Luminosity of a star primarily depends on its radius and surface temperature. Larger stars have more surface area to emit radiations and therefore, tend to be more luminous. Similarly, for stars of equal size, the one with the higher surface temperature will radiate more energy and hence, exhibits greater luminosity. The luminosity of a star can be represented as the product of its surface area (A=4πr2A=4\pi r^{2}) and radiant intensity (I=σT4I = \sigma T^{4}).

L=4πr2σT4…(21.1)L = 4\pi r^{2} \sigma T^{4} \quad \ldots (21.1)

Equation (21.1) is called Stefan-Boltzmann law. It states that;

The total output power of a star is proportional to the product of its surface area and the fourth power of its temperature. where σ\sigma is Stefan-Boltzmann constant. Its value is 5.67×10−8 W m−2 K−45.67 \times 10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}, TT is the surface temperature and rr is the radius of the star.

(Fig. 21.1: The Sun)

Tidbit!

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The Sun shines with an immense luminosity of about 4×10264\times 10^{26} watts that is equivalent to billions of nuclear explosions happening every second.

Do you know?

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Stefan-Boltzmann's law helps astronomers estimate the size and brightness of stars just by knowing their temperature.

Example 21.1

The luminosity of a star is given as 2.5×1026 W2.5 \times 10^{26}\,\mathrm{W}, and its surface temperature is 6000 K6000\,\mathrm{K}. Estimate the radius of the star.

Solution

Luminosity of the star L=2.5×1026 WL = 2.5 \times 10^{26}\,\mathrm{W} Surface temperature of the star T=6000 KT = 6000\,\mathrm{K} Radius of the star r=?r = ? Using Stefan-Boltzmann law; L=4πr2σT4L = 4\pi r^{2} \sigma T^{4} or r=L/(4πσT4)r = \sqrt{L / (4\pi\sigma T^{4})} =2.5×1026 W/(4×3.14×5.67×10−8 W m−2 K−4×(6000K)4)= \sqrt{2.5 \times 10^{26}\,\mathrm{W} / (4 \times 3.14 \times 5.67 \times 10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}} \times (6000\mathrm{K})^{4})} r=5.2×108 mr = 5.2 \times 10^{8}\,\mathrm{m}

Brain Teaser

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If a star becomes twice as hot, how many times more energy will it radiate?

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