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If the stars Betelgeuse and Rigel each have the same luminosity, but the temperature of Betelgeuse is lower than Rigel, which star has the greater surface area?

User Achennu
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Answer:

Betelgeuse

Step-by-step explanation:

Stars luminosity is given by the Stefan–Boltzmann law, since the emitted energy is treated as black body radiation:


L=\sigma AT^4

Here
\sigma is the Stefan–Boltzmann constant, A is the surface area and T is the temperature. According to this formula, if both stars have the same luminosity, then the lower temperature star must have a greater surface area.

User Andrew Daniel
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