Calculating Surface Area of Stars Based on Luminosity and Temperature

Which star has the greater surface area if Betelgeuse and Rigel have the same luminosity but Betelgeuse is cooler?

A. Betelgeuse

B. Rigel

Answer:

The star with the greater surface area in this scenario is Betelgeuse.

To determine which star has the greater surface area when Betelgeuse and Rigel have the same luminosity but Betelgeuse is cooler, we need to consider the relationship between luminosity, surface area, and temperature of stars.

The Stefan-Boltzmann Law states that Luminosity (L) equals the product of surface area (A), Stefan-Boltzmann constant (σ), and temperature (T) raised to the fourth power: L = A * σ * T⁴. Since Betelgeuse and Rigel have the same luminosity, we can set their luminosity equations equal to each other as A1 * T1⁴ = A2 * T2⁴.

Given that the temperature of Betelgeuse is cooler than Rigel (T1 < T2), Betelgeuse must have a larger surface area (A1) to compensate for its lower temperature in order to maintain the same luminosity. Therefore, Betelgeuse has the greater surface area compared to Rigel in this scenario.

In conclusion, when two stars have the same luminosity but different temperatures, the cooler star will have a larger surface area. This is due to the need for the cooler star to emit more of its energy at longer wavelengths, requiring a larger surface area to maintain the same luminosity level. In the case of Betelgeuse and Rigel, Betelgeuse would have the greater surface area because of its cooler temperature.