Answer :
Using the Stefan-Boltzmann law and the given temperatures, we can determine that the hotter star at 9000 K is 81 times more luminous than the cooler star at 3000 K, because luminosity is proportional to the fourth power of the star's temperature.
To determine how much larger the luminosity of the hotter star is compared to the cooler star, we use the Stefan-Boltzmann law, which states that luminosity (L) is proportional to the fourth power of the temperature (T), expressed in the formula L = 4πR²σ[tex]\text T^4[/tex], where R is the radius of the star and σ is the Stefan-Boltzmann constant. Since the sizes and distances of the stars are the same, we can compare their luminosities based on their temperature alone.
Let's compare the two stars in the student's example:
Star A: 3000 K
Star B: 9000 K
Using the proportionality, [tex]L_A / L_B = (T_A / T_B)^4[/tex], we can determine the relative luminosity.
Plugging in the temperatures:
[tex]L_A / L_B = (3000 / 9000)^4\\L_A / L_B = (1/3)^4\\LA / LB = 1/81[/tex]
Therefore, the hotter star (Star B) is 81 times more luminous than the cooler star (Star A).
The luminosity of the hotter star is approximately 81 times larger than that of the cooler star.
The luminosity of a star is directly related to its temperature according to the Stefan-Boltzmann law, which states that the luminosity of a star is proportional to the fourth power of its temperature. In this case, the temperature of the hotter star is 9000 K, while the temperature of the cooler star is 3000 K.
To calculate the ratio of their luminosities, we can use the formula:
Luminosity ratio = (T₂ / T₁)⁴
where T₂ is the temperature of the hotter star and T₁ is the temperature of the cooler star.
Substituting the given values, we have:
Luminosity ratio = (9000 K / 3000 K)⁴
= (3)⁴
= 81
Therefore, the luminosity of the hotter star is approximately 81 times larger than that of the cooler star.
Learn more about Luminosity
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