Final answer:
Star B, with a surface temperature of 12,000 K, is 16 times more luminous than Star A, which has a temperature of 6000 K, due to the Stefan-Boltzmann law that relates luminosity to the temperature to the fourth power for stars of the same size.
Step-by-step explanation:
The question pertains to the stellar characteristics that determine a star's luminosity. To understand how much more luminous one star is compared to another, we turn to the Stefan-Boltzmann law, which tells us that a star's luminosity (L) is proportional to the fourth power of its surface temperature (T) and also proportional to its surface area. The formula is given by L = 4πR²σT⁴, where R is the radius of the star and σ is the Stefan-Boltzmann constant.
Considering two stars of the same size and the same distance from us, if Star A has a surface temperature of 6000 K and Star B has a surface temperature of 12,000 K (twice that of Star A), we can calculate the ratio of their luminosities. Since both stars have the same radius, we can ignore the surface area component and focus solely on the temperature. The luminosity of Star B in relation to Star A will be:
(12,000 K / 6000 K)ࠚ = 2ࠚ = 16
Hence, Star B is 16 times more luminous than Star A, solely due to its higher surface temperature, since they have identical sizes and distances from us.