Answer:
The radius of the red supergiant is [tex]2.58x10^{11}m[/tex].
Explanation:
The radiant power establishes how much energy an observer or a detector can get from a luminous source per unit time and per unit surface area.
[tex]R_{p} = \frac{L}{4\pi r^2}[/tex] (1)
Where [tex]R_{p}[/tex] is the radiant power received from the source, L its intrinsic luminosity and r is the distance.
The Stefan-Boltzmann law is defined as:
[tex]R_{p} = \sigma \cdot T^{4}[/tex] (2)
Where [tex]R_{p}[/tex] is the radiant power, [tex]\sigma[/tex] is the Stefan-Boltzmann constant and T the temperature.
Then, equation 2 can be replaced in equation 1
[tex]\sigma \cdot T^{4} = \frac{L}{4\pi r^2}[/tex] (3)
Notice that L is the energy emitted per second by the source.
Therefore, r can be isolated from equation 3.
[tex] r^2 = \frac{L}{4\pi \sigma\cdot T^{4}}[/tex]
[tex] r = \sqrt{\frac{L}{4\pi \sigma\cdot T^{4}}}[/tex] (4)
The luminosity of the Sun can be estimated isolating L from equation 3.
[tex]L = (4\pi r^2)(\sigma \cdot T^{4}) [/tex]
but, [tex]r = 696.34x10^{6}m[/tex] and [tex]T = 5778K[/tex] for the Sun.
[tex]L = 4\pi (696.34x10^{6}m)^2(5.67x10^{-8} W/m^{2} K^{4} )(5778K)^{4}) [/tex]
[tex]L = 3.85x10^{26} W[/tex]
The following expression can be used to find the luminosity of the star:
[tex]\frac{L_{star}}{L_{sun}} = 10000[/tex]
[tex]{L_{star}} = (3.85x10^{26} W)(10000)[/tex]
[tex]{L_{star}} = 3.85x10^{30}W[/tex]
Finally, equation 4 can be used.
[tex] r = \sqrt{\frac{3.85x10^{30}W}{4\pi (5.67x10^{-8} W/m^{2} K^{4})(3000K)^{4}}}[/tex]
[tex]r = 2.58x10^{11}m[/tex]
Hence, the radius of the red supergiant is [tex]2.58x10^{11}m[/tex].
