Answer:
Sirius A is 1.608 times the size of the Sun.
Explanation:
The radiant flux 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 is 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 is 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 = 5800K[/tex]
[tex]L_{Sun} = 4\pi (696.34x10^{6}m)^2(5.67x10^{-8} W/m^{2} K^{4} )(5800K)^{4}) [/tex]
[tex]L = 3.90x10^{26} W[/tex]
To find the luminosity of Sirius A, the following can be used:
[tex]\frac{L_{SiriusA}}{L_{sun}} = 23[/tex]
[tex]{L_{SiriusA}} = (3.90x10^{26} W)(23)[/tex]
[tex]{L_{SiriusA}} = 8.97x10^{27}W[/tex]
Finally, equation 4 can be used to determine the radius of Sirius A.
[tex] r = \sqrt{\frac{8.97x10^{27}W}{4\pi (5.67x10^{-8} W/m^{2} K^{4})(10000K)^{4}}}[/tex]
[tex]r = 1.12x10^{9}m[/tex]
So, Sirius A has a radius of [tex]1.12x10^{9}m[/tex]
[tex]\frac{1.12x10^{9}m}{696.34x10^{6}m} = 1608[/tex]
Hence, Sirius A is 1.608 times the size of the Sun.
