Answer:
b) Betelgeuse would be [tex]\approx 1.43 \cdot 10^{6}[/tex] times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is [tex]\approx 1.37 \cdot 10^{-5} }[/tex] the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
[tex]B = \displaystyle{\frac{L}{4\pi d^2}}[/tex] (1)
where [tex]B[/tex] is the brightness, [tex]L[/tex] is the star luminosity and [tex]d[/tex], the distance from the star to the point where the brightness is calculated (measured). Thus:
b) [tex]B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}[/tex] and [tex]B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}[/tex] where [tex]L_{Sun}[/tex] is the Sun luminosity ([tex]3.9 x 10^{26} W[/tex]) but we don't need to know this value for solving the problem. [tex]ly[/tex] is light years.
Finding the ratio between the two brightness we get:
[tex]\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }[/tex]
c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is [tex]1.581 \cdot 10^{-5}\ ly[/tex]. Then
[tex]\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }[/tex]
Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.
