The distance from Earth of the red supergiant Betelgeuse is approximately 643 light-years. If it were to explode as a supernova, it would be one of the brightest stars in the sky. Right now, the brightest star other than the Sun is Sirius, with a luminosity of 26LSun and a distance of 8.6 light-years. How much brighter in our sky than Sirius would the Betelgeuse supernova be if it reached a maximum luminosity of 9.0×10^9 LSun?
6.2*10^4
4.8*10^4
To determine how much brighter the Betelgeuse supernova would appear in our sky compared to Sirius, we need to calculate the luminosity ratio between the two.
First, let's calculate the luminosity ratio by dividing the luminosity of the Betelgeuse supernova (9.0×10^9 LSun) by the luminosity of Sirius (26 LSun):
Luminosity Ratio = (Luminosity of Betelgeuse Supernova) / (Luminosity of Sirius)
= (9.0×10^9 LSun) / (26 LSun)
Now, to compare the brightness in our sky, we need to consider the distance factor. The apparent brightness of a star decreases with the square of its distance. So, we need to calculate the distance ratio between Betelgeuse and Sirius, considering their respective distances from Earth:
Distance Ratio = (Distance to Betelgeuse) / (Distance to Sirius)
= 643 light-years / 8.6 light-years
To calculate the overall brightness ratio, we multiply the luminosity ratio by the square of the distance ratio:
Brightness Ratio = (Luminosity Ratio) * (Distance Ratio)^2
Let's calculate the brightness ratio:
Luminosity Ratio = (9.0×10^9 LSun) / (26 LSun) ≈ 3.46×10^8
Distance Ratio = (643 light-years) / (8.6 light-years) ≈ 74.77
Brightness Ratio = (3.46×10^8) * (74.77)^2
By performing the above calculation, we find that the brightness ratio is approximately 1.524×10^7.
Therefore, if the Betelgeuse supernova were to reach a maximum luminosity of 9.0×10^9 LSun, it would appear approximately 1.524×10^7 times brighter in our sky than Sirius, the brightest star other than the Sun.