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. Part A How much brighter in our sky than Sirius would the Betelgeuse supernova be if it reached a maximum luminosity of 1.1×1010 LSun? Express your answer using two significant figures.

The first step in order to solve the problem is to define our values, from this we can proceed to find the apparent brightness relationship for the two objects,

So,

[tex]B_i = \frac{L}{4 \pi R^2}[/tex]

B=Apparent Brightness

L= Luminosity

R=Radius

Then,

[tex]\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{\frac{L_{Betelgeuse}}{4\pi (D_{Betelgeuse})^2}}{\frac{L_{Sirius}}{4\pi (D_{Sirius})^2}}[/tex]

[tex]\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{\frac{1.1*10^{10}*L_{sun}}{4\pi (643)^2}}{\frac{26L_{Sun}}{4\pi(8.6)^2}}[/tex]

[tex]\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{1.1*10^{10}}{643^2}*\frac{8.6^2}{26}[/tex]

[tex]\frac{B_{Berelgeuse}}{B_{Sirius}} = 15682.3[/tex]

[tex]B_{Betelgeuse} = 15682.3B_{Sirius}[/tex]

Under this relationship we can conclude that in the case of a Supernova, in our sky Betelgeuse Supernova would be [tex]1.5 * 10 ^ 4[/tex] times brighter than Sirius seen from the earth to the sky.