Question

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.

Answer 1

Answer:

1.5 Times.

Explanation:

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,

$B_i = \frac{L}{4 \pi R^2}$

B=Apparent Brightness

L= Luminosity

R=Radius

Then,

$\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{\frac{L_{Betelgeuse}}{4\pi (D_{Betelgeuse})^2}}{\frac{L_{Sirius}}{4\pi (D_{Sirius})^2}}$

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

$\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{1.1*10^{10}}{643^2}*\frac{8.6^2}{26}$

$\frac{B_{Berelgeuse}}{B_{Sirius}} = 15682.3$

$B_{Betelgeuse} = 15682.3B_{Sirius}$

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