Sure! Let's go through the steps to determine how much brighter the Betelgeuse supernova would be in our sky compared to Sirius.
1. Identify the given values:
- Luminosity of the Betelgeuse supernova: [tex]\( 1.0 \times 10^{10} \)[/tex] times the luminosity of the Sun ([tex]\( L_{sun} \)[/tex]).
- Luminosity of Sirius: 25.4 times the luminosity of the Sun ([tex]\( L_{sun} \)[/tex]).
2. Understand the question:
- We are asked to find out how many times brighter the Betelgeuse supernova would be compared to Sirius.
3. Calculate the brightness ratio:
We need to compare the luminosity of the Betelgeuse supernova to Sirius by dividing the luminosity of the supernova by the luminosity of Sirius.
[tex]\[
\text{Brightness ratio} = \frac{\text{Luminosity of Betelgeuse supernova}}{\text{Luminosity of Sirius}}
\][/tex]
Substituting the given values into the equation:
[tex]\[
\text{Brightness ratio} = \frac{1.0 \times 10^{10} L_{sun}}{25.4 L_{sun}}
\][/tex]
4. Simplify the ratio:
[tex]\[
\text{Brightness ratio} = \frac{1.0 \times 10^{10}}{25.4}
\][/tex]
This gives:
[tex]\[
\text{Brightness ratio} \approx 393700787.40157485
\][/tex]
5. Express the result using two significant figures:
- The calculated brightness ratio is approximately [tex]\( 3.937 \times 10^8 \)[/tex].
- To express this with two significant figures, we round it to [tex]\( 3.9 \times 10^8 \)[/tex].
So, the Betelgeuse supernova would be approximately [tex]\( 3.9 \times 10^8 \)[/tex] times brighter than Sirius when expressed in two significant figures.
