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 26LSunLSun 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 1.2×1010 LSunLSun?
Express your answer using two significant figures.
The Correct Answer and Explanation is :
To find how much brighter the Betelgeuse supernova would appear in the sky than Sirius, we need to use the inverse square law for brightness. The apparent brightness ( b ) of a star depends on its luminosity ( L ) and the distance ( d ) to the observer. The formula for this relationship is:
[
b = \frac{L}{d^2}
]
Where:
- ( L ) is the luminosity of the star,
- ( d ) is the distance to the star.
Step 1: Calculate the brightness of Sirius
Sirius has a luminosity of ( L_{\text{Sirius}} = 26 L_{\odot} ) and a distance of 8.6 light-years. Using the inverse square law:
[
b_{\text{Sirius}} = \frac{L_{\text{Sirius}}}{d_{\text{Sirius}}^2}
]
Substitute the known values:
[
b_{\text{Sirius}} = \frac{26 L_{\odot}}{(8.6)^2} = \frac{26 L_{\odot}}{73.96}
]
[
b_{\text{Sirius}} \approx 0.351 L_{\odot} \, \text{per light-year squared}
]
Step 2: Calculate the brightness of the Betelgeuse supernova
The luminosity of the Betelgeuse supernova is given as ( L_{\text{Betelgeuse}} = 1.2 \times 10^{10} L_{\odot} ), and its distance is 643 light-years. Using the same formula:
[
b_{\text{Betelgeuse}} = \frac{L_{\text{Betelgeuse}}}{d_{\text{Betelgeuse}}^2}
]
Substitute the known values:
[
b_{\text{Betelgeuse}} = \frac{1.2 \times 10^{10} L_{\odot}}{(643)^2} = \frac{1.2 \times 10^{10} L_{\odot}}{413,649}
]
[
b_{\text{Betelgeuse}} \approx 28,997 L_{\odot} \, \text{per light-year squared}
]
Step 3: Compare the brightness
To compare how much brighter the Betelgeuse supernova would be than Sirius, we take the ratio of the two brightnesses:
[
\frac{b_{\text{Betelgeuse}}}{b_{\text{Sirius}}} = \frac{28,997 L_{\odot}}{0.351 L_{\odot}} \approx 82,600
]
Thus, the Betelgeuse supernova would appear about 82,600 times brighter in the sky than Sirius.
Final Answer:
The Betelgeuse supernova would be 82,000 times brighter than Sirius (rounded to two significant figures).