<p>The distance to betelgeuse is poorly known, so we don't actually know how bright it is with much accuracy measurements of its parallax by satellite give a distance of 197 parsecs <em>+/- 45 parsecs</em> (1 parsec is 3.26 light years). The absolute magnitude (the brightness if it were 10 parsecs distant) is estimated to be -5.85. This is based on both the distance and models of how bright red supergiants "should" be.</p>

<p>To convert from absolute (M) to apparent (m) magnitudes at distance D one can use the formula:
$$m-M = 5 ((\log_{10}{D}) - 1)$$
And at a distance of Sirius (2.54 parsecs) that gives a brightness of -8.7, bright enough to cast shadows, as bright as a crescent moon</p>

<p>A supernova would be a lot brighter. A typical supernova has an absolute magnitude of between -15 and -20, with the low end more likely. Assuming an absolute magnitude of about -16, gives an apparent magnitude of about -19. This would be a bright as interior lighting. </p>

<p>After the supernova, there might be a neutron star which, if it was like the crab pulsar, might have an absolute magnitude of about 4, and at the distance of sirius appear to have a magnitude of about 1, just a little less bright than Betelgeuse is now.</p>

CC BY-SA 3.0
2016-04-07T21:01:39.073Z
2016-04-07T21:01:39.073Z
James K
<p>Given a distance of $d_1 = 640~\mathrm{ly}$ for Betelgeuze and $d_2 = 8.6~\mathrm{ly}$ for Sirius the difference in the apparent magnitude for Betelgeuze at distances $d_1$ and $d_2$ that is given by</p>

<p>$$m_1 - m_2 = 5~\mathrm{mag} \cdot log_{10}(\frac{d_1}{d_2})$$</p>

<p>is $9.4~\mathrm{mag}$. With magnitude $m_1 = 0.45~\mathrm{mag}$ we get $m_2 = -8.9~\mathrm{mag}$ for Betelgeuze at the distance of Sirius.</p>

<p>This is about $860$ times brighter than Sirius ($-1.46~\mathrm{mag}$). It would be visible during daytime</p>

CC BY-SA 3.0
2016-04-07T19:19:14.180Z
2016-04-07T19:28:36.657Z
2016-04-07T19:28:36.657Z
aventurin
<p>How large and bright would Betelgeuse appear if it were as close to Earth as Sirius, before and after it goes supernova?</p>

<distances><astrophysics><size><luminosity><red-giant>
What would be the apparent magnitude of Betelgeuse if it were as close to Earth as Sirius?
Justin Waters
# What would be the apparent magnitude of Betelgeuse if it were as close to Earth as Sirius?

(Asked by: username_0 on 2016-04-07)

How large and bright would Betelgeuse appear if it were as close to Earth as Sirius, before and after it goes supernova?

**Tags:** distances, astrophysics, size, luminosity, red-giant


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## 2 answers

### Answer by username_1 on 2016-04-07. Score: 4 | Accepted answer

The distance to betelgeuse is poorly known, so we don't actually know how bright it is with much accuracy measurements of its parallax by satellite give a distance of 197 parsecs *+/- 45 parsecs* (1 parsec is 3.26 light years). The absolute magnitude (the brightness if it were 10 parsecs distant) is estimated to be -5.85. This is based on both the distance and models of how bright red supergiants "should" be.

To convert from absolute (M) to apparent (m) magnitudes at distance D one can use the formula:
$$m-M = 5 ((\log\_{10}{D}) - 1)$$
And at a distance of Sirius (2.54 parsecs) that gives a brightness of -8.7, bright enough to cast shadows, as bright as a crescent moon

A supernova would be a lot brighter. A typical supernova has an absolute magnitude of between -15 and -20, with the low end more likely. Assuming an absolute magnitude of about -16, gives an apparent magnitude of about -19. This would be a bright as interior lighting.

After the supernova, there might be a neutron star which, if it was like the crab pulsar, might have an absolute magnitude of about 4, and at the distance of sirius appear to have a magnitude of about 1, just a little less bright than Betelgeuse is now.


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### Answer by username_2 on 2016-04-07. Score: 3

Given a distance of $d\_1 = 640~\mathrm{ly}$ for Betelgeuze and $d\_2 = 8.6~\mathrm{ly}$ for Sirius the difference in the apparent magnitude for Betelgeuze at distances $d\_1$ and $d\_2$ that is given by

$$m\_1 - m\_2 = 5~\mathrm{mag} \cdot log\_{10}(\frac{d\_1}{d\_2})$$

is $9.4~\mathrm{mag}$. With magnitude $m\_1 = 0.45~\mathrm{mag}$ we get $m\_2 = -8.9~\mathrm{mag}$ for Betelgeuze at the distance of Sirius.

This is about $860$ times brighter than Sirius ($-1.46~\mathrm{mag}$). It would be visible during daytime


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