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


2 Answers 2


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.


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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