# What would be the apparent magnitude of Betelgeuse if it were as close to Earth as Sirius?

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

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