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If Betelgeuse were to go supernova would the sky be twice as bright, or day time extended, depending on what time of year it happened in.

Basically when it does supernova how bright and large will it appear in the sky, and how do you work out the apparent brightness and size for something that hasn't happened yet?

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Most models of type II supernovae expect that the explosion has a peak luminosity of around $10^8$ to $10^9$ solar luminosities. This gives it a peak absolute magnitude of -15.16 to -17.16. At a distance of 200 parsecs, this is an apparent magnitude of -8.652 to -11.15.

Okay, so how bright is that? Well, the full moon has an apparent magnitude usually around -12.75. This means the total luminosity from the explosion of Betelgeuse would be something like $1/1000$ times the luminosity of the full moon. But the human eye sees things logarithmically, so we might expect that it looks like its about a third as bright as the full moon.

This matches up roughly to descriptions of SN 1006 which was a supernova 1,000 years ago. Observers at the time described a slowly growing ball in the sky about a quarter the brightness of the moon.

As for your second question, how we predict the brightness: We have our estimates about the brightness of the supernova mostly from computer simulations. Supernovas occur when large stars run out of fuel and their core collapses, causing all the mass to suddenly fall inward. A series of complicated nuclear reactions take place as the matter rebounds in a huge shockwave. The parameters of this whole process are quite nasty to work out in detail and the results depend somewhat (but less than you might think) upon the size and composition of the star. We've figured out those parameters mostly by running lots and lots of computer simulations with different parameters and figuring out which sets of parameters match up best with observations of actual supernovae.

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  • $\begingroup$ Would another factor be angular diameter? The Moon after all has an angular diameter of approximately 30 degrees which, to us, would likely make it seem rather bright. $\endgroup$
    – Nick Bedford
    Commented Feb 17, 2011 at 4:54
  • $\begingroup$ @Nick Bedford Actually, it works in the opposite direction - the brightness figure I quoted for the moon is the total, integrated over the entire surface, so a given point on the moon looks less bright than a -12.75 magnitude star. Somewhat counterintuitively, the angular diameter doesn't affect the total number of photons your eyeball receives. $\endgroup$
    – spencer nelson
    Commented Feb 17, 2011 at 5:20
  • $\begingroup$ Another way of thinking about this is that "magnitude" is a measure of luminosity, which is energy per second. It doesn't care about surface area of the object. That's a different quantity, the surface brightness. As you spread out an object over the sky, it will appear to get dimmer, not brighter, since the total energy available (the luminosity) stays the same, but it is divided over a larger area of your field of view. See en.wikipedia.org/wiki/Surface_brightness $\endgroup$
    – spencer nelson
    Commented Feb 17, 2011 at 5:23
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    $\begingroup$ In other words, the supernova would be less bright overall than the full moon, but far more piercingly brilliant, since it would still appear as a single point. $\endgroup$
    – Keenan Pepper
    Commented Feb 17, 2011 at 14:52
  • $\begingroup$ I would expect it look like the diamond ring during a total solar eclipse, with shadow bands on the ground and stuff. I think it will be bright enough to hurt your eyes. $\endgroup$
    – Calmarius
    Commented May 10, 2018 at 11:27

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