Due to its recent dimming and apparent change in physical size, there has been speculation that the red supergiant star Betelgeuse may be on the verge of exploding in a supernova. If that were to occur – and it really is a question of when not if – it will likely become visible in the daytime sky for a few weeks, and may shine brighter than the Full Moon at night. It could happen before I finish writing this sentence, or it could be 100,000 or even a million years in the future. Any of that is “soon” in astronomical terms. We just do not know. To be honest, I don’t really expect it in my lifetime, but who knows? It certainly prompts some speculation.
Some observers (not I) have claimed to be able to discern the visible disk of Jupiter with the unaided human eye. At closet approach, Jupiter spans about 50 seconds of arc. I have never made this observation myself this but I do not challenge the claim. It would be just on the verge of discernability.
So, my question is, when Betelgeuse goes supernova, how long after the explosion would a nebula likely be large enough to be just visible to the unaided human eye from Earth, at least by observers of exceptional ability?
I made a few quick, simplistic calculations that ignore a number of potentially important factors, using comparison values for just one other supernova. So don’t hold me to this… they’re just guesses.
In the case of Betelgeuse, you have to make some assumptions since we do not know some basic facts with certainty. Because of its particular characteristics, it is difficult to determine the exact distance. Estimates range from about 400 to about 800 light years for the distance to Betelgeuse. I will assume 640 light years, a commonly cited figure.
Assuming also that the supernova is a type II, how big would that nebula need to be to be seen as such by the human eye?
To find the diameter of an object of 60 arc seconds (1 arc minute, just slightly larger than Jupiter at its largest apparent size from Earth) at the distance of Betelgeuse, use the small angle formula.
Angle/206265 = d/D
d = (Angle/206265) * D
where Angle is in seconds of arc, 206265 is the number of arc seconds in a Radian, d is the actual size (diameter) and D is the distance, with both d and D in the same units. Here we use arc seconds for angle and Astronomical Units (AU) for distance. So if D is the distance to the object (Betelgeuse, 640 light years (ly) or about 40473241 AU) and the angle is 60 arc seconds, the diameter of the object (nebula) would be
d = (60/206265) * 40473241 AU = 11773 AU
Assuming a Solar System (SS) diameter out to Pluto of 80 AU, we have 11773/80 = 147. Thus the nebula would need to be about 150 times the diameter of the SS.
So according to my very crude and tentative calculations, a sufficiently illuminated distended object would need to be roughly 150 times the current diameter of the Solar System out to Pluto, to subtend an angle of about 60 seconds (1 minute) of arc.
But how long would this take? According to published information on type II SN1987A (https://file.scirp.org/pdf/IJAA_2012123110412746.pdf), that object expanded to about .39 parsecs (pc) or about 80442 AU in less than 25 years. Given the expansion rate of SN1987A, which may or may not be typical, that would imply to me that it would take only a few years to get to 150 Solar System (SS) diameters (.06 pc if taken out to the orbit of Pluto). However, I have no idea how long any distended nebula would be overshadowed by the brilliance of the central supernova.
1) The image is a modification of an uncredited artist’s conception from JPL/NASA found on this page: https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA16885
”Image credit: NASA/JPL-Caltech”
2) Type II supernovae involve the explosion of a single supergiant star. Other types (famously the Type Ia) involve the interaction of an old white dwarf star with a companion star.
3) Others would consider the SS as extending only to the orbit of the last known major planet, Neptune, while others consider it extending to the largest known TNO (“Trans Neptunian Object”) or KBO (“Kuiper Belt Object”) or to the Oort cloud or the Heliopause or some other preferred boundary. I chose the more traditional orbit of Pluto out of convenience.
4) Please feel free to correct me if you see an error, false assumption or other problem.