If Betelgeuse exploded, how bright would it become?
This supergiant star is our stellar neighbor, and it's about to go supernova. What will it look like from our perspective?
There is one famous star that I'm sure you've seen in the sky. It's called Betelgeuse, and it can be found in the constellation Orion, where it marks Orion's right shoulder. If you want to call it “Beetlejuice“, I don't mind – unless you repeat that word three times. [каламбур, основанный на похожести слов Betelgeuse и Beetlejuice / прим. перев.].
But something is going on with it. The red supergiant has dimmed repeatedly over the past few years, which could mean it's ready to go supernova very soon — and by “soon,” we mean within the next 10,000 years. In fact, since it's about 500 light years away, it's possible it's already gone supernova and we just don't know it yet. It could show up as early as tomorrow.
One thing is for sure: If Betelgeuse explodes, it will be the brightest supernova ever seen by humans. How bright? Will you be able to see it in the daytime? Will it be dangerous? I'll show you how to figure it all out using just some basic physics.
What is a supernova?
The core of most stars is made up of hydrogen and helium, the two lightest elements, but only the positively charged nuclei of these atoms, because it is too hot for electrons. Under the influence of enormous gravity and temperature, these cores can fuse heavier elements, releasing enormous amounts of energy in the process. (It is through nuclear fusion that our Sun produces its energy.)
In a stable star like our Sun, there is a balance between two opposing forces. The mass of all the matter in the star creates a gravitational force that tends to collapse the star. However, this is countered by a force pushing the core outward, so the size of the star remains fairly constant, even though it is not a solid object like a planet.
But as the star ages, it gradually uses up its hydrogen and helium and begins to produce heavier elements—carbon, oxygen, silicon, and finally iron. And that's where it ends—it takes more energy to fuse elements heavier than iron than it produces, so the star essentially runs out of fuel and collapses in on itself.
In some cases, this collapse can be very violent — so violent that it quickly increases the pressure and temperature in the star's core. And then the star explodes. There's a big bada-boom. Well, a big quiet bada-boom, since explosions don't make a sound in the vacuum of space.
But it also releases a huge amount of light energy. For comparison, our Sun has a luminosity, or power output, of 3.8 x 10^26 watts. The supernova observed in 2015 (ASASSN-15h) had a peak luminosity of about 2 x 10^38 watts. That’s more than 500 billion Suns. That’s crazy. Oh, you didn’t see it? That’s because it was in another galaxy. Astronomically speaking, Betelgeuse is in our backyard.
Brightness and luminosity
Long ago, the Greek philosopher Hipparchus divided stars into six groups based on how bright they appear in the night sky. From this, we developed a classification scheme called “apparent magnitude,” where a magnitude 1 star looks very bright, while a magnitude 6 star you'll likely be unable to see due to light pollution. Betelgeuse is in the first group.
To be clear, this isn't the star's actual luminosity — it's how bright it appears from Earth, which depends on (1) the amount of light it's emitting, and (2) how far away it is. Oh, and (3) it's based on how the human eye sees objects, which isn't a linear thing either. An object with a magnitude of 1 has 100 times more light (in watts per square meter) than an object with a magnitude of 6.
There may be objects even brighter than magnitude 1, and they will have negative values. For example, the planet Venus is the brightest object in the night sky other than the Moon, with an average brightness of about -4.1 depending on its position.
The influence of distance
So here's how we can proceed: if we have an object's intrinsic luminosity (how much light energy it produces), we can calculate the luminous intensity at a given distance (how much light is coming into that point). And then we can translate that into a scale of magnitude to describe how bright it appears to the human eye.
For example, we have a light bulb that emits 20 watts of light. This is its luminosity. If the light is emitted equally in all directions, and you are standing at a distance of r meters, you can imagine that all the light is spread over the area of a sphere of radius r. As the distance increases, the light spreads over a larger sphere, so there is less of it at any one point. Since the surface area of a sphere (A = 4πr^2) is proportional to the square of the radius, we call this the inverse square law.
Thus, we can write the intensity (I) at a given location as a function of luminosity (L) and distance (r):
This means that if we halve the distance, the light will be four times more intense. Thus, distance has a huge impact on the apparent brightness of an object.
This is going to be epic
Okay, now let's apply this to the Betelgeuse supernova. We can start with a luminosity of 2 x 10^38 watts, like the supernova in 2015. For the distance, I'll use 500 light years. (Surprisingly, we don't have an exact distance to this star, but this is a pretty good estimate. If you're interested, here methodswhich we use to measure distances in astronomy).
Of course, to use our formula, we need to convert this distance from light years to meters. Since light travels at 300 million meters per second, I get a distance of 4.73 x 10^18 meters. This gives the light intensity received by the Earth as 0.711 watts per square meter.
Now, to convert this to magnitude, we need a star to use as an example. Let's take Sirius. It is one of the brightest stars in the sky, with an intrinsic luminosity of 25.4 times that of the Sun, and is 8.79 light years away. This gives us a magnitude (mr) of -1.46. The magnitude (m) of Betelgeuse's supernova would then be calculated as:
After doing the math, I got a magnitude of -18.5, which, holy cow, is just mind-blowing. If our guesses about the luminosity and distance aren't too far off, this will be the brightest object in the night sky. For comparison, the full moon has a magnitude of -12.6, so this supernova will be quite visible even during the day.
To the naked eye, the supernova will appear as a single point of light because, although the star is closer than many others, it is still quite far away. You won't see a disk as big as our Sun or Moon, but it will be the brightest point you've ever seen in the night sky, and it will likely remain there for weeks.
But is it dangerous? Well, it's still a lot less bright than the Sun, which has a magnitude of -26.8. So you won't get sunburned, but you probably shouldn't look at it with an optical telescope. I guess you could take a selfie with the supernova in the background. Your grandkids will want to see that.
