A hydrogen bomb detonated against your eyeball
November 28, 2013…would deliver less energy to your retina than a supernova observed from a distance of one astronomical unit (AU; the distance from the Earth to the sun). How much less? From this XKCD What If:
Which of the following would be brighter, in terms of the amount of energy delivered to your retina:
A supernova, seen from as far away as the Sun is from the Earth, or
The detonation of a hydrogen bomb pressed against your eyeball?
Applying the physicist rule of thumb suggests that the supernova is brighter. And indeed, it is … by nine orders of magnitude.
That rocked me back on my heels. And it got me thinking: how far away would one have to be for a supernova to be only as bright as an h-bomb pressed against one’s eyeball?
Radiated energy is subject to the inverse-square law, by which intensity of radiation is inversely proportional to the square of the distance. So the answer to my question is the square root of billion in AU, which is 31,623 AU, which is almost precisely half a light year. (BTW, Google will translate AU to light years for you!)
So if you’re close enough to a supernova that the light takes six months to reach you, it will still be like being nuked at point-blank range.
How far away from a supernova do you need to be to be safe? According to this article, even at a distance of 3000 light years, a supernova could still wreck the ozone layer of an Earth-like world.
Even more suprisingly (to me, anyway), the 1006 and 1054 supernovae apparently left detectable chemical traces on Earth, despite being 7200 and 6500 light years away, respectively. From farther down in the same article:
Gamma rays from a supernova would induce a chemical reaction in the upper atmosphere converting molecular nitrogen into nitrogen oxides…. In 2009, elevated levels of nitrate ions were found in Antarctic ice, which coincided with the 1006 and 1054 supernovae.
Amazing. The 1054 supernova is near and dear to my heart. Its visible remnant, the Crab Nebula, is also catalogued as Messier 1. I have observed it dozens of times, most notably during my nearly-annual Messier Marathons. I had no idea that it had literally left its mark on Earth.
So, here’s something to be thankful for this Thanksgiving: there are no particularly good supernova candidates close enough to Earth to pose a serious threat. All of the contenders are not massive enough yet (if they’re white dwarfs) or too far away, or won’t blow for millennia, or some combination of the above. So you can tuck in with abandon. We could still be annihilated at any moment by death from space–just ask the folks in Chelyabinsk–but it probably won’t come in form of a supernova.
Hat tip to Mike.
10 Minute Astronomy 
“How far away from a supernova do you need to be to be safe? According to this article, even at a distance of 3000 light years, a supernova could still wreck the ozone layer of an Earth-like world.”
What I’m about to say is clearly nonsense. But I don’t understand why it’s wrong. Help me out.
A supernova 3000 light-years away is 6000 times as far as one that’s half a light-year away, which we agreed is like an H-bomb to the eyeball. Since supernova and H-bomb radiation both go by inverse squares, the 3000 light-years-away supernova should be about as devastating as an H-bomb 6000 times as far away as zero meters from my eyeball. But that is still zero meters.
Huh.
Yeah, there’s a weird Math Thing there.
Since we’re voyaging into realms of epic pedantry, I must note that the XKCD guy specified energy delivered to the retina by an h-bomb pressed against the eyeball, meaning presumably the cornea, which is just under an inch away on average (24mm). But that’s no help, because 6000*24mm=144000mm, or 144m, and that’s just crazy. For a decent-sized h-bomb, being 144m from the center of the blast would put you a mile or so inside the fireball; for all practical purposes, you’d still be at ground zero.
So maybe the eyeball thing is just a rhetorical flourish, and what he really meant by that was the max radiation flux across a 24mm diameter circle placed at the center of an h-bomb explosion. But that doesn’t solve your problem, because that’s still zero meters.
We also can’t assume that the hypothetical observer’s retina is separated from the center of the blast by the size of the bomb itself. Even if we assume an unrealistic distance of a full meter, which would only have applied to the earliest and most massive h-bombs, 6000 times that far is only 6 km, still well within the “pretty much instantly burst into flame” radius, which obviously would not happen with a supernova 3000 light years away (otherwise Earth would have been sterilized by the supernovae 6500 and 7200 light years away).
For the record, it’s 2:15 AM and I absolutely should not be committing acts of math in public.
That said, I am trying to figure out how to come at this from the other direction. Basically, we’re trying to figure out how far you have to be from an h-bomb detonation for it to look like a supernova does from 3000 light years out. According to the inverse-square law, if X is 1/9 as bright as Y but 1/3 as far away, they will look equally bright. So if an h-bomb is one billionth as bright as a supernova, it has to be 1/31623 as far away to be equally bright. And 1/32163 of 3000 light years is 0.093 light years, or 5881 AU. But that’s clearly nuts, Neptune is 30 AU away and it’s a speck, no way is an h-bomb 200 times farther out going to even show up. I mean, the collision of Comet Shoemaker-Levy 9 with Jupiter made fireballs the size of Earth, and they weren’t visible without telescopes, but supernovae 6000 light years away are bright enough to read by.
So I can’t make the numbers work that way either. I’m stumped. And tired. Here’s hoping someone with more math chops will show up and save the day.
I think this may be because of the more fundamental problem of inverse-square laws: that they tend to infinity as distance tends to zero. If I clap my hands 20 m away from you, a certain amount of energy reaches you; if I close in to 10 m and clap, the distance is halved and energy goes with 1/d^2, so you’ll get four times the energy. But if I stand right next to you and clap at distance d = 0, the energy is infinite. Which is clearly nonsense.
So what’s going on here?
(late reply, but)
I think the problem is that the inverse-square law of radiation applies to a point source, and when you are inside the H-bomb fireball, it can no longer accurately be approximated as a point source — you’re getting light from all directions.
Maybe?
Reblogged this on Alexander Riccio.
Some empirical evidence might help to clarify. 😏
The original post on XKCD mixes up power and energy.
“Which of the following would be brighter, in terms of the amount of energy delivered to your retina”
Brightness is like power and conventionally measured in Watt (J/s). Energy is measured in J. This is important because H-bomb and supernova durations differ by 12 orders of magnitude.
The supernova (10^44 J) at 1 AU (1.5 x 10^11 m) delivers 3.5 x 10^20 J/m^2 on the earth, but only 2 x 10^14 W/m^2 or so over its 20 day duration. On the retina (10^-3 m^2) this is 3.5 x 10^17 J and 2 x 10^11 W.
An H-bomb makes about 10^17 J in 1 microsecond, or 10^23 W in total.
So the energy delivered by the supernova is higher, but the brightness is lower. The talk of nine orders of magnitude is not supported by these numbers.