I know there are questions pertaining to "What would happen if all nukes were exploded at X location?", however, I can't seem to find a question asking this: What would happen if all of the Earth's nuclear weapons simultaneously exploded where they are now?
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$\begingroup$ I think it's safe to say that the answer is "death and devastation on a large scale". Probably tectonic instability, nuclear winter, and radiation clouds which would make life really miserable to the survivors. $\endgroup$– AndreiROMCommented Jul 11, 2016 at 15:09
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$\begingroup$ I am interested on how much thrust it would provide. 15,000 missiles with a wide range. 2 Tsar bombas would destroy the world. $\endgroup$– MuzeCommented Feb 13, 2018 at 20:15
2 Answers
This fictional scenario may be less catastrophic than the one cited where the weapons are used to maximum destructive effect in a nuclear strike and retaliation. If they detonate where they are stored, that means underwater or underground in a smaller number of locations.
Unless they all go precisely simultaneously, the first to go will destroy the other warheads in its vicinity without exploding them. Also the yield of a warhead depends on a precisely timed sequence. Any departure from this will reduce the yield, quite possibly to a mere few tons ( a "fizzle").
Even in the worst case where the explosive yield is the designed maximum possible, the number of fireball locations will be far smaller than with the actual use of the weapons in anger. So there will be less consequential fires and less soot induced global cooling.
For an underwater detonation there will be no fire. Most of the energy will vaporize water which will soon condense and fall as rain. For an underground detonation a greater amount of energy will be absorbed by rock.
There is some bad news. The fallout will be terrible. An underground burst will be worse than a ground burst. (I am assuming they are not buried deep enough to contain the fireballs). Most if not all of the bombs are in the Northern hemisphere and global air circulation tends to delay mixing between hemispheres. Fallout radiation decays rapidly. New Zealand will be least worst affected?
Putting these various things together ought to allow you enough convincing handwavium to justify whatever degree of disaster your plot requires.
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$\begingroup$ Chalk up another reason to move to New Zealand $\endgroup$– KysCommented Jul 11, 2016 at 19:34
Andrei has it right, but I'm not certain about any tectonic problems. Ploughshares.Org cites that at this time there are 15,375 nuclear weapons remaining in the world, a number down by almost 60% since the peak proliferation in the 1980s. While reduced, the number remaining is far in excess of what would be needed to end life as we understand it.
Your question poses them exploding simultaneously. While this may or may not be possible, the resulting devastation would definitely include the following, as quoted from "Bulletin of the Atomic Scientists", 1984. With the 'worst case' scenario, and detonation of 16,000 weapons with approximately 10,000 total megatons of energy released, 63% of those as surface bursts, the resulting particulates, smoke, soot and airborne fallout would reduce ground level temperatures to approximately -50 C for a year or more.
The initial detonations would reduce the population by 2 to 3 billion as urban areas are hit and irradiated, burned or vaporized. The subsequent collapse of infrastructure in the following hours would increase that toll by another 1-2 billion as over the first week, untreated radiation sickness, thirst and starvation settle in with near complete darkness. The remaining populations' survival rate depend on a great many variables, chief among them, location and proximity to shielded space to protect from cold, radiation and disease.
I highly recommend "Bulletin of the Atomic Scientists", 1984, available at Amazon, it answers in scholarly, scientific depth the exact details of many such scenarios.
As you might expect, there are no detonation scenarios where life just goes on...
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$\begingroup$ but way fewer than 63% would be surface burst. Most nuclear warheads are stored in bunkers or silos or in submarines a few hundred feat underwater. Granted the result would still kill billions $\endgroup$ Commented Jul 11, 2016 at 16:41
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$\begingroup$ @sdrawkcabdear True, but you will have a hard time convincing the nuclear capable governments to give you all the information you need about their storage facilities to make for a better guess ;-) You'd certainly not get to know how many nukes are currently loaded in subs. I'd say its pretty clear the surface-detonation case is a worst-case-scenario, but its what we've got to work with. $\endgroup$ Commented Jul 11, 2016 at 17:11
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$\begingroup$ @Cort, sdrawkcabdear; you're both right; the question did specify an 'in-place' detonation, however since we don't know ( and, no they're not telling) exactly where those locations are, I went with a citation based on a potential launch scenario, most likely to nearly match the question. It's a scary read, but very informative. It only takes one nuclear weapon to ruin a lot of days, the math behind the energy of 10,000+ megatons left me speechless... Or as it was once put: "A strange game. The only winning move is not to play. How about a nice game of chess? " $\endgroup$– JoeCommented Jul 11, 2016 at 17:26
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$\begingroup$ Weapons detonating underwater or inside silos and bunkers are by definition "ground bursts", and the pulverized soil/vapourized water will become large quantities of radioactive fallout. Since most weapons storage, missile silos and patrolling submarines are well away from the population, the bulk of effects will be determined by where the wind blows fallout plumes. $\endgroup$ Commented Jul 11, 2016 at 18:46
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$\begingroup$ The real problem here is ground bursts are dirty. If you're in the northern hemisphere and lack a fallout shelter you probably die. The southern hemisphere would be pretty much untouched, though, other than by the world economic collapse. (But that very well might bring down their societies.) $\endgroup$ Commented Jul 11, 2016 at 22:01