What if you dropped on in the middle of Antarctica or Greenland. How much of the ice would melt?
How big a bomb? Airburst or ground burst? If airburst, at what altitude? If groundbusrt, over what thickness of ice? Summer or winter? Under what prevailing weather conditions? Need more specific question…
That said: Not nearly as much melting as you might think… The blast would leave quite a big footprint, but both those places are really cold, and anything that melted but didn’t vaporize outright would freeze again in fairly short order.
A 100 megaton device (really, really big) releases 4 x 10[sup]17[/sup] Joules of energy.
The average temperature at the South Pole is -60 [sup]o[/sup]C.
To boil a kilogram of water from -60 takes:
60 x 2.1 kJ = 126 kJ (to heat to 0 [sup]o[/sup]C), plus
333 kJ (to melt), plus
100 x 4.2 kJ = 420 kJ (to heat to 100 [sup]o[/sup]C), plus
2500 kJ (to boil)
Total 3.4 MJ/kg.
4 x 10[sup]17[/sup] J / 3.4 x 10[sup]6[/sup] J/kg = 120 x 10[sup]9[/sup] kg = 120 MT.
And so it works out that a 100 MT bomb can vaporise just over 100 MT of water.
Assuming, of course, that all that energy gets transfered directly to the ice… No reflecting, none lost into the atmosphere, etc.
Good thing we don’t have any 100MT bombs in inventory then, yes?
I was assuming undergroung detonation.
BTW, what difference does it make that the U.S. never made a 100 MT bomb? The Russians certainly did.
Anyway, a better answer to the original question would be to consider what happens after the steam re-condenses and cools back to 0 [sup]o[/sup]C, whilst melting further surrounding ice.
The maximum quantity of liquid water that could be formed is:
4 x 10[sup]17[/sup] J / ((126 + 333) kJ/kg x 1000 J/kJ) = 870 x 10[sup]9[/sup] kg = 870 MT
Of course, you couldn’t really expect such even temperature distribution, so the actual answer is “less than 870 MT of water using a 100 MT device”.
To put the 870 MT of water into perspective, it’s almost 1 km[sup]3[/sup]. Lake Mead, behind the Hoover Dam, holds 35 km[sup]3[/sup].
If you had this specific info could you really give a factual answer?
Somewhat related to this post: I suggest reading “The Firecracker Boys”, by Dan O’Neill. This is the true recounting of the AEC’s plans to detonate several nuclear devices in Alaska in order to create a harbor. Fairly chilling, to say the least.
Or a fairly close approximation; observe Desmostylus rather excellent posts. Mind you, there’s a lot more to it in the real world, but Demo’s math, given his assumptions, is flawless.
I’d have chosen a much more likely weapon size, say one to two megatons, and I’d be more inclined to assume complete penetration of the ice layers (unless we’re talking about really thick ice shets!), and thus greater ‘wasted’ energy. Call Demo’s case an upper limit, using a 1MT weapon, using average weapons energy distribution, and borrowing his second equation:
4 x 10[sup]15[/sup] J / ((126 + 333) kJ/kg x 1000 J/kJ) = 870 x 10[sup]9[/sup] kg x .35 (thermal pulse fraction of total bomb energy) x .5 (surface burst, assumed sperical energy distribution, ice not compeletely penetrated) = 1.52 MT of melted ice.
Mind you, this too is probably over-large just as Demo noted, as I’ve assumed that the thermal pulse is evenly distributed and that all thermal energy is absorbed completely with maximum efficiency, none being wasted in superheating already vaporized ice, for instance.
Slipped an exponent up there: “870 x 10[sup]7[/sup] kg”
thanks for coming back, I just have no idea if people can figure this out or its just an opinion
Well, to a large degree, lacking emperical evidence (no one’s done this), call it a semi-scientific opinion.
Figure it out with perfect accuracy? Nope. Can it be figured with better accuracy. Yes. You just need a supercomputer (some of the fastest computers in existence are used to model nuclear explosions in great detail so they don’t need to go out and actually blow them up anymore).
What Tranq and Demo gave you were upper limits of ice turned to water with a given bomb size. You can take those numbers to the bank. In short, you won’t get more liquid water than what they stated. However, the ‘real’ number is likely to be a good deal less than what they said as the perfect thermal distribution, perfect thermal absorbtion, etc., just won’t happen in real life. If you want a more specific answer that includes these and a ton of other variables you will need some pretty sophisticated modelling equipment. The math involved is going to be very hairy and the amount of math you will need to do will require that supercomputer if you hope to get an answer in your lifetime.
Their numbers are still useful though if you are wondering how much damage a nuke could cause by melting ice as you know it can’t be worse than what they said.
For comparison, would anyone happen to have the tonnage of ice in the Antarctica and Greenland icecaps? I couldn’t turn it up quckly. I suspect that any reasonable A-weapon detonation would barely make a dent.
I don’t have a definitive answer but if someone wants to do the math:
Mean thickness of Antarctic Ice Sheet: 2.16 kilometers
Land size (I don’t think this includes ocean coverage): 13 million square kilometers (twice the size of Australia)
The Antartic Ice Sheet is estimated to hold 80% of the world’s fresh water.
Whether the calculations are done or not it is plain to see the ice cap there is freakin HUGE. It would take a LOT of nukes to melt it all and one truly would be an insignificant dent.
Not sure about Greenland…
http://pumas.jpl.nasa.gov/PDF_Examples/02_10_97_1.pdf (warning, .PDF)
Those numbers should be rounded, of course:
Greenland - 2.6 million km[sup]3[/sup]
Antarctica - 29 million km[sup]3[/sup]
Using Desmostylus’s figure, I get 2.88 x 10[sup]16[/sup] tons of water in the Antarctic ice sheet (after allowing for the coume difference between liquid water and water ice).