How does overpressure convert radiation to something else?
I would have thought the radiation would be same.
But definitely it'll have trouble cooling down in vacuum
Other way around. The high-energy EM radiation emitted by the nuclear reaction is slowed and absorbed by the surrounding atmosphere, heating it and, through Boyle's Law, creating the massive overpressure that generates the blast effects in an atmospheric nuclear detonation. In a vacuum, there's nothing to absorb that radiation, so it doesn't get converted into a fireball and blast wave, except by the relatively negligible mass of the bomb case, internal components, and unreacted nuclear materials (only a small percentage of the fission and/or fusion fuel actually is involved in the chain reaction before the bomb blows itself apart too much for the reaction to continue).
This is why it's easy to differentiate an atmospheric nuclear burst from any other sort of massive explosion on Earth; the nuclear burst has a distinctive "double-flash" signature, where there's an initial burst of
violently intense visible light from the ongoing nuclear reaction, then a drop-off in light levels followed by them rising again to a second peak (significantly lower than the first) as the fireball forms and rises to its peak temperature through mechanical compression of the shock front.
In a case like this, of nukes going off inside some sort of structure on the lunar surface... well, since it's unlikely that the five devices would have initiated
exactly simultaneously, the most likely result I see is as follows: The first device goes off inside the base, where it
initially acts like an atmospheric burst, with the radiation release heating and compressing the air inside it, but that's a very limited amount and it's not enough to fully absorb the radiation pulse like the massive volume of Earth's atmosphere can, so much of the radiation continues on unabated. Over the course of a few dozen nanoseconds (one light-nanosecond is near enough to one foot/30 centimeters for these purposes), the radiation will have fully heated and compressed all the air in the base, and have also massively heated all the solid and liquid materials in it, as well. Within 100 nanoseconds, Rita and her minions will all be dead. To paraphrase Randall Munroe, they won't have died
of anything, per se, they'll simply have stopped being biology and started being physics. Meanwhile, the metal and/or stone structure of the base itself is starting to rapidly sublimate away, but on a time scale a few orders of magnitude slower than the air and liquids. Likewise, the other nukes will have their bomb cases starting to sublimate, but given the thermal resilience and neutron reflectivity of the materials used for nuclear bomb cases, that probably won't be an issue for them before their own timers reach zero. The high-energy radiation pulse, having passed through the outer walls of Casa Rita, now splits into two different cases. Any of the photons (since that's what they are) on trajectories that take them above the local horizon will now expand unimpeded into space, dispersing according to the square-cube law as they do; depending on the relative positions of sun and base, and if there's a solar storm ongoing, auroral effects may be seen near the zero point. The photons heading
below the local horizon, meanwhile, will pass into the lunar regolith (i.e., surface) and start heating that, resulting in a thermal pulse through the upper layers that sees much of the nearby regolith begin to sublimate into vapor and generate a pressure wave that starts excavating a crater. (Please note that I am basing this on boosted fission weapons; you can probably double the timeframes for fission-fusion-fission devices known as "hydrogen bombs," due to the extra steps.)
Within the first microsecond or so, the nuclear reactions within the first bomb to go off have run their course, as the vaporized and superheated nuclear assembly blows itself apart and drops the radiation levels below prompt criticality. Within a few more microseconds, the pressure effects of the heated air inside the base will be enough to rupture the exterior and allow a very tiny fireball to form as the plasma that was the contents of the base rapidly begins to expand into the surrounding vacuum, the superheated solid material vaporizing and continuing to expand it briefly before the energy levels drop low enough that the plasma recondenses into gas and drops to a low enough temperature that blackbody radiation is no longer in the visible spectrum.
Around this time, another of the weapons will likely go off. At this point, with little but very thin plasma/vapor surrounding it above the horizon, the upper half of the emitted radiation will reheat the plasma (making it light up again, but not doing much else except boosting its dispersal speed) and then shoot off into space to interact with, at most, the solar wind; the lower half will first heat the vaporized regolith, boosting its pressure (and thus expansion speed), and then add another pulse of energy to the regolith, sublimating more of it and making the crater bigger. You can then repeat this step three more times for the remaining weapons, which means that in under a second, all aspects of what could be seen as a fireball would have cooled and recondensed into particulates that will fall out over the next several hours; with the lack of wind on the lunar surface, these will only travel as far as their horizontal velocity will allow before impact, likely limiting the significant fallout area to less than a kilometer around the zero point. The primary radioactive fallout will be, as on Earth, actual bomb debris, rather than regolith transmuted into radioisotopes by the radiation flux, but I'm not familiar enough with the radiochemistry of lunar surface material to say exactly what would happen.
So within a few minutes, we'd end up with a red-hot glass-walled crater and some radioactive dust in the area, as described. We would also have diplomatic phone lines burning up worldwide as everyone frantically tried to figure out who the hell just violated the Outer Space Treaty and the Moon Treaty by setting off nuclear weapons there, and how in hell they
got them there without anyone noticing (no matter how classified, you
can't hide a space launch with current technology).
I'm not sure how long it would take for the crater's glass to cool to the point of solidifying (it might well still be liquid a few minutes after the bursts), much less cool down to the point of no longer glowing with blackbody radiation, but I'd say that OL's/Billy's estimate of it being safe for someone in an ordinary space suit to walk through within five years is reasonable; remember, it's the
immediate radiation levels that are elevated in an exoatmospheric burst compared to an atmospheric one. Once the nuclear reactions are over, that radiation pulse is done and expanding at the speed of light, leaving the area, never to return. Lasting radiation is in the form of transmuted radioisotopes found in the crater itself, and the fallout that comes primarily as bomb debris that's now mixed with other materials kicked up by the mechanical actions from heating the surrounding matter, and those isotopes tend to have half-lives measured in days or hours; within a few years, there will have been enough halving cycles to lower radiation levels to well below what EVA suits are designed to protect against.
As for
why someone would want to explore that crater? Well, the world's nuclear powers do have only a limited data set regarding the effects of nuclear weapons, and the more data they can harvest, the better the models they can develop from them to allow for
simulations to better predict weapons performance and effects. This is
particularly true of exoatmospheric nukes; the first exoatmospheric tests were conducted in 1958, and there was a limited window of time before a testing moratorium went into effect to conduct further tests. Even after the moratorium broke down in 1962, it was only another year before the Limited Test Ban Treaty prohibited further exoatmospheric testing, so there's
very limited data there... and
no data regarding how an exoatmospheric burst on a planetary surface would behave (plus nobody's ever experimented with what the effects of a nuke on lunar regolith are, given how small a sample we have of it on Earth!). So all of the nuclear powers' defense hierarchies would consider examining the results of this to be of
vital importance... and honestly, the
civilian physicists would really want to get a look at it, too. The geologists might also be interested, since there's a chance that they could get a much better look at the strata of the lunar surface than they got from core samples taken on Apollo missions (if nothing else, this would definitely go deeper into the surface!).