Suppose there is a radioactive source on the ground, say a rod of cobalt 60. Since it’s a point source, the intensity of the radiation drops with the inverse square of the distance, and you could fly at high altitude (several miles up) over it safely.
But what if a large area had been recently hit with many ICBMs and thus the whole area was lethally radioactive, were you in line of sight within a reasonable distance of any patch of ground.
If you fly higher, that just means there are more rays coming from the ground surface that intersect with you. So the intensity does *not *drop with inverse square, and you get fried.
Where this comes up is in B-52 bombing missions. If the B-52s are ordered to attack a target that has already been hit with dozens of ICBM warheads, and they have to fly overhead to drop their bombs (maybe they are fresh out of air launched cruise missiles), wouldn’t the crew be killed by the radiation?
The radiation you’re describing can be considered as a whole bunch of point sources. The B-52s are receiving radiation from each, proportional to the inverse square of the distance.
It’s obviously still going to be true that flying higher reduces the received radiation energy.
The basic idea is correct. If the area you are flying over is large in comparison with your flight height, the intensity of radiation is constant, no matter what your flight height. There is no inverse square.
Indeed this effect is important and often overlooked.
Inverse square only works for a source with limited extent compared to your distance from it. So a small number of point sources, or sources that you are further away from than their extent. If you are close to an extended source (which can be large number of point sources) it works differently. If the extended source is a line source, the intensity falls off linearly with distance. And if the source is an area, it doesn’t fall off at all.
Eventually of course, all real life sources have a finite extent, and the simple rule fails.
I like to point out two real world examples. Road noise from busy highways only falls off linearly with distance, and rather like radiation, atmospheric absorption of noise is an important part of how the intensity falls with distance.
A fire front in a wild fire. Forest fires (or bushfires here in Oz) can form a vertical wall of fire and the heat intensity within a few hundred feet of the fire front is for all useful intents identical to that found in the fire itself. Hence the fire can propagate very quickly by radiative heating alone, and a person caught in the open can die very quickly even though the fire front itself has not reached them. Houses can catch fire simply from the radiative heat coming in through windows and setting furnishings alight.
So, for the OP, of the extend of contaminated land was say 3km on a side, and you were within say 1km of the ground, you would be within the near field, and indeed, absent any atmospheric adsorption, the intensity would be identical to that found at ground level. However, as beowulf notes, the atmosphere does absorb radiation. At 1km the air will have adsorbed a very significant amount of the radiation.
Except that you’re not actually worried about particles of radiation hitting you. You’re worried about specks of dust containing radioactive material drifting up and getting into your lungs. The most dangerous radioactive materials are the alpha sources (or to a lesser extent, the beta): They’re harmless if they’re outside of you, but do serious damage if they get inside you. Unless you have some seriously weird weather patterns, then, it won’t be long after the blasts (weeks, say) that a high-altitude plane would be perfectly safe.
I’d be most concerned about dust at my altitude and relatively unconcerned about direct radiation emitted from radionuclides sitting on the ground.
FV’s math is 100% valid. But at “high altitude” per the OP an aircraft’s line of sight to the ground is on the order of 150 miles in diameter = 17,000 square miles. The seriously hot region at any single ground burst is a few square miles tops. Unless you’re overlying a moonscape that absorbed literally multiple thousands of ICBMs in a neat grid pattern you’re not going to experience the complete collapse of the inverse square law. You won’t experience much of a collapse at all until/unless you’re right down close at low altitude.
Ground-based alpha particles are zero threat. They won’t penetrate more than a foot of atmosphere. Good bet you’re flying higher than that. Ground based beta and gamma emitters are a different matter. I struggled to find a reference for atmospheric beta attenuation, but my vague recollection is ground based beta sources, and the consequential bremsstrahlung, will be pretty thoroughly attenuated from high altitude.
Especially when fresh. Naturally, emission intensity and half-life are inverses of one another. The stuff really spewing out radiation burns itself out quickly in a matter of minutes or hours. The stuff trickling out radiation occasionally, even highly penetrating gamma rays, remains an energetic emitter for centuries.
It doesn’t matter if the alpha and beta will be stopped by the atmosphere, because they’ll certainly be stopped by the skin of your airplane.
And the atmosphere isn’t all that transparent to gamma rays, either. There’s a reason that all gamma-ray astronomy is done from satellites and sounding rockets.
There was a real-life analog to what you describe. In the wake of the Chelyabinsk (Khyshtym) explosion in the late 1950s and continuing into the 1960s, the region was heavily radioactive, and the authorities told you to drive through it rapidly in your car with the windows rolled up. (They apparently didn’t tell you to detour around it – life in the USSR at its best)
The contamination was a matter of public knowledge in the Soviet Union, but was not talked about and even publicly denied in the west until Zhores Medvedev wrote his book Nuclear Disaster in the Urals.
Ignoring the complicating issues such as radioactive rays from the ground versus radioactive particles in the air versus atmospheric attenuation and so on the basic problem of radiation versus geometry is one encountered in radiative heat transfer. The specific relevant concept is view factor, a mathematical term that describes, with a single number derived from a surface’s geometry and your distance to it, how much of the radiation emitted from that surface is intercepted by you. Any heat transfer textbook will have an appendix with a long list of arbitrary geometries and their associated view factors.
I think this page describes the view factor you’re looking for, with A1 as the radioactive patch of ground, A2 as the aircraft being irradiated (A2 << A1), and c as the cruising altitude. I don’t feel like plugging all those variables into Excel just now - if you’re bored, you can do so - but in looking at the picture, it seems obvious that if the radioactive zone A1 is finite, then a higher cruising altitude c is better. If the radioactive zone A1 is infinite (or just very large, such that a >> c), then your cruising altitude c doesn’t matter.
Note that the portion of radiation emitted from the ground that is intercepted by the plane is the product of the view factor and the size of A1 (since total radiation from A1 scales with A1).
What are some realistic values for A1 and c?
Cruising altitude (c) appears to be on the order of 50,000 feet, so say c = 10 miles.
What about A1? You’ve posited a “large area recently hit with many ICBMs.” I’m not a nuclear warrior, so I can’t speak with authority on this, but AIUI, you don’t carpet-bomb the entire countryside with nukes - you target specific entities, such as cities or missile silo complexes. I would guess a city that’s been hit with 2-4 warheads might be 10x10 miles, and maybe there’s a fallout zone that extends downwind for another ten miles. Under these conditions it seems clear that altitude matters (and probably also your approach path, and time over target).
Add in the complicating factors of atmospheric attenuation as described upthread, and it seems likely to me that the bomber’s aircrew would not be killed by radiation.
What is different is the crew will have oxygen masks on during at least the tactical part of the mission. Which will be feeding them ambient cabin air until/unless a pressurization problem occurs.
I don’t know what degree of pressurization a, say, B-52 has. It may well be less potent than an airliner. With the effect that up near the max altitude of the bomber the masks are required and the mask is feeding a mix of bottled pure oxygen in with the cabin air. That was certainly how it worked in fighters.
A difference is a bomber may be at high altitude for many hours and be supporting a crew of 6-ish people. That’s going to need a lot of bottled oxygen. So they may indeed be pressurized enough that at even very high altitude cruise no canned O2 is consumed.
Later model bombers aircraft (e.g. B-2) probably carry LOX instead of scuba tanks. That’s also current fighter technology.
There was/is(?) also an experimental system called OBOGS that’s been tried on various aircraft to extract and concentrate breathing oxygen out of the exterior atmosphere. Depending on how that tech works exactly it might have … shortcomings in a fallout situation vs. carrying pristine oxygen from home.