A few sequential questions:

  • Is it possible for a nuclear explosion to be small enough to produce a 250-ml (one cup) mushroom cloud?
  • If so, how much uranium would that take?
  • How close to the explosion could one be (in normal clothing) not to suffer from burns or excessive radiation exposure?
  • 8
    $\begingroup$ Also: mushroom clouds happen because of high temperatures in a big explosion. They're not specific to nuclear explosions. $\endgroup$
    – Dan
    Aug 9, 2013 at 6:09

3 Answers 3


Mushroom clouds are formed in explosions (not necessarily nuclear - see picture) as a result of the Rayleigh-Taylor instability. For given density contrast $\frac{\rho_{cold}-\rho_{hot}}{\rho_{cold}+\rho_{hot}}$ between the hot cloud and the cold atmosphere, the timescale $t_{RT}$ for this instability scales with the length scale $L_{RT}$ according to:

$$t_{RT} \approx \sqrt{\frac{L_{RT}}{g}\frac{\rho_{cold}+\rho_{hot}}{\rho_{cold}-\rho_{hot}}}$$

With $g$ the gravitational acceleration. For $\frac{\rho_{cold}-\rho_{hot}}{\rho_{cold}+\rho_{hot}}\approx 0.1$ and $L_{RT} = 1 km$ we find $t_{RT} \approx 30 s$. For a cup sized ($L_{RT} = 0.05 m$) explosion with the same density contrast we find $t_{RT} \approx 0.2 s$.

The 'mushroom' has disappeared before it really takes shape. Bottom line is that you need fairly large explosions to observe a mushroom.

enter image description here

  • $\begingroup$ ... or you need a very fast explosion that's observed at high time resolution! $\endgroup$ Sep 4, 2013 at 11:44

Let's try to answer your question in terms of fluid mechanics. A mushroom cloud is quite interesting, because of the very large scale phenomena, down to the vary small scales, which are called Kolmogorov microscales.

The mushroom cloud is quite complicated, as there are probably supersonic velocities, buoyancy effects and of course combustion playing important roles. Lets ignore all these effects, and assume ideal isotropic turbulence as a first order approximation.

The large scale we call $L$, and the small scale, the Kolmogorov scale $\eta$. Then we are interested in the fraction of these scales, as this is determines whether you'll qualify it as a mushroom or not.

$\eta$ can be approximated by


Here, $\nu$ is the kinematic viscosity and $\epsilon$ the dissipation rate of turbulent kinetic energy. Under our assumptions, we can take the dissipation rate at the small scale equal to the large scale production, thus


Here, $U$ is the large scale velocity. And thus

$$\frac{L}{\eta}=\frac{U^3 L^3}{\nu^3 }^{\frac{1}{4}}=Re^{\frac{3}{4}},$$

where $Re=\frac{UL}{\nu}$ is the Reynolds number.

For the large mushroom, with assuming $U=100 m/s$, $L=1km$ and $\nu=10^-5 m^2/s$ (air viscosity), I would get $L/\eta=3\times 10^7$, while for a small mushroom, with $L=0.1m$, and probably a lower velocity, $U=1m/s$, the range of scales would be $L/\eta=10^3$, which would look completely different. You could also argue the other way around: What velocity $U$ do I need to get the same scale separation? This means Reynolds similarity and you would end up with $U=10^6 m/s$ in the small cup! In all cases it would be gone instantly.


In addition it is worth to know that to make a nuclear explosion, you need enough fissile material. Sufficient mass is called "the critical mass".


As you can see here, you need at least about 15 kg of uranium. This costs a lot. Moreover you would need plenty of gear to initiate explosion (because you firstly have to assemble everything and then, using explosives rise the temperature to lower the critical mass parameter).

Such nuclear device would be bigger than explosion itself.


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