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We read in Atomic Archive

6.08

All crater dimensions resulting from a surface burst of yield W kilotons are related approximately to those given above for a yield of 1-kiloton by the factor W03[sic].

6.10

Optimum values, like all crater dimensions, are approximately proportional to W^0.3.

6.72

The best empirical fit to crater data indicates that, for a given scaled depth of burst, i.e., actual depth divided by W^0.3, both the radius and depth vary approximately as W^0.3, where W is the weapon yield.

And so on and so forth.

But surely the gravitational energy needed to lift the material increases with the fourth power, since not only does the mass cube with dimensions, the distance it has to be lifted also increases?

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W is a measure of the energy released. As you say, $W \approx mgh \approx \rho g r^4$. This is approximate because the shape isn't a perfect cube, and some energy goes into radiation, churning the debris and such.

This means $r \approx W^{0.25}$

If $ h < r$, then it is closer to $r \approx W^{0.33}$

So it is reasonable that $r \approx W^{0.3}$ is a rough approximation.

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  • $\begingroup$ "If h<r , then it is closer to r≈W0.33" why? $\endgroup$ Commented Aug 23 at 1:56
  • $\begingroup$ I meant if h is closer to constant than linear in r, then $W \propto r^3$ $\endgroup$ Commented Aug 23 at 3:09
  • $\begingroup$ But why would h be constant? $\endgroup$ Commented Aug 23 at 3:59
  • $\begingroup$ Maybe it isn't. Maybe $W \propto h^{1/2}$. Or maybe the depth of the crater is less than you would expect. The empirical results are that $r \propto W^{0.3}$, which is equivalent to $W \propto r^{3.3}$. This is an indication that some of the energy goes into something other than lifting dirt. But I cannot say what. $\endgroup$ Commented Aug 23 at 12:27

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