How Does a Nuclear Explosion Affect Shock Wave Radius?

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SUMMARY

The discussion focuses on the mathematical relationship governing the radius of a shock wave produced by a nuclear explosion. The derived formula is r = k * E^(1/5) * ρ^(-1/5) * t^(2/5), where E represents energy, ρ is the initial air density, and t is time. Key conclusions include that the exponents a, b, and c are determined through dimensional analysis, leading to the equations a + b = 0 and 2a - 3b = 1. This analysis clarifies the absence of mass in the final expression for radius, confirming that the radius is a function of length only.

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eliassiguenza
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Homework Statement


In a nuclear explosion there is a very quick release of energy in a small region of space. This produces a spherical shock wave, with the pressure inside the shock wave being thousands of times greater than the initial air pressure.
From wave theory it is found that the only properties of the explosion and the medium that the wave travels through that may determine how the radius R of this shock wave grows with time t are:
the Energy E (in kg.m2/s2) released in the explosion the initial air density p0 (in kg/m3) the time t (in s) after the explosion
Given that k is a dimensionless constant, find the dependence of the radius of the shock wave on the initial air density, the energy of the nuclear explosion and the time after the explosion (i.e. determine a, b, and c in the
relation r = k* E ^a*ρ^b*t^c).


Homework Equations


The answer is provided, but I don't understand several things...How can some one say m^a+b? since there was two m's at the beginning to start with I thought it would be 2m^a+b .. also how did they arrive to the conclusion that a+b = 0? why! i don't get it! ='(
Please help me out I'm lost...

The Attempt at a Solution


Rr = k E^a ρ^b t^c
[L] = [mL2 t^-2]^a [mL^-3]^b[T]^c
[L] = m^a [L^2]^a [T^-2]^a m^b [L ^-3]^b T^c
[L] = m^ a+b [L^ 2]^a-3b T c^-2a
so 2a-3b = 1
and a + b = 0
so a = -b
and c – 2a = 0
Then -2b – 3b = 1 or -5n = 1
and b = -1/5
(1) and a = 1/5
(1) c- 2/5 = 0 so c = 2/5
(1) so r = k*E1/5*ρ-1/50*t2/5
 
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eliassiguenza said:
The answer is provided, but I don't understand several things...How can some one say m^a+b? since there was two m's at the beginning to start with I thought it would be 2m^a+b .. also how did they arrive to the conclusion that a+b = 0? why! i don't get it! ='(
Please help me out I'm lost...

In unit analysis, mundane constants are ignored. So while energy might be given by m*v2/2 in "real life", in terms of units it's [M][L]2[T]-2.

a+b = 0 because the result has no mass term; radius is pure length [L].

P.S. You might find it handy to use the X2 menu button in the header of the reply or posting window in order to produce nice-looking exponents for your equations.
 

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