How Does a Nuclear Explosion Affect Shock Wave Radius?

AI Thread Summary
A nuclear explosion generates a spherical shock wave characterized by a rapid energy release, leading to a pressure significantly higher than the surrounding air. The shock wave's radius, R, is influenced by the explosion's energy (E), the initial air density (ρ0), and the time (t) post-explosion, with a relationship expressed as r = k * E^a * ρ^b * t^c. Through dimensional analysis, it is determined that the exponents must satisfy specific equations, leading to the conclusions that a + b = 0 and c = 2/5. The confusion arises from the interpretation of mass terms and the simplification of units, which ultimately shows that the radius is a function of energy, density, and time without a mass component. Understanding these relationships is crucial for accurately modeling the shock wave's behavior.
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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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