Dimensional analysis - atomic bomb explosion radius

[vector]

Homework Statement

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An atomic explosion can be approximated as the release of a large amount of energy $E$ from a point source. The explosion results in an expanding spherical fireball bounded by powerful shock wave. Let $R$ be the radius of the shock wave and assume that $R=f(E,T,\rho_0, p_0)$ where $t$ is the elapsed time after the explosion takes place, $\rho_0$ is the ambient air density and $p_0$ is the ambient air pressure. Using dimensional analysis show that $R=\frac{Et^2}{\rho_0}g(\pi_1)$ where $\pi_1$ is a dimensionless variable. Choose it so that it involves $E$ to a negative exponent. What is $\pi_1$?

Homework Equations

$R=f(E,T,\rho_0, p_0)$
$R=\frac{Et^2}{\rho_0}g(\pi_1)$ where $\pi_1$ is a dimensionless variable

The Attempt at a Solution

The problem is that the relationship does not appear to be homogeneous. If $\pi_1$ is dimensionless, then the units of $\frac{Et^2}{\rho_0}$ must be the same as the units of $R$. However,

$[E] = ML^2 T^{-2}$
$[t] = T$
$[\rho_0] = ML^{-3}$
$[p_0] = ML^{-1}T^{-2}$
$[R]=L$, where M is express in kilograms, T is in seconds, L is in meters.

What is it that I'm not getting right here?

On the other hand, I expressed $\pi_1$ as follows: $\pi_1 = \frac{p_0 R^3}{E}$. Is this correct?

Thank you!

 

[majid313mirzae]

The Attempt at a Solution

The problem is that the relationship does not appear to be homogeneous. If $\pi_1$ is dimensionless, then the units of $\frac{Et^2}{\rho_0}$ must be the same as the units of $R$. However,

$[E] = ML^2 T^{-2}$
$[t] = T$
$[\rho_0] = ML^{-3}$
$[p_0] = ML^{-1}T^{-2}$
$[R]=L$, where M is express in kilograms, T is in seconds, L is in meters.

What is it that I'm not getting right here?

On the other hand, I expressed $\pi_1$ as follows: $\pi_1 = \frac{p_0 R^3}{E}$. Is this correct?

Thank you!

yes, it's correct. The dimension $p_0 R^3$ is same for E.
$ML^{-1}T^{-2}$ * $L^{3}$ = $ML^2T^{-2}$

 

[vector]

yes, it's correct. The dimension $p_0 R^3$ is same for E.
$ML^{-1}T^{-2}$ * $L^{3}$ = $ML^2T^{-2}$

Thanks. But do you think the expression given in the problem statement is correct at all? The problem is that $R$ has units of $L$, but $\frac{Et^2}{\rho_0}$ has units of $L^{-5}$.

 

[majid313mirzae]

Thanks. But do you think the expression given in the problem statement is correct at all? The problem is that $R$ has units of $L$, but $\frac{Et^2}{\rho_0}$ has units of $L^{-5}$.

unit of $Et^2$ is $ML^2T^{-2} * T^2$ then $ML^2$
so unit of $R$ is $\frac{ML^2}{ML{-3}}$ = $L^5$

 

[vector]

Sorry, I meant $L^5$. So then the expression is not quite homogeneous. Namely, the units of $R$ are $L$, but the units of $\frac{ET^2}{\rho_0}$ are $L^5$. Am I correct?

 

[majid313mirzae]

yes , you are. Below file can be useful
http://dspace.mit.edu/bitstream/handle/1721.1/42045/228875559.pdf?sequence=1

 

[vector]

Thanks. So, the $\frac{ET^2}{\rho_0}$ should be raised to the power of 1/5, shouldn't it?

 

[majid313mirzae]

Thanks. So, the $\frac{ET^2}{\rho_0}$ should be raised to the power of 1/5, shouldn't it?

yes, it's right