Terminal velocity of a rocket using dimensional analysis

AI Thread Summary
The discussion revolves around using dimensional analysis to derive the terminal velocity formula for a rocket, incorporating variables such as air density, gravity, rocket area, mass, and drag coefficient. The initial attempts at dimensional analysis led to confusion regarding the role of mass, as the derived equations seemed to imply it had no effect. Clarifications were provided, indicating that the constant of proportionality, often determined experimentally, is essential for establishing the correct formula, which is v = √(mg/(1/2)ρC_dA). The conversation also touched on the refinement of dimensional analysis by considering directed dimensions, ultimately leading to a consistent expression for terminal velocity. Understanding these concepts is crucial for accurately applying dimensional analysis in physics.
Enthu
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I started taking MIT's online physics course a couple of days ago, and since I have no physics background at all, I'm getting a bit confused with dimensional analysis.
I'm trying to find a formula for the terminal velocity of a rocket, using air density ρ, gravity g, area of rocket affected by air A, mass m and the drag coefficient Cd. I have reached the following equation of dimensional analysis:
LT-1 = [M]V\cdot[LT-2]W\cdot[ML-3]X\cdot[L2]Y

I'm pretty sure that is completely wrong. I tried solving for the coefficients and ended up with V = 1 X = 1/2 W = 1 Y = 1/2, and I'm certain that is wrong as well.

Can somebody help me with this? I just can't seem to get a correct formula.
Also, as far as I understood it, from the equation above it is implied that mass should have no effect, but the actual formula for the terminal velocity of a rocket does have mass in it.

--this isn't a homework/coursework question, I'm doing this out of my own curiosity--
 
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Well, you have 4 variables and just 3 equations. Possible fix: Use the gravitational force of the rocket, m*g, as single variable.

Anyway:
Solving for M gives V=-X (note the sign).
Solving for T gives W=1/2
The equation for L is now 1=W-3X+2Y, which can be simplified to 1/2 = -3X+2Y.
If m and g can appear as m*g only, this implies V=W and therefore V=1/2, X=-1/2. This allows to determine Y=-1/2.

As final formula:
v \propto \sqrt{\frac{mg}{\rho A}}
 
Ah, thanks. I get it now. Any idea how the actual formula got to v = \sqrt{\frac{mg}{\frac{1}{2}ρC_dA}} (i.e. where's that 0.5 from)?
 
Enthu said:
Ah, thanks. I get it now. Any idea how the actual formula got to v = \sqrt{\frac{mg}{\frac{1}{2}ρC_dA}} (i.e. where's that 0.5 from)?

That is the constant of proportionality.

If, for example,

v \propto \sqrt{\frac{mg}{C_d\rho A}}

then to introduce an equality sign, we have to introduce a constant,

v = k \sqrt{\frac{mg}{C_d\rho A}}

Usually, this constant is evaluated by experiments. In this particular case k came out to be \sqrt{2} You cannot get it by dimensional analysis because the number has no dimension by itself.
 
Infinitum said:
That is the constant of proportionality.

If, for example,

v \propto \sqrt{\frac{mg}{C_d\rho A}}

then to introduce an equality sign, we have to introduce a constant,

v = k \sqrt{\frac{mg}{C_d\rho A}}

Usually, this constant is evaluated by experiments. In this particular case k came out to be \sqrt{2} You cannot get it by dimensional analysis because the number has no dimension by itself.

Oh, that's something I must have missed going through my papers. Thanks.
 
I know this is not a "standard" analysis, but you can say that the x, y, and z distances have different dimensions. In other words, you have 5 fundamental variables m, t, Lx, Ly, and Lz. Then, assuming the z direction is the direction of motion (up):
v\propto L_z/t
\rho\propto\frac{m}{L_x L_y L_z}
g\propto L_z/t^2
A\propto L_x L_y
and you get
\frac{L_z}{t}\propto\left(m\right)^a\left(\frac{L_z}{t^2}\right)^b\left(\frac{m}{L_xL_yL_z}\right)^c\left(L_xL_y\right)^d

You can solve to get a=b=1/2 and c=d=-1/2, and get v\propto\sqrt{\frac{m g}{A \rho}} same as above. Check out http://en.wikipedia.org/wiki/Dimensional_analysis#Huntley.27s_extension:_directed_dimensions for an explanation of this refinement of dimensional analysis

(Note - the area is Lx Ly because the plane of the area is perpendicular to the direction of motion)
 
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