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Terminal velocity of a rocket using dimensional analysis

  1. Jun 1, 2012 #1
    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[itex]\cdot[/itex][LT-2]W[itex]\cdot[/itex][ML-3]X[itex]\cdot[/itex][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--
     
  2. jcsd
  3. Jun 1, 2012 #2

    mfb

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    Staff: Mentor

    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:
    [tex]v \propto \sqrt{\frac{mg}{\rho A}}[/tex]
     
  4. Jun 1, 2012 #3
    Ah, thanks. I get it now. Any idea how the actual formula got to [itex]v = \sqrt{\frac{mg}{\frac{1}{2}ρC_dA}}[/itex] (i.e. where's that 0.5 from)?
     
  5. Jun 1, 2012 #4
    That is the constant of proportionality.

    If, for example,

    [tex]v \propto \sqrt{\frac{mg}{C_d\rho A}}[/tex]

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

    [tex]v = k \sqrt{\frac{mg}{C_d\rho A}}[/tex]

    Usually, this constant is evaluated by experiments. In this particular case k came out to be [itex]\sqrt{2}[/itex] You cannot get it by dimensional analysis because the number has no dimension by itself.
     
  6. Jun 2, 2012 #5
    Oh, that's something I must have missed going through my papers. Thanks.
     
  7. Jun 2, 2012 #6

    Rap

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    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):
    [tex]v\propto L_z/t[/tex]
    [tex]\rho\propto\frac{m}{L_x L_y L_z}[/tex]
    [tex]g\propto L_z/t^2[/tex]
    [tex]A\propto L_x L_y[/tex]
    and you get
    [tex]\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[/tex]

    You can solve to get a=b=1/2 and c=d=-1/2, and get [tex]v\propto\sqrt{\frac{m g}{A \rho}}[/tex] 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)
     
    Last edited: Jun 2, 2012
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