1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    2016 Award

    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

    User Avatar

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Terminal velocity of a rocket using dimensional analysis
  1. Terminal velocity (Replies: 4)

Loading...