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Thread: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

  1. #1
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    Question Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Here is the context of the problem:
    Calculate the size of the iron meteor that would have its initial velocity reduced by only 10% by
    drag in Earthís atmosphere.

    I was able to prove the first part:
    Write down a differential equation for the deceleration of the meteor du/dt.
    du/dt=3*Cd/(4Rmpm)*u2 p0*exp(−z/h)

    However have no idea about the next:

    You should find an equation of the form du/dt = . . ., where the stuff on the RHS depends
    only on the density of air (which depends on height in the atmosphere z), the
    internal density of the meteor, the radius of the meteor, the meteorís velocity, and dimensionless
    constants. You can assume (check, if you like) that the extra acceleration
    provided by gravity over this distance is negligible, since the meteorís velocity is already
    terrifyingly large.
    Convert your equation to a linear, first-order differential equation. That is, you should
    have an equation which reads something like dy/dz = f (z)y, where f (z) is some function
    of z. (Hint: begin by noting that dt = dz/v, and substitute y = v2.)
    Solve this equation, imposing boundary conditions as appropriate. (You will probably
    end up using an integrating factor.)
    Hence, calculate the critical radius and mass of the meteorite such that the velocity at
    z = 0 is 0.9 times the initial velocity.

    Please Help
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Not sure what hep you need here. Your equation looks fine. You have a DE in this form, and combining a bunch of constants into a single constant K = \frac {3 C_D}{4 \rho_m} :

     \frac {du}{dt} = \frac K R u^2 e ^{-z/h}

    Now substitute y = u^2, and dy = 2u du; the left hand side becomes:

    \frac {dy}{2u dt}

    Since u = dz/dt, this is further simplified:

    \frac {dy \ dt}{2dz\ dt}  = \frac 1 2 \frac {dy}{dz}

    So now you have your DE:

    \frac {dy}{dz} = \frac {2K}{R} y e^{-z/h}

    Can you take it from here?
    Last edited by ChipB; Dec 13th 2016 at 11:07 AM.
    Thanks from HallsofIvy and lac220
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Thanks for your help but do you mind integrating that for me. It's been ages since I last had to do this and I've completed forgotten how to solve this type of integral.
    Would be a great help for my exam.
    Cheers
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    I'll get you started on the integral: rearrange the equation I gacve you to get:

     \frac {dy}y = \frac {2K} R e^ {-z/h} dz

    and integrate:

     \int \frac {dy}y = \frac {2K} R \int e^ {-z/h} dz

    Do you recognize the integral on the left? It has to do with the natural logarithm. And on the right recall that \int  e^{ax} dx = \frac 1 a e^{ax}. Don't forget you will need to include a consant of integration. Show us your attempt from here and we'll help if you get stuck.
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz-codecogseqn.gif

    then

    Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz-codecogseqn2.gif

    Is this right or have I made a mistake? (K=2K^)

    What boundary conditions could I use to remove the constant?
    Last edited by lac220; Dec 14th 2016 at 05:11 AM.
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  6. #6
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Looks fine, though I would have written it as

     v^2 = Ce^{\frac {-hK}{R_m}(e^{-z/h})}

    You can determine C if you know a boundary condition, such as initial velocity at a certain height z. Did they define the value 'h' for you? You could set the boundary condition of v_0 = 10000m/s at z = h = 100,000 m, and then determine the value for C.

    Also please note that the value for the constant K that I used should have included the  \rho_0 term:

     K = \frac {3 C_d \rho_0}{4 \rho_m}
    Last edited by ChipB; Dec 14th 2016 at 06:24 AM.
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Yes they defined z=h=80,000m and v_0=30,000m/s.

    This gets me a value of C in terms of Rm, however when I try and solve for Rm I'm unable to separate it out.

    Thanks again for helping me out.
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Try taking the ratio of V_final over V_initial = 0.9:

     \frac {V_f}{V_i} = 0.9 = \frac {e^{[ \frac {-K 80000}{R} e^{-1}]}}{e^0}

    Now you can solve for R.
    Thanks from lac220
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    Re: Fluid Mechanics + Atmosphere + Meteor. Help convert du/dt into dy/dz

    Got it now thank you!!
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