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A system with varying mass - a rocket

  1. Jan 26, 2005 #1


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    In Fundamental of Physics 6th ed. by Halliday, Resnick, Walker pg. 181-183, the section is about systems with varying mass: a rocket.
    It is written:
    Is it okay if I change M+dM with M-dM because I know that the rocket loses its mass and -dM with dM because the exhaust product has mass in positive quantity?
    So that figure (b) becomes
    {~~~~~~~~~~~~~~~~~~~~}--System boundary
    {. dM .......... |--------\ .......... }
    { <<<< --> U |. M-dM . > ---> v }
    { ................ |--------/ .......... }
    ------------------------------------------- x
    And dM is not a negative quantity anymore because I've taken account of it in the figure (the rocket loses its mass and the exhaust products gain mass).

    Next it is written:
    If I'm allowed to change M+dM with M-dM and -dM with dM then Eq. 9-38 becomes
    M * v = dM * U + (M - dM) * (v + dv) ... (Eus-1)

    Then it is written:
    If I do Eq. Eus-1 with little algebra it becomes
    dM * vrel = M * dv ... (Eus-2)
    Dividing each side by dt gives us
    (dM/dt) * vrel = M * (dv/dt)
    I replace dM/dt by R not -R because I've taken account of this decrement in rocket mass since the beginning.
    Am I right?
    Therefore, I got the first rocket equation without any problem that is R * vrel = M * a.

    Now I run into a problem when finding the velocity of the rocket
    It is written:
    My problem is that when I use Eq. Eus-2 as below
    dv = vrel * (dM/M)
    Integrating each side leads to
    vi S vf dv = vrel * [Mi S Mf (1/M) dM]
    And evaluating the integrals then gives
    vf - vi = vrel * ln( Mf / Mi ) ... (Eus-3)
    Yup! This is the problem. I cannot replace vrel by -vrel because I don't have any reason to do that.
    Because Mf is always less than Mi, Eq. Eus-3's right hand side always has a negative value and that means vf is less than vi (the rocket is slowing down)
    Mmmm... do you think I've done a big mistake by changing M+dM by M-dM and -dM by dM? Why is it wrong?
    If I'm not wrong then is there something that I've taken account into my calculation when I calculate the second rocket equation? What is it?

    Thank you very much for you help! ^_^
    I really appreciate it! ^^v
  2. jcsd
  3. Jan 27, 2005 #2
    a very long typing, how long did you take to type all this? :confused:
    I am not gonna type this long... my typing is slow :rolleyes:

    I think you have already noticed your mistake is on the sign only...... you changed the M+dM to M-dM' (i added a prime in your dM to make thing looks nicer)
    Here is your problem:
    when you doing the integral since dM' = -dM, the limit of the integral changed:
    [tex] \int_{b}^a dM' = \int_{b}^a (-dM) = -\int_{b}^a dM=\int_{a}^b dM [/tex]

    the negative sign in Eus-3 is due to the integral... vrel is fine....
    if you carry a prime notation all the way in your calculatioin... you will see this problem had never been taken care of..... even if you changed the sign a couple times
    Last edited: Jan 27, 2005
  4. Jan 27, 2005 #3


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    Hi Ho! ^_^

    Mmm... about an hour ^^;;

    Ummm... I know that this is all about the sign but what make me wonder actually is not the sign but the author's reason.
    Why did he use M+dM when he knew that the rocket loses its mass?
    If I were the author I would use M-dM because I know that the rocket loses its mass.
    Here -dM is really a -dM not -dM' because the rocket loses its mass.
  5. Jan 27, 2005 #4
    I used the prime notation because wanna distinguish your dM from the auther's dM. sure the dM is the rocket mass... I can't say your dM=-dM.... that doesn't make sense mathematically and conceptaully... One small question, do u understand my reasoning above?

    if the auther use M-dM... while he do the integral in eq 9-43, he have to make the same argument------the limit of the integral must flip------ that will make this complicate calculation more complicated.....

    let me know do you understand my reasoning.....
    Last edited: Jan 27, 2005
  6. Jan 27, 2005 #5


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    Mmmm... yup! I know exactly that dM is not equal to -dM
    And I know your reason why you used dM'.
    Wait a minute, I think I see the light.

    Could you please check my story below?
    A rocket initially has mass M and velocity v relative to an inertial reference frame.
    During dt, it ejects exhaust products with mass dExhaust.
    Therefore, at the end of dt, the rocket has velocity v+dv and mass M-dExhaust.
    Whereas, the exhaust products have velocity u and mass dExhaust.
    Then using the conservation of momentum,
    M * v = dExhaust * u + (M-dExhaust) * (v+dv)
    M * v = dExhaust * (v+dv-vrel) + (M-dExhaust) * (v+dv)
    M * v = dExhaust * (v+dv) - vrel * dExhaust + M * v + M * dv - dExhaust * (v+dv)
    0 = -vrel * dExhaust + M * dv
    vrel * dExhaust = M * dv ... (1)
    vrel * (1/M) * dExhaust = dv
    Okay! So, dExhaust = -dM because ... ?
    Could you please tell me why is dExhaust equal to -dM?
    If I continue, I'll find the second rocket equation without any problem.
    But I'll screw up the first rocket equation.

    If dExhaust = -dM then ...
    Back to equation (1)
    vrel * dExhaust = M * dv
    Dividing each side by dt gives us
    vrel * (dExhaust/dt) = M * (dv/dt)
    Because dExhaust = -dM,
    vrel * (-dM/dt) = M * (dv/dt)
    -vrel * R = M * a
    Argh, I screw up the first rocket equation because R is positive.

    Do you like Final Fantasy 7? ^^;;;

    Ah, my problem arose when I saw M+dM in my text book.
    I don't have any problem when the author wrote v+dv because I know the rocket will gain additional speed by ejecting something.
    Hehehe... why didn't I think that the rocket will slow down? Because in TV I see the rocket will gain speed ^^;;;
    But I don't think that the rocket will gain additional mass.
    It ejects something so it loses its mass.
    Therefore, I have a problem with M+dM.
    Last edited: Jan 27, 2005
  7. Jan 27, 2005 #6
    The equations work perfectly well if you assume dM > 0. In fact, I prefer the derivation your way. We assume that initially the rocket is moving with velocity [itex]v[/itex] and the fuel is ejected with velocity [itex]U[/itex], so
    P_1 & = P_2 \\
    Mv & = (M - \mathrm{d}M)(v + \mathrm{d}v) + U\,\mathrm{d}M.

    Expanding this out gives

    [itex]Mv = Mv + M\,\mathrm{d}v - v\,\mathrm{d}M - \mathrm{d}M\,\mathrm{d}v + U\,\mathrm{d}M[/itex]

    The vector term [itex]- \mathrm{d}M\,\mathrm{d}v[/itex] is negligible, so

    [itex]Mv & = Mv + M\,\mathrm{d}v - v\,\mathrm{d}M - \mathrm{d}M\,\mathrm{d}v + U\,\mathrm{d}M.[/itex]

    Substituting [itex]U = v_\mathrm{ex} + v[/itex] and solving for [itex]M\,\mathrm{d}v[/itex], before dividing both sides by [itex]\mathrm{d}t[/itex] gives

    M\frac{\mathrm{d}v}{\mathrm{d}t} = - v_\mathrm{ex}\frac{\mathrm{d}M}{\mathrm{d}t}.

    Separating the variables and integrating gives

    v_\mathrm{f} - v_\mathrm{i}= v_\mathrm{ex} \ln\frac{M_\mathrm{i}}{M_\mathrm{f}}.
  8. Jan 27, 2005 #7
    sorry, I don't play Final Fantasy.....

    it is easy... because the text book define the mass of rocket is M+dM... the rocket is lossing mass... so dM is negative.... but you define the rocket's mass as M - dExhaust.... the dExhaust is positive.... therefore

    dExhaust = -dM

    dExhaust equal to a negative (negative number) => dExhaust is positive... simple logic...

  9. Feb 2, 2005 #8


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    Hi Ho! ^_^

    Although I'm still confused, thank you very much for all of your helps! ^^v
    I really appreciate it! ^^
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