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Homework Help: Rocket flight

  1. Aug 13, 2009 #1

    State the fundemental equation of motion for a particle of variable mass.

    A rocket of initial mass m0 is fired vertically, under the influence of a uniform gravitational field, and expels propellant at a constant relative velocity c downwards. The propellant is completely consumed after a time T, leaving the rocket with a mass mT. Calculate the velocity and position at time t < T. Show that the velocity at time T is independent of the rate of consuming the propellant.


    I don't know how to go about solving this. I vaguely recall the principle behind rocket flight; the reaction force against the expulsion of propellant is what drive the rocket upwards. The resultant force will be something like F = f(c) - mg, m of course being a variable and f being some function of c. I have no worked examples for this type of problem to refer to, so I need someone to show me the steps involved.
  2. jcsd
  3. Aug 13, 2009 #2


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    The fundamental equation is Newton's second law [itex]\textbf{F}_{\text{net}}=\frac{d\textbf{p}}{dt}[/itex]....for constant mass, [itex]\frac{d\textbf{p}}{dt}=m\frac{d\textbf{v}}{dt}[/itex]....when the mass varies with time, you need to use the product rule....what do you get then?
  4. Aug 14, 2009 #3
    F=m(dv/dt)+v(dm/dt), which makes a delightful amount of sense.

    What then?
  5. Aug 14, 2009 #4


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    Force = (rate of) change of momentum …

    so how much momentum is the rocket chucking out every second? :wink:
  6. Aug 14, 2009 #5


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    Now apply this to the rocket....what external force(s) are applied to the rocket?...that gives you a value for [itex]\textbf{F}_{\text{net}}[/itex]. What is the mass of the rocket as a function of time [itex]m(t)[/itex]? What is [itex]\frac{dm}{dt}[/itex]?

    Plug those values into your fundamental equation [itex]\textbf{F}_{\text{net}}=m\frac{d\textbf{v}}{dt}+\textbf{v}\frac{dm}{dt}[/itex] and you end up with a first order differential equation for the velocity of the rocket [itex]\textbf{v}[/itex]...you know how to solve such an equation right?
  7. Aug 14, 2009 #6
    In this example, just gravity
    m(t)=m0 + dm/dt
    dm/dt= (mT-m0)/T (this will be a negative value)

    F=dv/dt(m0 + (mT-m0/T)) + v(mT-m0/T)

    I passed an exam on differential equations a few months ago, so in theory, yes.

    If I want to find the point at which the propellant is consumed I would set F=0? But we already know that, it's T. Hold on I'm getting confused again..

    it's chucking out mass at a rate of (m0-mT/T) units per unit (second?), with a velocity of c..

    so c(m0-mT/T)?

    Oh oh, Fnet=c(m0-mT/T) - g(m0+(mT-m0/T))?

    How would I eliminate mass to get an expression for position/velocity? Divide through by m0+(mT-m0/T)? Seems awfully messy...
  8. Aug 14, 2009 #7


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    Well, that's one force...and that'll be [itex]\textbf{F}_{\text{gravity}}=m(t)\textbf{g}[/itex]...but there is also a reaction force from the ejection of propellant [itex]\textbf{F}_{\text{reaction}}=-\textbf{F}_{\text{propellent}}[/itex] (Newton's 3rd Law).

    To calculate [itex]\textbf{F}_{\text{propellent}}[/itex], use Newton's 2nd law again:


    You mean [itex]m(t)=m_0+\frac{dm}{dt}t[/itex] right?

    Careful, it only ejects mass at this rate for [itex]0\leq t \leq T[/itex], for [itex]t>T[/itex], dm/dt=0....so you can write

    [tex]m(t)=\left\{ \begin{array}{lr}m_0+\frac{m_T-m_0}{T}t, & 0\leq t \leq T \\ m_T, & t>T\end{array}[/tex]

    When the propellant is consumed, there will still be a gravitational force of [itex]m_T\textbf{g}[/itex] won't there?

    Close, the net force on the rocket should be [itex]F=-m'(t)c-m(t)g[/itex] upward

    Set it equal to [itex]\textbf{F}_{\text{net}}=m(t)\frac{d\textbf{v}}{dt}+\textbf{v}\frac{dm}{dt}[/itex]
  9. Aug 14, 2009 #8
    That comes out as reaction= (mT-m0)(c/T + dv/dt)

    so the whole thing is Fnet=dv/dt(mT-m0)+c/T(mT-m0)+(m0+((mT-m0)/T))g





    I'm still not getting it :(

    I'll keep working at this. If I do figure it out I'll make a post saying so.
  10. Aug 14, 2009 #9
    hold up, I've got to here:


    I've got something equal to 0 at least.
  11. Aug 14, 2009 #10


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    Close, you're missing a negative sign (from the Newton's third law equation) and the propellent is ejected at a constant speed, so dv/dt=0 for the propellent. Therefor Freaction=-(mT-m0)c/T=-m'(t)c

    Well, you have two equations for Fnet:

    (1) [tex]\textbf{F}_{\text{net}}=m(t)\frac{d\textbf{v}}{dt} +\textbf{v}\frac{dm}{dt}[/tex]


    (Choosing k to be the upward unit vector)

    Set them equal to eachother and you will get an inhomogeneous, linear, first order differential equation for [itex]\text{v}_z(t)[/itex] (the other components of [itex]\text{v}[/itex] have trivial equations and solutions)

    Remember, you've already calculated the values of m(t) and m'(t), so you can just plug them into your differential equation....what does your DE become for 0<t<T ? How about for t>T ?
    Last edited: Aug 14, 2009
  12. Aug 15, 2009 #11

    How do I go about solving that? General solution and particular integral?

    I'll look back over my differential equations, but can you show me the steps from here?

    Well.. from this point the propellant force dissapears completely, as does dm/dt, and the mass becomes the constant mT so:

    dv/dt(mT)=-mTg => dv/dt=-g

    Thanks for all the help so far.
  13. Aug 20, 2009 #12
    bump because I still haven't figured this out
  14. Aug 23, 2009 #13


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    You really need to be much more careful with your [itex]t[/itex]'s and [itex]T[/itex]'s; the above expression is not correct.

    Once you have the correct expression, manipulate it to get it into the form [itex]\frac{dv}{dt}+f(t)v=g(t)[/itex]. From there, you should be able to solve it without much trouble.

    Post you work if you get stuck.
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