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Homework Help: Net force on a rocket

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data

    I am trying to derive an equation for the net force, or thrust, acting upon a rocket. The equation I get is different from the standard equation given in most textbooks, so I want to know where I am going wrong.

    2. Relevant equations

    [itex] F = d(mv)/dt [/itex]

    I also make use of the principle of conservation of momentum.

    3. The attempt at a solution

    I will deal with the scenario whereby the only force acting upon the rocket is due to the ejected exhaust. At time t=0, let the rocket have mass m and velocity v with respect to an inertial reference frame. At this instant, its momentum, p1 is given by

    p1 = mv

    Let the ejected exhaust have a speed v e with respect to the rocket. Since the exhaust is ejected opposite to the direction of motion, its relative velocity is –ve. Let vf be the velocity of the exhaust relative to the inertial reference frame. Then,

    vf =v – ve

    After an infinitesimal amount of time, dt, let the mass of the rocket be m – dm. So, the mass of the ejected exhaust is dm. Also, the rocket’s velocity becomes v + dv. The total momentum of the system p2 is given by

    p 2 = (m - dm)(v +dv) + (dm)( vf)
    = (m – dm)(v+ dv) + (dm)( v – ve)
    = mv + mdv – (ve)(dm)

    The change in momentum of the system, dp, is given by

    dp = mdv – (ve)(dm) =0

    mdv = (ve)(dm)
    The change in momentum of the rocket is given by

    [itex] (m – dm)(v+ dv) – mv = mdv – vdm [/itex]

    The rate of change of momentum of the rocket, or net force, Fnet is given by
    Fnet = (m)(dv/dt) – v(dm/dt)
    = ((ve) – v)(dm/dt)
    = (-vf)(dm/dt)

    However, most textbooks simply define the net force to be (m)(dv/dt). How can this expression be used to find the force on the rocket if its mass is changing? And what is wrong with my derivation?
    I would appreciate if someone could clarify. Thanks in advance.
    Last edited: Apr 17, 2013
  2. jcsd
  3. Apr 17, 2013 #2


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    Your net force F accelerates the rocket of mass m (at that instant in time), and F=ma leads to F=m*(dv/dt). As m is a function of time, you can write this as F=m(t) * (dv/dt).
  4. Apr 17, 2013 #3


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    Yes, that's the rate of change of the momentum, but it's not the thrust. Consider a cart with a load of sand rolling along on the level, no frictional losses. Sand is falling out through a crack. The cart's momentum is changing, but there's no thrust. If you look at the whole system (rocket+exhaust) you have ## (mdv – vdm) + (v-v_e)dm = 0##, ##mdv/dt = v_edm/dt = thrust##
  5. Apr 17, 2013 #4
    Thanks to both of you for your replies.

    With regard to the cart example, doesn't Newton's 2nd law state that the rate of change of momentum of a body is directly proportional to the resultant force acting upon it? So how can the cart undergo a change of momentum if there is no net force acting upon it? Wouldn't that violate the second law?
  6. Apr 17, 2013 #5


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    The cart's velocity remains the same, but the mass of the sand within the cart is being reduced over time. Instead of a cart, imagine a rocket in space, firing two thrusters on the opposite sides of the rocket, so that there is no net force on the rocket. There are two plumes of fuel being emmitted from the rocket, with a net force being applied to each plume (outwards). There is no net force on the rocket and it's remaining fuel, but its momentum decreases because the mass of the on board fuel decreases over time. The total momentum of the rocket and the plumes of fuel is constant (assuming no external forces).
  7. Apr 17, 2013 #6
    Sorry I am still a bit confused.

    For the case whereby the rocket is emitting two plumes of fuel in opposite directions, I understand your point about the momentum of the rocket changing even though there is no net force acting upon it. But how do we reconcile this with Newton's 2nd law, which requires the presence of a resultant force if an object's momentum is changing?
  8. Apr 17, 2013 #7


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    Ah, but what is "the object"? The point of my cart example is that your intuitive notion of what constitutes the object is inappropriate for this law. You must either consider the cart and the sand it loses in time dt as separate objects or parts of the same object - you can't switch view as time goes from t to t+dt.
  9. Apr 18, 2013 #8

    Ok I think I finally understand. Thanks for all the replies.

    So just to check that I have the right understanding, I will briefly redo the relevant part of the derivation considering the rocket and the exhaust emitted in time dt as separate objects.

    By conservation of momentum,

    mdv = (ve)(dm)

    where m is the total mass of the rocket and the emitted exhaust. The change in momentum of the rocket alone is given by

    [itex] (m – dm)(v+ dv) – (m - dm)v = (m - dm)(dv) [/itex]

    Thus, the rate of change of momentum, or net force, is given by

    Fnet = (m - dm)(dv/dt) = (m)(dv/dt) - (1/dt)(dm)(dv)

    Thus, the net force is approximately (m)(dv/dt) as the product of dm and dv is extremely small.


    Fnet = (m)(dv/dt) = (ve)(dm/dt)

    Is this derivation now correct?
  10. Apr 18, 2013 #9


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    Looks ok to me.
  11. Apr 18, 2013 #10
    Ok thanks a lot.
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