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Rocket in empty space

  1. Dec 27, 2011 #1
    I'm reading Spivak's mechanics book and I'm stumped on some of the math in his derivation of an analytic expression for the motion of a rocket in empty space.

    Let [itex]\mathbf{v}(t)[/itex] be the rockets velocity, let [itex]\mathbf{q}(t)[/itex] be the velocity at which fuel is ejected from the rocket, and let [itex]m(t)[/itex] be it's mass. Suppose [itex]m'(t) = -k[/itex] for some constant k.

    From a stationary frame of reference, the ejection velocity of the fuel is [itex]\mathbf{v}(t) + \mathbf{q}(t)[/itex]. In a small time interval, the amount of fuel ejected is [itex]m(t) - m(t+h)[/itex], and its ejection velocity is approximately [itex]\mathbf{v}(t) + \mathbf{q}(t)[/itex]. Hence the total momentum is
    [tex] \mathbf{p}_{fuel}(t) = [m(t) - m(t+h)] \cdot (\mathbf{v}(t) + \mathbf{q}(t))[/tex]

    Spivak claims the time derivative at time t of the momentum of the expelled fuel is
    [tex] \lim_{h\rightarrow 0} \frac{m(t)-m(t+h)}{h} ( \mathbf{v}(t) + \mathbf{q}(t) = - m'(t) \cdot ( \mathbf{v}(t) + \mathbf{q}(t) ) [/tex]

    Here's where I'm confused. I'm not sure how he calculates the time derivative of the momentum of the ejected fuel. I'm assuming he uses the formula
    [tex] \lim_{h\rightarrow 0} \frac{\mathbf{p}(t) - \mathbf{p}(t+h)}{h}[/tex]

    But when I do it this way, I get that [itex]\mathbf{p}_{fuel}(t) = m'(t) \cdot (\mathbf{v}(t) - \mathbf{p}(t)[/itex] (I'm not even sure if my evaluation of the limit is correct). Perhaps the negative sign denotes the momentum of the ship, but nonetheless, I would really appreciate if someone could explain/show this step to me.

    Next Spivak sets his time derivative of momentum equal to the time derivative of [itex] -m(t) \mathbf{v}(t) [/itex]. I'm a bit confused here as well. I know that the time derivative of momentum is equal to the force, and the second law tells us that the sum of the forces is equal to the product of mass and acceleration. So again it looks like he's using the negative sign to denote the velocity of the ship? Again, could someone explain this part to me as well?

    At any rate, in case you were curious, Spivak determines that
    [tex]m(t) \frac{d \mathbf{v}(t)}{dt} = \frac{dm(t)}{dt} \mathbf{q}(t)[/tex]
    Last edited: Dec 27, 2011
  2. jcsd
  3. Dec 27, 2011 #2
    A real number that is sufficiently small.
  4. Dec 27, 2011 #3

    D H

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    Staff Emeritus
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    I don't have Spivak, but from your description I can see what he is doing. He is defining a system comprising the rocket plus the cloud of exhaust that the rocket has expelled. This is presumably a chemical reaction, making the total mass of the rocket+exhaust cloud system constant (chemical reactions conserve mass). The system is isolated, so the total linear momentum of the rocket+exhaust system is also constant.

    Let [itex]m_r(t)[/itex] and [itex]m_e(t)[/itex] be the masses of the rocket and exhaust cloud. Define the changes in mass over some time interval [itex](t,t+\Delta t)[/itex] as
    \Delta m_r(t;\Delta t) &= m_r(t+\Delta t) - m_r(t) \\
    \Delta m_e(t;\Delta t) &= m_e(t+\Delta t) - m_e(t)
    By conservation of mass, [itex]\Delta m_e(t;\Delta t) = -\Delta m_r(t;\Delta t)[/itex] and thus
    [tex]\Delta m_e(t;\Delta t) = m_r(t) - m_r(t+\Delta t)[/tex]
    (This apparently backwards delta is where you appear to have gotten confused.)

    For a sufficiently small period of time, the exhaust speed relative to some inertial observer will be nearly constant (your [itex]\mathbf v(t) + \mathbf q(t)[/itex]), and thus the momentum transferred to the exhaust cloud will be approximately [tex]\Delta m_e(t;\Delta t)(\mathbf v(t) + \mathbf q(t)) = (m_r(t) - m_r(t+\Delta t))(\mathbf v(t) + \mathbf q(t))[/tex]
    Dividing by [itex]\Delta t[/itex] and taking the limit as [itex]\Delta t \to 0[/itex] yields
    [tex]\dot{\mathbf p}_e(t) = -\dot m_r(t) (\mathbf v(t) + \mathbf q(t))[/tex]

    Since the total momentum of the system is constant, we must have
    [tex]\dot{\mathbf p}_r(t) = - \dot{\mathbf p}_e(t) = \dot m_r(t) (\mathbf v(t) + \mathbf q(t))[/tex]
    From [itex]\mathbf p_r(t) = m_r(t)\mathbf v(t)[/itex], we also have
    [tex]\dot{\mathbf p}_r(t) = \dot m_r(t) \mathbf v(t) + m_r(t)\dot{\mathbf v}(t)[/tex]
    Equating these two expressions for [itex]\dot{\mathbf p}_r(t)[/itex] yields
    [tex]\dot m_r(t) (\mathbf v(t) + \mathbf q(t)) = \dot m_r(t) \mathbf v(t) + m_r(t)\dot{\mathbf v}(t)[/tex]
    Canceling the common term [itex]\dot m_r(t) \mathbf v(t)[/itex] from the left- and right-hand side yields
    [tex]\dot m_r(t)\mathbf q(t) = m_r(t)\dot{\mathbf v}(t)[/tex]
  5. Dec 27, 2011 #4

    Philip Wood

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    Gold Member

    re your first question... Spivak is simply treating (v(t) + q(t)) as a constant, and replacing the expression for the limit of [itex]\Delta[/itex]m/[itex]\Delta[/itex]t by the rather simpler m[itex]\acute{}[/itex].

    I find Spivak's derivation, though correct, so cumbersome in its notation, as to make an easy piece of Physics seem difficult. I'd do this...

    momentum of rocket at time (t + [itex]\Delta[/itex]t) + Momentum of gases ejected in time [itex]\Delta[/itex]t = momentum of rocket at time t.

    So (m + [itex]\Delta[/itex]m)(v + [itex]\Delta[/itex]v) + (-[itex]\Delta[/itex]m)(v + q) = mv

    Note that [itex]\Delta[/itex]m is negative. Multiplying out and cancelling those terms that cancel:

    (m + [itex]\Delta[/itex]m)[itex]\Delta[/itex]v - ([itex]\Delta[/itex]m)q = 0

    So (m + [itex]\Delta[/itex]m)([itex]\Delta[/itex]v/[itex]\Delta[/itex]t) - ([itex]\Delta[/itex]m/[itex]\Delta[/itex]t)q = 0

    In the limit as [itex]\Delta[/itex]t[itex]\rightarrow[/itex]0, [itex]\Delta[/itex]m[itex]\rightarrow[/itex]0, and we have

    m(dv/dt) - q(dm/dt) = 0
  6. Dec 27, 2011 #5
    Thanks DH and Philip. Can't believe I didn't think of dividing both sides by change in t then letting the change go to zero! Although I probably wouldn't of appealed to the relation [itex]p_r'(t) = -p_e'(t)[/itex].
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