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Maximum kinetic energy of a rocket in free space.

  1. Jul 8, 2011 #1
    I'm returning to grad school for physics this fall, but I've been out of the classroom for thirteen years. (teaching music, of all things) I'm working my way through practice problems for the placement exam, but I don't have any way to verify correct answers for the problems unless I find something similar online. To be fair, I usually do. But this one, surprisingly, I can't find a direct analog for.

    1. The problem statement, all variables and given/known data

    A rocket starts from rest in free space by emitting mass at a constant rate [itex]\alpha[/itex] and with ejection velocity u relative to the rocket. At what fraction of the original rocket mass is the kinetic energy of the rocket a maximum. (Assume the mass of the empty rocket to be negligible)

    2. Relevant equations

    Just kinetic energy and momentum, as far as I can tell. Forgive me, I'm not great with LaTex yet, so I'm trying to minimize typing it here.

    3. The attempt at a solution

    Using conservation of momentum, reach the basic rocket equation:

    M dv = u dm

    where M is the inital mass of the rocket. Separate variables and integrate:

    v(t) = u ln (M / m(t))

    where m(t) is the mass of the rocket at time (t)

    The Kinetic Energy, T(t), of the rocket is:

    1/2 (m(t) ) v(t)^2

    Substitute m(t) = M - [itex]\alpha[/itex]t and the value of v(t) from above to get:

    T = M u^2 ln (M/m(t)) - [itex]\alpha[/itex] t u^2 ln (M/m(t))

    The kinetic energy should be at a maximum when the time derivative of the KE is zero, so differentiate (and this is the most likely place I screwed up) the above equation with respect to time to get:

    [itex]\frac{dT}{dt}=ln(\frac{M}{m(t)})[/itex]

    Setting that equal to zero, you find only one extremum in T at m(t) = M, or at the point of total fuel consumption. Given that the KE of a massless rocket is zero, I'm at a loss.

    Just thinking about this in the abstract, it seems that the velocity over the entire period of thrust, the mass should decrease, and the KE, as the product of the mass and the square of the velocity (and 1/2), should increase to a maximum and then decrease to zero as all the mass moves to the ejecta.

    What am I missing here?
     
  2. jcsd
  3. Jul 8, 2011 #2
    I think that one of the mistakes you made is when you substitute the expression for v(t):
    you should have [itex](ln (m/M))^2[/itex] instead of [itex]ln(m^2/M^2)=2ln(m/M)[/itex]
     
  4. Jul 9, 2011 #3

    Filip Larsen

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    Gold Member

    Welcome to PF!

    You may also want to find maximum kinetic energy using kinetic energy expressed as function of mass, Tm, since dTm/dm = 0 is a bit easier to solve by hand than dTt/dt = 0. Once you have a mass for maximum T you can easily insert that in your mass-time equation to find the corresponding time.
     
  5. Jul 9, 2011 #4
    Thanks to both of you for the help. It's clear that my facility with logarithms has eroded greatly. After making the correct substitution into the time differential, I discovered that it was indeed just as nasty as Filip had suggested. The mass differential was simpler, but still a bit of a chain rule slog.

    T = 1/2 m(t) v^2 = 1/2 m(t) u (ln (M/m(t)) )^2

    dT/dm = 1/2 u [ ln^2(M/m) - 2 ln (M/m) ]

    Setting this equal to 0 yields the relatively simple looking:

    ln(z)^2 = - ln^2(z) where z = (M/m(t))

    Plugged it into Alpha and got a solution right away, but it took me a good half an hour of head scratching to understand it. I just about kicked myself when I got it - but that's always the way, isn't it? I hope my speed increases with practice - this was one of twelve questions on a three hour test.

    Here's the final answer.

    ln^2(z) + ln(z) + ln(z) = 0
    ln(z)(ln(z) + 2) = 0

    So there are two extrema - one at z = 1, and the other at z = e^-2

    Looking at this, I think I lost a sign somewhere, since it implies that m / M at max KE is greater than one (e^2).
     
  6. Jul 9, 2011 #5

    Filip Larsen

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    Gold Member

    I concur with your 2nd equation, but from that you should get

    ln2(z) - 2 ln(z) = ln(z) (ln(z)-2) = 0

    which implies that z = e2 and thus m = M/e2 is a solution for dT/dm = 0. I guess you are correct about loosing a sign along the way :smile:.
     
  7. Jul 9, 2011 #6
    Oh... right. (bangs head on desk)
     
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