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Rocket Motion

  1. Jan 26, 2009 #1
    A rocket is accelerating upwards from rest in a uniform gravitational field g. Notation: m(t) is the mass of the rocket plus remaining fuel, m0 is the initial total mass, vex is the exhaust speed (relative to the rocket), and k is the rate, in kg/s, at which fuel is consumed. By considering momentum changes in a short time dt, show that mdv/dt = kv(ext) - mg(upward direction is positive). Assuming k is constant, find the speed v as a function of the
    remaining mass m. Show that, if k is very large, it agrees with the result derived in class (without gravity).

    Ok, so i have already proved the first part.

    P(t)=mv and P(t+dt) = (m+dm)(v+dv)-dM(v-v(ext))


    The a couple of steps later we get: -mgdt=mdv+(ext)(dm/dt)

    so we get -mg=mdv/dt -kv(ext), where -k=dm/dt

    and therefore mdv/dt=kv(ext) - mg.

    Ok, so I'm a little stuck on the second part. Not really sure where to start off. Any hints?
  2. jcsd
  3. Jan 26, 2009 #2
    Assuming that k=dm/dt is constant isolate dv then integrate. You have to assume the exhaust velocity, v(ext) is constant.
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