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Find the terminal speed of a 750-kg rocket

  1. Apr 10, 2005 #1
    Find the terminal speed of a 750-kg rocket that starts from rest carrying 2600kg of fuel and that expels its exhuast gases at 1.8 k/ma

    here's the formula im using: [tex]v_f = v_i + v_{ex}*ln{\frac{M_i}{M_f}}[/tex]

    well since it's at rest, initial velocity is zero. so...

    [tex]v_f = 1.8*ln(750/2500)[/tex] which turns out to be 2.2 km/s. but the book says the answer is 2.7km/s
     
  2. jcsd
  3. Apr 10, 2005 #2
    I believe the Mi and Mf mean mass initial and mass final, check those, the initial mass will be the mass of the rocket and its fuel and the final mass will be just the rocket...I think :uhh: Guess I'm a little tired again today :zzz:
     
  4. Apr 10, 2005 #3

    Pengwuino

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    Gold Member

    Yah that is correct. Mi is mass initial (overall mass) and Mf is mass final (overall mass - expelled gas)
     
  5. Apr 10, 2005 #4
    If I denote the relative velocity of the ejected fuel with respect to the rocket as [itex]\u[/itex] then the variable mass equation I use is

    [tex]M\frac{dv}{dt} = F_{ext}-u\frac{dM}{dt}[/tex]

    In case of a rocket in the gravitational field (I guess this is what you want?), [itex]F_{ext} = Mg[/itex] so here,

    [tex]M\frac{dv}{dt} = Mg-u\frac{dM}{dt}[/tex]

    At terminal speed, the acceleration of the rocket is zero....
     
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