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Chain question

  1. Dec 11, 2003 #1
    rocket propulsion/momentum

    A 1520 kg rocket has 4860 kg of fuel on board. The rocket is coasting through space at 94 m/s and needs to boost its speed to 348 m/s. It does this by firing its engines and ejecting fuel at a relative speed of 807 m/s until the desired speed is reached. How much fuel is left on board after this manuever?
     
    Last edited: Dec 11, 2003
  2. jcsd
  3. Dec 12, 2003 #2
    Apply the Conservation of Momentum in inertial reference frame with which the rocket is moving
     
  4. Dec 13, 2003 #3

    ShawnD

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    Re: rocket propulsion/momentum

    this work is all wrong
     
    Last edited: Dec 14, 2003
  5. Dec 13, 2003 #4

    Doc Al

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    Re: rocket propulsion/momentum

    You have to treat the rocket as a system of variable mass. As the fuel is ejected, it exerts a thrust [tex]v_{rel}\frac{dM}{dt}[/tex] on the rocket. Give it a shot (you'll need a little calculus).

    ShawnD, note that 807 m/s is the relative speed: it is not the speed of the exhaust with respect to the ground.
     
  6. Dec 13, 2003 #5
    Yes Shawn D u have to apply momentum conservation w.r.t ground
     
  7. Dec 13, 2003 #6

    ShawnD

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    Ok then lets try this again. I'll change the frame of referance to the ship so that my initial is 0.

    initial momentum:
    p = (1520 + 4860) * 0
    p = 0

    relative momentum of the fuel:
    p = -807x

    new momentum of ship:
    p = (1520 + (4860 - x )) * (348 - 94)
    p = (1520 + 4860 - x) * 254
    p = (6380 - x) * 254
    p = 1620520 - 254x


    final = initial:
    1620520 - 254x - 807x = 0
    1620520 - 1061x = 0
    x = 1527kg of fuel spent



    look better?
     
  8. Dec 13, 2003 #7

    Doc Al

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    No. The ship is a non-inertial frame. Its speed (and mass) are changing. See my previous post. (I'll post a solution tomorrow.)
     
  9. Dec 13, 2003 #8

    ShawnD

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    You souldn't have to factor that in. The velocity of the fuel is always the same speed relative to the ship.
     
    Last edited: Dec 13, 2003
  10. Dec 14, 2003 #9

    Doc Al

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    Well, no. The ship is speeding up. As each bit of fuel is ejected, its speed relative to the ship is the same. But the speed of the ship is different for each bit.
     
  11. Dec 14, 2003 #10

    ShawnD

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    Al can you please explain what you mean? I said the fuel is always the same speed relative to the ship. You said no, then you said it's relative speed is the same (what I said).
     
  12. Dec 14, 2003 #11

    Doc Al

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    Sure. Each bit of fuel, as it is ejected, moves at the same speed relative to the ship. For example, at time t=1 a bit of fuel m1 is ejected. It moves at 807 m/s relative to the speed of the ship at time t=1. But the ship is accelerating. At time t=100 (say) the bit of fuel m1 is no longer moving at 807 m/s relative to the ship, since the ship is now moving faster. So... all those bits of fuel are moving at different speeds. Make sense?
     
  13. Dec 14, 2003 #12

    ShawnD

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    You are looking at fuel it has already spent, I'm looking at fuel it is currently spending.
     
    Last edited: Dec 14, 2003
  14. Dec 14, 2003 #13

    Doc Al

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    How to solve this problem

    Yep. It's meant for a calculus-based college course, not high school. For your amusement, here's how I would solve this problem:

    Let M stand for the mass of the ship+remaining fuel, v for the speed of the ship. Each bit dM of fuel ejected exerts a force on the ship:
    [tex]v_{rel}\frac{dM}{dt}=Ma=M\frac{dv}{dt}[/tex]

    Rearranging,
    [tex]v_{rel}\frac{dM}{M}={dv}[/tex]

    Integrating from the initial speed&mass to the final,
    [tex]v_{rel}\ln{\frac{M_f}{M_i}=v_f-v_i[/tex]

    Plugging in some numbers:
    [tex]\ln{\frac{M_i}{M_f}=\frac{348-94}{807}[/tex]

    Finally,
    [tex]\Delta M= 1723[/tex]

    So, of the original 4860 kg of fuel, 3137 kg is left. (Unless I made an arithmetic error.)

    But the real question is: Why is this thread titled "chain question"?
     
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