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Question about conservation of energy/mass

  1. Aug 3, 2009 #1
    I've been thinking about this problem today. I tried reasoning my way through it and I haven't been able to. I might be totally missing something obvious, so if so please feel free to laugh and point it out, but for me right now this is confusing. It deals with the conservation of energy/mass:

    Let's say that a 10,000 kg spaceship is at rest in outer space. The crew decides that they want to then begin moving, so they fire their engines. They accelerate at 500 m/s^2 for 10 seconds.

    F = mA
    F = (10,000 kg)(500 m/s^2)
    F = 5,000,000 N

    xt = x0 + v0*t + (A*t^2)/2
    xt = 0 + 0 + ((500 m)*(10 m/s^2)^2)/2
    xt = 25,000 m

    W = Fd
    W = (5,000,000 N)(25,000 m)
    W = 1.25 x 10^11 J

    Their engines expended 1.25 x 10^11 joules of energy doing this process (I'm assuming 100% efficiency for simplicity, but even if it weren't it shouldn't change the nature of my problem).

    vt = v0 + A*t
    vt = 0 + (500 m/s^2)(10 s)
    vt = 5,000 m/s

    KE = (m*v^2)/2
    KE = ((10,000 kg)(5,000 m/s)^2)/2
    KE = 1.25 x 10^11 J

    Since there was 100% efficiency, it makes sense that the energy that the ship used would then be converted to its kinetic energy. This is conservation of energy, in that the energy stored as fuel is now the ship's kinetic energy.

    Now the crew of the ship decides they want to stop moving after reaching their destination. The ship then chooses to fire their rockets in the same fashion, except the opposite direction: 500 m/s^2 for 10 seconds.

    The force applied to the ship by the engines is negative (because the acceleration is in the opposite direction as the burst tomorrow) and the work done on the ship is negative. Therefore, after the 10 seconds the ship's velocity will be 0 and therefore kinetic energy will be 0. However, during the burn, the engines were still expending energy as they were fired 1.25 x 10^11 J.

    One would think that because the ship is reducing its kinetic energy, the engines would actually be taking that energy back in and storing it for use later. However, in this example, the engines are actually expending more energy.

    So now to conclude, the ship now has 0 joules of KE and over the process has expended 2.5 x 10^11 joules of energy. What has happened to this energy from a standpoint of conservation of energy? Where is it? I don't understand how it was conserved, because although in the first half it would had been transferred to KE, once the ship decelerated I see it was being "lost forever".

    What am I missing?
  2. jcsd
  3. Aug 3, 2009 #2


    User Avatar

    You need to think about what the 'engines' are doing.

    Consider the case of some kind of thrusters. These work by converting stored chemical potential energy in the fuel into kinetic energy.

    The fuel is ignited in the thrusters, causing the fuel to be expelled from the thrusters and the spaceship to be propelled forwards (conservation of momentum/Newton's 3rd law). This increases the kinetic energy of the fuel and of the spaceship.

    The fuel is ignited in the thrusters and directed such that it is expelled outwards in the direction of travel. This causes a backward acceleration (i.e. deceleration) of the spaceship (again conservation of momentum/Newton's 3rd law). This increases the kinetic energy of the fuel and decreases the kinetic energy of the spaceship. However this time the fuel has more kinetic energy (because the spaceship was already travelling in the direction it was expelled), which balances the loss in kinetic energy of the spaceship. Energy remains conserved.

    Of course in reality the expelled fuel has mass, so you would need to use the full version of Newton's second law... but that's a separate issue :smile:
  4. Aug 3, 2009 #3
    Ah, that makes sense. Thanks!
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