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Conservation of energy - rocket hovering above moon

  1. Jun 13, 2012 #1
    Suppose a rocket is hovering above the moon at a constant height, burning just enough propellant to stay at that height, and not move upwards or downwards.

    Viewed from an inertial frame, we have. Initial energy (just before ignition) :

    1. Gravitational potential energy between rocket and the moon = I1
    2. Chemical energy stored in fuel = I2


    Assume kinetic energies of rocket and moon are both 0 initially.

    ----

    Final energy:

    1. Gravitational potential energy between rocket and the moon = F1
    2. Kinetic energy (or heat) of the burning propellant = F2
    3. Kinetic energy of the rocket = F3
    4. Kinetic energy of the moon (it is gravitationally accelerated towards the rocket) = F4

    If there is to be conservation of energy, (I1+I2) should equal (F1+F2+F3+F4).

    But I2 = F2 + F3, because chemical energy of the fuel is converted into kinetic energy of the rocket and of the propellant. And since the rocket is still hovering at the same distance from the moon, I1 = F1.

    So where does the additional term, F4 come from? Is energy conserved?
     
  2. jcsd
  3. Jun 14, 2012 #2
    1,When the exhaust hits the surface of moon, it will prevent moon moving toward the rocket;
    2,The mass of rocket continues to decrease, the lost part of fuel will lose its potential energy ;
    3,Another better example similar:
    A helicopter hovering aloft in the atmosphere of the earth ,the airflow moving downwards will diffuse in the open air but not hit the surface of the earth.
    It's sure the earth will move towards the helicopter with a extremely tiny velocity.
    Unfortunately analysts used to ignore this point.
     
    Last edited: Jun 14, 2012
  4. Jun 14, 2012 #3
    If you manage to let the exhaust not hit the moon, then the kinetic energy of the moon comes from the fact that both the moon and the rocket start to move (the rocket is hovering at constant distance from the moon), so the kinetic energy of the exhaust will be less in the frame where the moon was initially at rest.
     
  5. Jun 14, 2012 #4
    Right, viewed in the rest frame, both rocket and moon start to move with kinetic energies R and M respectively. There's also the kinetic energy of the exhaust, E.

    After all the fuel has been burned, the Δkinetic energy = R+M+E, and Δpotential energy = -C where C is the potential energy of the fuel at the start. So by conservation of energy,

    R+M+E = C (1)

    But... Since the exhaust doesn't hit the moon, C is converted into only R and E, so

    C = R+E (2)

    So there's a contradiction, between (1) and (2). Seems like there should be an additional term on the right hand side of (1), equal to M. But where does it come from?
     
  6. Jun 14, 2012 #5

    A.T.

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    Gold Member

    This is wrong. The mass of the rocket has changed.
     
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