- #1
bob900
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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.
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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?
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?