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Suppose I have a rocket thats exerts a force (with negligible loses in mass), which cancels out the downward force of gravity. Clearly my rocket could be moving at a constant velocity (ignore air resistance) or not at all. Nothing seems fishy yet, but what about this: Clearly, no work is being done, but energy is still being expended to keep the rocket in place. So how can the energy expended be found (please no answers of the form: bond enthalpies of fuel etc., I am sure there is an explanation for this in Newtonian kinematics/energetics)?

Because if we try to use the formula $\mathrm{Work} = F \cdot d$, then $d$ could be zero or anything at all, which doesn't tell us anything useful. Another similar scenario would be that of two rocket exerting forces in opposite directions, clearly energy is being expended but no work so what to do?

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For a normal rocket engine, the power is going to two things: Heating the exhaust from the temperature it was stored in the rocket at, and the kinetic energy of the exhaust. Eventually this kinetic energy is dissipated in friction with the atmosphere and the ground and then it too turns to heat.

Work is being done, just not on what you wanted.

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  • $\begingroup$ Yeah exactly, you're doing work on a gas in order to stay stationary. $\endgroup$ – Brandon Enright Nov 13 '13 at 21:43
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Clearly, no work is being done,

This is the flaw in your reasoning as, in fact, work is being done.

Don't forget that, when the rocket engine is operating, mass is accelerated and exhausted from the engine.

You're thinking only of the rocket proper but the rocket, loaded with propellants, forms a system with a center of mass.

So, if you consider the center of mass of the rocket/propellent/exhaust system, you'll find that it is accelerated and thus work is being done.

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  • $\begingroup$ @GeorgeE.Smith, I take "negligible loss of mass" to mean that the dm/dt term in the thrust equation is "small enough" that the loss of mass, compared to the mass of the rocket / propellent system, is negligible. But recall that thrust also depends on the velocity of the expelled mass and so, in this thought experiment, the exhaust velocity and thus, thrust can be whatever it needs to be $\endgroup$ – Alfred Centauri Nov 19 '13 at 12:10

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