GIThruster wrote:Remember, this is a maximum. In other words, the energy gained by an accelerating spacecraft cannot exceed the energy provided to its thrusters in any reference frame, assuming the thrusters use internally-carried propellant. This is because the loss of mass due to expenditure of propellant becomes a large kinetic energy sink as the initial velocity increases.
If that were true, then this case would not apply in the case of a thruster on a swing arm, and it does; so obviously the propellant is not the salient explanation.
If you put the whole propellant tank on the end of the arm with the thruster, what I said above would apply. In the case where the propellant is continuously fed in from somewhere else, the mechanism is a bit different, but the principle is the same if you think about it, and it gives you essentially the same answer.
Okay, time for Part 2:
Take a pair of thrusters (to avoid undesirable bending moments on the shaft) mounted tangentially on a flywheel which rotates at a rate such that the speed of the outer edge is V. Propellant is fed to the thrusters continuously from an external source. They have an exhaust velocity v_exh and a combined mass flow rate mdot, resulting in a jet power (the effective input to this mechanical system) of
P_jet = 0.5*mdot*v_exh^2
(Note that this is after inefficiencies in the thrusters; ie: P_jet = η*P_in, where P_in could be electrical, chemical, nuclear thermal...)
The outputs are the exhaust energy rate and shaft power:
P_exh = 0.5*mdot*(V-v_exh)^2 = 0.5*mdot*(V^2 - 2*V*v_exh + v_exh^2)
P_sh = F*V = mdot*V*v_exh
This results in a total output power of
P_exh + P_sh = 0.5*mdot*(V^2
- 2*V*v_exh + v_exh^2)
+ mdot*V*v_exh = 0.5*mdot*(V^2 + v_exh^2)
which looks like a violation of conservation. In fact, for V > 0.5*v_exh, P_sh > P_jet and the device looks like a free energy machine.
But wait - where's mdot coming from? Of course it has to be pumped up to the kinetic energy it has when fed into the thrusters. So the pumping power is
P_pump = 0.5*mdot*V^2
This is the theoretical minimum (no losses) and comes directly out of the useful work produced by the device, in this case the shaft power:
P_sh - P_pump = mdot*V*v_exh - 0.5*mdot*V^2
which is quadratic in V, exhibiting a maximum at
mdot*v_exh - mdot*V = 0
or
V = v_exh. Sound familiar?
The maximum in question is of course
P_sh[@V=v_exh] = mdot*v_exh^2 - 0.5*mdot*v_exh^2 =
0.5*mdot*v_exh^2 = P_jet
So you cannot actually get more power out of this system than you put in. It is at most (for non-electric thrusters) a silly-looking heat engine.
The total power output, including the energy of the exhaust and with pumping power accounted for, is
P_out = P_exh + P_sh - P_pump = 0.5*mdot*(
V^2 - 2*V*v_exh + v_exh^2)
+ mdot*V*v_exh - 0.5*mdot*V^2
P_out = 0.5*mdot*v_exh^2
or exactly the same as the input power.
So what we have here is that the kinetic energy lost by expending propellant in the spacecraft case is replaced by the pumping power necessary to supply propellant in the flywheel case. It's the same thing in two different forms.
If you had used the thrust/power FOM for a stationary ion thruster
This sounds distressingly like something you said earlier, that was wrong:
GIThruster wrote:If you take any thrust efficiency figure of merit in force/input power (N/W) and allow the thruster to accelerate at any rate over an arbitrarily long period of time, you can show that all thrusters will appear to violate conservation at some time in the future. This is because thrust efficiencies are stationary. Energy and thus power are not invariant so GR requires anyone doing this calculation will seem to uncover a conservation violation. I think it was chris who stated the issue quite well, as well as one of the folks over at NBF (the physicist there, not GoatGuy), that power in is linear, but power out would be quadratic, so this MUST yield a conservation violation. The remedy is to note that the dynamical relationship for stationary thrust can only be used while the thruster is stationary.
This is not true, as I demonstrated in the previous example. I used "stationary" values for input power and thrust: η*P = 0.5*mdot*v_exh^2 and F = mdot*v_exh. Those are invariant. And I got conservation.
Okay, only the four-force is Lorentz invariant. But the three-force is Galilean invariant, and for a model problem in Newtonian mechanics that's all you need.
Now, it is possible to use an accelerating reference frame and get the right answer (you have to assume a fictitious force, which naturally enough does fictitious work), but the fact that the thruster itself is accelerating has nothing to do with this... as long as the acceleration is insignificant compared with that of the reaction mass being accelerated in the thruster, which is generally a pretty good assumption. Relax that assumption and you just get a slight acceleration dependence of the thruster performance parameters, which complicates the math greatly but gives you basically the same answer - energy conservation.
And you don't need any form of relativity to do the math in the non-inertial reference frame. It's still Newtonian. You
never need GR to get energy conservation in a model problem. It's built into Newtonian mechanics and will always be observed if you use the equations properly.
BTW, if you really feel up to it, you can do a very similar calculation using the active mass of an M-E thruster, and calculating the action of the mass using it as propellant. After all, M-E thrusters are not technically "propellantless" but rather they use a "recycled propellant" -- the active mass of the thruster.
No, that's
exactly the mistake GoatGuy made. His control volume was too small.
Do you understand the concept of a "black box"? It doesn't matter what's going on inside; it's what it does that matters. In the case of an M-E thruster, you can turn it on, then off again, and you've got a spacecraft of exactly the same mass as before moving at a different velocity, meaning its kinetic energy is (in general) different. No part of the spacecraft can serve as a 'sink' for that energy difference because it's all exactly the same as it was before the thruster was used, with the exception of whatever energy transfer happened to operate the thruster. That energy transfer is invariant (e.g. chemical energy from a fuel cell) and thus cannot be matched with the spacecraft's kinetic energy change except by arbitrarily selecting a reference frame.
To get conservation with an M-E thruster, you need to consider the distant mass that it is ultimately reacting against. Without that you really do get an apparent conservation violation no matter what you do.
Also note, that if the exclusion of propellant were an explanation for a real conservation violation, then both linear and circular electric motors would go overunity and they do not.
You can't just consider the motor and its payload; you have to account for everything that might experience a force as a result of its operation.
In the case of an electric car, the reaction mass is the Earth. And it's true - without accounting for the energy gained/lost by the Earth, you
will get an apparent conservation violation in the general case. The fact that you can usually analyze an electric car without reference to the Earth's kinetic energy is due to the Earth being much more massive than the car, so its change in velocity (and thus kinetic energy) is negligible during the operation of the car
and its local surface velocity can be used as a reference frame for the calculation. This is only
almost right, but it's close enough. Use a different reference frame and you
do need to account for Earth's motion - for instance, if you're driving alongside a car and it begins to brake, it appears to be accelerating by doing negative work. What's actually happening is that the Earth is doing work on
it - from your perspective.
GIThruster wrote:The claim that energy conservation is the result of using propellant is hand-waving.
No, it's called Newton's Third Law. And it's a beautiful thing, when you finally get it...
More to come, when time allows.