Does a Rocket Engine Exert Same Force Regardless of Velocity?

[Hunter235711]

This is a problem I have never really understood. If a rocket engine is fired in a vacuum, does it exert the same force regardless of its velocity? If so, I don't know how to get around the following issue.

Say that the chemical energy stored in a particular rocket engine is 100 Joules (I am just making these numbers up). Also, assume the rocket exerts a 10 Newton force for 3 seconds when fired. Now, if this rocket engine were mounted on a spaceship traveling at 1 meter per second it would release 3seconds*1meter/second*10Newtons = 30 Joules of energy when fired. No problem here, engine is about 30% efficient. However, now assume the spaceship is traveling at 100 meters per second. Now the same engine is fired and releases 3seconds*100meters/second*10Newtons = 3000 Joules! But this is impossible because the total chemical energy liberated by the rocket engine was only 100 Joules.

What is wrong with my logic? Thanks for your help!

Hunter

 

[rcgldr]

If you include the energy added to the ejected fuel, and the fact that the mass of the fuel remaining on the rocket decreases over time, then the output power consumed in accelerating the rocket, on board fuel, and ejected fuel, for a given throttle setting remains the same regardless of the rockets velocity relative to some inertial frame, including the rockets initial state.

Note that the force that accelerates the rocket forwards also accelerates the on board fuel, and the acceleration of the ejected fuel is relative to the rocket. Wiki article for rocket equation:

http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Getting back to your question:

Say that the chemical energy stored in a particular rocket engine is 100 Joules (I am just making these numbers up). Also, assume the rocket exerts a 10 Newton force for 3 seconds when fired. Now, if this rocket engine were mounted on a spaceship traveling at 1 meter per second it would release 3seconds*1meter/second*10Newtons = 30 Joules of energy when fired. No problem here, engine is about 30% efficient. However, now assume the spaceship is traveling at 100 meters per second. Now the same engine is fired and releases 3seconds*100meters/second*10Newtons = 3000 Joules! But this is impossible because the total chemical energy liberated by the rocket engine was only 100 Joules.

The unknown here is how much energy was added to the ejected fuel. Continuing with your example numbers, assume that the exit velocity of the fuel is 100 meters per second (with respect to the rocket) and all of it is released in a single instantenous impulse. If the rocket is moving at 100 meters per second, then the ejected fuel is decelerated from 100 meters per second to 0 meters per second, reducing it's kinetic energy to zero. So the increase in the rocket's energy is offset by the decrease in the ejected fuels energy.

After all the fuel is consumed, the total mechanical energy of rocket and the ejected fuel will equal the initial chemical potential energy of the fuel, minus the losses such as the chemical potential energy converted to heat and light.

 

[Bandersnatch]

@rcgldr: while what you said is very true, I think you'd overlooked the undelying issue about the work concept. He appears to be confused primarily about the independence of work from the choice of a reference frame. That he chose a rocket engine is just adding to the confusion.

@Hunter,

The reason you get crazy results is the same reason you'd get them if you were to ask why a force doing work on a rocket that is farther away(r₁) imparts it with more energy than if the rocket were close by(r₂). After all, r₁>r₁, so F*r₁>F*r₂!

Maybe that's enough of a clarification for you, if not, then read on.

The main issue here is with haphazard use of equations. When you see a variable, you need to understand exactly what it represents.

By definition, a force does work when it causes displacement of the application point along the direction of the force vector. The difference in position Δr has to be the result of the force. Any displacement not connected with the application of the force does not contribute to the work done.
If it weren't so, the amount of work done would vary with the choice of reference frames, as is the case in your example.

So the velocities of 1m/s and 100m/s are of no importance to our considerations. They are merely artifacts of reference frame choice. When writing W=Fr, the displacement ought to be only due to the result of the force, not due to initial velocites.

Say, you've got a moving system(initial velocity V₀, initial position r₀), on which work is being done along the direction of motion.
W=FΔr
the displacement due to force only is the difference between the total displacement, and the displacement the system would undergo without any forces acting(purely due to its inital velocity).
Δr=r_{total}-r_{V0}
total displacement, from Newton's second equation of motion, is the initial position on the coordinate grid, plus the distance the system travels due to its inital velocity, plus the distance covered as a result of the force
r_{total}=r_0+V_0t+\frac{1}{2}at^2
while with the absence of forces:
r_{V0}=r_0+V_0t
after deducting the two, we get
Δr=\frac{1}{2}at^2
The displacement due to force is shown thus to be indepent of the initial velocity, as well as the inital position. If it doesn't matter what the initial velocity is, then we may just as well treat the system as if it were initially motionless and centered in our coordinate system(V₀=0; r₀=0) to make our thinking easier.

 

[Oldfart]

Isn't the simple answer that the question that the answer to a physics experiment will be the same in any frame of uniform motion?

 

[rcgldr]

He appears to be confused primarily about the independence of work from the choice of a reference frame.

In some cases the amount of work performed does depend on the choice of reference frame, but in this case, the issue is inclusion of both the ejected fuel as well as the rocket, and the fact that the mass of the rocket and remaining on board fuel decreases as fuel is ejected from the rocket. As mentioned above, the increase in energy of the rocket and ejected fuel plume over some period of time will be the same regardless of the choice of reference frame.

For an example where the reference frame affects the amount of work done, take the case of an ideal wing. Using the ideal wing as a frame of reference, it diverts the relative air flow without changing the speed of the air flow, so the energy of the air is not changed. Using the air as a frame of reference, the air initially has zero mechanical energy, but after the wing passes by, the air is moving downwards (corresponding to lift) and forwards (corresponding to drag). It has a non zero "exit velocity" (the velocity where the affected air's pressure returns to ambient), and from the air's frame of reference the wing performed work on the air, increasing it's kinetic energy.

 

[Lsos]

As others are saying, rockets can be hard to get your head around because you can get your frames of reference confused. “Normally” your frame of reference is relative to the earth, but with a rocket it’s very easy to switch this to something else since the rocket doesn’t actually push off the earth, but off of the mass that it’s carrying with it. Essentially, a rocket is carrying its “earth”…its frame of reference, with it.

That is also why you can’t really say that rocket fuel has “100 joules”, because the energy in the rocket fuel depends on what it’s relative to. Relative to the earth, 1 kg of rocket fuel in orbit has MUCH more energy than 1 kg of rocket fuel sitting on the launch pad. Relative to itself, the energy content is the same.

Due to ½mv², it gets exponentially harder to go faster and faster. Nevertheless, given the same fuel a rocket in orbit can accelerate just as much as a rocket sitting on the launch pad (disregarding gravity). The discrepancy is due to the fact that you had to invest A LOT of energy to get that rocket and its fuel into orbit in the first place.

 

[rcgldr]

That is also why you can’t really say that rocket fuel has “100 joules”, because the energy in the rocket fuel depends on what it’s relative to.

This is also solved if the fuel ejected from the rocket is taken into consideration. The rocket fuel has a fixed amount of chemical potential energy, and as the fuel is burnt, the chemical poential energy is converted into mechanical kinetic energy in the form of an ejected fuel plume and the rocket with it's remaining on board fuel, plus the energy lost to heat and light, regardless of the frame of reference.

 

[QuantumPion]

Yes, a rocket gains more energy by firing its thrusters while moving at high relative speed (e.g. periapsis) than at low speed. This is known as the Oberth Effect.

http://en.wikipedia.org/wiki/Oberth_effect

 

[rcgldr]

Yes, a rocket gains more energy by firing its thrusters while moving at high relative speed (e.g. periapsis) than at low speed. This is known as the Oberth Effect. http://en.wikipedia.org/wiki/Oberth_effect

The key point I was making above is mentioned in that wiki article: It may seem that the rocket is getting energy for free, which would violate conservation of energy. However, any gain to the rocket's energy is balanced by an equal decrease in the energy the exhaust is left with. So the additional increase in the rockets energy is offset by a smaller increase or a decrease in the energy of the exhaust.

 

[bahamagreen]

The OP's scenario is a rocket in space... it may help to compare that to a rocket traveling through the atmosphere.

A rocket in air may reach an air speed where that air speed is equal and opposite to the speed of the thrust matter (eg, thrust exhaust speed is 2000 m/s and the rocket reaches a speed of 2000 m/s).

At that point, the rocket is basically decelerating its fuel from 2000 m/s to 0 m/s and depositing it at rest wrt the ambient air. This is the peak efficiency of the rocket.

If the rocket goes faster than this, the exhaust deposited will have a residual motion wrt the ambient air in the same direction as the rocket... the rocket's vapor trail will "follow" the rocket somewhat as it is decelerated by the ambient air, and the degree of that is a loss from peak efficiency.

This is easier to see in the ambient air example, because the air tends to decelerate the exhaust, and that effort is easy to see as wasted potential. Both the rocket and an external observer would agree on the speed at which the exhaust is being deposited at rest wrt the air.

In space, there will also be a transition from exhaust being deposited with a net "backward" movement, to a net zero, to a net "forward" movement wrt an IRF within which the rocket is accelerating, but there is no "absolute" IRF for this, so the transition is subject to the choice of IRF.
From the rockets' point of view, thrust is thrust and each and every action / reaction has the same purchase.

 

[sophiecentaur]

This is a problem I have never really understood. If a rocket engine is fired in a vacuum, does it exert the same force regardless of its velocity?

Hunter

The answer to this is YES (no one seems to have made this point, explicitly). If you put a force meter between the rocket output nozzle and the vehicle, the measured force will be constant as long as the fuel is being delivered at the same rate.
Approaching things in the right order can help in the understanding. Momentum will always be conserved so you can hang onto that if you want a good answer to any problem of this sort. When the rocket is stationary wrt Earth, all the Chemical Energy is going into the fast ejected fuel and the rocket KE is zero. (0% efficiency, to start with). As the rocket speeds up, the Power transferred to it movement will increase steadily. The KE I refer to is the energy that the rocket would expend in a collision with a stationary body, stationary in an Earth reference frame.

Factors like the decreasing mass of the rocket plus fuel payload will affect the actual velocity.

 

[sophiecentaur]

The OP's scenario is a rocket in space... it may help to compare that to a rocket traveling through the atmosphere.

A rocket in air may reach an air speed where that air speed is equal and opposite to the speed of the thrust matter (eg, thrust exhaust speed is 2000 m/s and the rocket reaches a speed of 2000 m/s).

At that point, the rocket is basically decelerating its fuel from 2000 m/s to 0 m/s and depositing it at rest wrt the ambient air. This is the peak efficiency of the rocket.

The rocket isn't 'doing this'. Because of the pressure behind the rocket in an atmosphere, there will, I agree, be a bit more thrust but most of the excess pressure is vented in the vastly greater solid angle which doesn't include the angle subtended by the rocket. Most of the work done is on expanding the volume of air in the region behind. It's not a piston in a cylinder. Only a hover vehicle uses the pressure 'back' to any serious effect.

 

[bahamagreen]

sophiecentaur,

I don't understand anything you wrote... could you elaborate?

*except the turtles, I know about that.

 

[sophiecentaur]

sophiecentaur,

I don't understand anything you wrote... could you elaborate?

*except the turtles, I know about that.

You are saying that, because the rocket gases are slowed to zero (ish), all that force is somehow 'going back' onto the rocket. The only extra force that the rocket is getting as a result of the atmosphere is the excess pressure times the area of the back end of the rocket. (I could have put it that way in the first place and it would probably have made more sense). The rocket is not "decelerating its fuel". The fuel is not connected, once it has left the nozzle.

 

[voko]

By definition, a force does work when it causes displacement of the application point along the direction of the force vector. The difference in position Δr has to be the result of the force. Any displacement not connected with the application of the force does not contribute to the work done.

There is no such definition. The work of a force over a displacement is the scalar product of that work with the displacement. No additional conditions such as "the displacement has to be caused by the force".

If it weren't so, the amount of work done would vary with the choice of reference frames

Which is does. Kinetic energy does that, too.

 

[bahamagreen]

sophiecentaur,

If the exhaust speed wrt the rocket is equal and opposite the air speed of the rocket, the air is not participating in the deceleration of the exhaust - the exhaust is being decelerated by the action inside the rocket and deposited at rest wrt the air.
The exhaust is fully decelerated at the time it leaves the rocket. The air has no role here except to establish a reference for the rocket's air speed.

The air only acts to decelerate the exhaust when it leaves the rocket when the rocket is moving wrt the air slower or faster than the speed of the exhaust wrt the rocket. That deceleration by the air is energy unavailable to the rocket.

I know this is a simplification, and maybe I was not clear, but the principle does not have any force going back into the rocket, nor any connection between the exhaust and the rocket after leaving the rocket. It is within the rocket that the deceleration of the fuel happens... which may be partial deceleration completed by the air after exit; the rocket only reacts to the exhaust decelerated by the rocket itself, that decelerated further by the air is not available as a reaction for the rocket.

 

[cjl]

Because of the pressure behind the rocket in an atmosphere, there will, I agree, be a bit more thrust.

This is exactly backwards - in an atmosphere, the ambient pressure effectively causes a thrust reduction equal to the exit area of the nozzle multiplied by the ambient pressure. A rocket generates the greatest thrust in vacuum (all other things equal).

 

[sophiecentaur]

This is exactly backwards - in an atmosphere, the ambient pressure effectively causes a thrust reduction equal to the exit area of the nozzle multiplied by the ambient pressure. A rocket generates the greatest thrust in vacuum (all other things equal).

I find this hard to square with ground effect and the operation of a hovercraft. But we may be at cross purposes here.

 

[D H]

I find this hard to square with ground effect and the operation of a hovercraft. But we may be at cross purposes here.

cjl is correct, sophiecentaur. Rockets aren't hovercraft. There is essentially no ground effect with rockets. Small rockets take off so fast that there is no time for a ground effect to set up. Big rockets are launched from a launch pad equipped with a flame trench that diverts the exhaust well away from the rocket. Once again, there's no ground effect.

A rocket generally generates the greatest thrust in vacuum. The thrust from a rocket is $F = \dot m v_e + A_e(p_e - p_{amb})$. The first term, $\dot m v_e$, is the momentum thrust from the exhaust gas across the plane where the exhaust detaches from the nozzle (typically this is the exit plane of the nozzle). The second term, $A_e(p_e - p_{amb})$, is additional thrust from the pressure difference at this plane. This second term acts to speed up (or slow down in the case $p_e < p_{amb}$) the exhaust after it has crossed the exit plane. Another way to express this is $F = \dot m v_{eff} \equiv m v_e + A_e(p_e - p_{amb})$.

To get a better handle on this pressure term, think of what happens when you blow up a balloon and then release it. The balloon flies away. The velocity of the about-to-escape air just inside the balloon is rather small, which in turn makes the $\dot m v_e$ term velocity small for a balloon. It's the pressure difference that yields most of the thrust in the case of a balloon.

 

[sophiecentaur]

I realize there's no ground effect with rockets and that the pressure under one on the launchpad is hardly affected by the presence of the ground (is the trench to increase performance or to remove the hot gases from the rocket's tail, though?). But I think you are using the assumptions about the rocket engine design to draw conclusions about the general case. You would have to (?) admit that the presence of the skirt on a hovercraft makes it work 'better' than a helicopter at that height. Won't the pressure be higher with than without a skirt, for a particular fan / rotor.
I do realize that the situation in a nozzle is not one which can be understood intuitively (as with most fluid dynamics). But, when you get down to it, doesn't the thing have to work by momentum conservation? (We had a similar long thread about how aircraft fly, in which the basic idea of momentum conservation was being ignored by a number of people, because the gas flow sums give a good enough approximation for an answer.) The pressure situation around the nozzle is an intermediate step which will govern the speed of the ejecta - hence the thrust. So, are you saying that the presence of an atmosphere is, effectively, reducing the speed of the ejecta? That would make sense to me, as long as there was some account taken of the 'downdraught' of the air that the rocket has disturbed (part of the total momentum transfer equation?). A rocket engine that works best in space would not work as well in an atmosphere but could you not design one which could be optimised in an atmosphere (a bit like improving on a jet by using a turbo fan)?

I think this thread has migrated a long way from the OP, which I don't think was really addressed at its basic level and leaped into higher things. (Rocketed away from us, in fact.)

 

[A.T.]

could you not design one which could be optimised in an atmosphere ?

http://en.wikipedia.org/wiki/Aerospike_engine

 

[sophiecentaur]

Nice.

 

[cjl]

I realize there's no ground effect with rockets and that the pressure under one on the launchpad is hardly affected by the presence of the ground (is the trench to increase performance or to remove the hot gases from the rocket's tail, though?).

Primarily to remove hot gases, but even if you did take advantage of some "ground-effect" or similar just off the launch pad, it really isn't representative of the general flight environment for a rocket (nor is it what the rocket is designed for).

But I think you are using the assumptions about the rocket engine design to draw conclusions about the general case. You would have to (?) admit that the presence of the skirt on a hovercraft makes it work 'better' than a helicopter at that height. Won't the pressure be higher with than without a skirt, for a particular fan / rotor.

Helicopters and hovercraft fly with fundamentally different mechanisms, and both are quite different from rockets, so I'm not really sure how much relevance this has to a discussion of rockets.

I do realize that the situation in a nozzle is not one which can be understood intuitively (as with most fluid dynamics). But, when you get down to it, doesn't the thing have to work by momentum conservation? (We had a similar long thread about how aircraft fly, in which the basic idea of momentum conservation was being ignored by a number of people, because the gas flow sums give a good enough approximation for an answer.)

Of course it has to work by momentum conservation. However, pressure can apply a force that's much easier to analyze without looking at the details of where every bit of momentum goes (since in a situation with unbalanced pressure with the object moving, the momentum transfer between the object and the surrounding fluid is nontrivial). As far as aircraft go, I don't really want to get into a very extensive discussion in this thread (it would drag it even farther off topic), but the "gas flow sums" as you call them (assuming you mean proper fluid mechanics) are just as good at getting an answer as a momentum balance is. Yes, the aircraft works by momentum conservation, but it also works by a pressure/force balance. The two aren't mutually exclusive - quite the opposite, they're two different ways of looking at the same overall physical phenomena, and therefore they should give the same answer (but one may be more convenient depending on your available equations and solution methods).

The pressure situation around the nozzle is an intermediate step which will govern the speed of the ejecta - hence the thrust. So, are you saying that the presence of an atmosphere is, effectively, reducing the speed of the ejecta? That would make sense to me, as long as there was some account taken of the 'downdraught' of the air that the rocket has disturbed (part of the total momentum transfer equation?).

Actually, in a supersonic nozzle, the pressure situation around the nozzle will not affect the flow in the nozzle at all (unless you have a severely overexpanded flow which separates from the nozzle). The exit speed of the gas is identical for an engine operating at sea level and in space. However, that doesn't account for everything. If you draw a control volume around your rocket engine (and let's assume that the fuel is inside the control volume for the moment, just to make things simpler), you need two things to figure out the net force on the engine. You need to know the momentum flux into and out of the system (which will be just the momentum flux out of the engine, since I'll assume we've drawn the control volume such that it doesn't include any ambient atmosphere, but rather its boundaries coincide with the outside boundaries of the engine), and you need to know of any external forces.

The force applied due to momentum will be the same from the ground to a vacuum, since (as I said earlier) in a supersonic nozzle, external disturbances cannot propagate upstream, and thus the nozzle flow is unaffected by ambient conditions. So, This term will simply give you a force equal to the mass flow rate out of the nozzle multiplied by the exhaust velocity.

However, the external forces will change from the ground to orbit. The only external force on our system will be pressure. The ambient pressure acting on most of the boundary of our system will vary from 1 atmosphere at the ground, all the way down to zero in orbit. At the nozzle though, the pressure will be whatever the exit pressure of the nozzle is (since, once again, external disturbances cannot propagate into the nozzle, so the exit conditions of the nozzle will be independent of the ambient conditions).

If you imagine this system as a box, with the exhaust coming out of the bottom surface, you have P_exit acting on the bottom of the box, and P_ambient acting on the other surfaces. All of the sides of the box will have their forces cancel out, since any pressure acting on one side will be counteracted by the pressure on the other. However, the pressure force on the top surface will be P_ambient * A[/sub]exit[/sub] (assuming the same area on top and bottom), while the pressure on the bottom surface gives a force of P_exit * A[/sub]exit[/sub]. Thus, if you define force to be positive in the direction of the rocket's motion, you get a so-called pressure thrust term equal to (P_exit - P_ambient)*A_exit.

With a bit of calculus knowledge, you should be able to convince yourself that this will still apply for an arbitrarily shaped rocket (and control volume) - the integrated pressure over the entire surface, assuming ambient pressure on every surface except the exit of the engines, will result in a pressure thrust equal to the difference in pressure between the engine's exhaust and ambient multiplied by the engine exit area.

All of this naturally leads to the relation DH stated above:

$F = \dot m v_e + A_e(p_e - p_{amb})$

The first term is typically called the momentum thrust, and the second term is called the pressure thrust. This leads nicely to your next question:

A rocket engine that works best in space would not work as well in an atmosphere but could you not design one which could be optimised in an atmosphere (a bit like improving on a jet by using a turbo fan)?

Interestingly, as it turns out (using a fair amount more fluid mechanics than I feel like going into at the moment), the maximum thrust for a given rocket engine (keeping all other things except the nozzle constant) occurs when the nozzle is sized such that P_e = P_amb - thus eliminating the pressure thrust term entirely. This is obviously impossible in space, but this is why rockets designed for space use very high expansion nozzles with very low exit pressures, while ones used at low altitudes use correspondingly smaller nozzles. In an atmosphere, however, you can design a nozzle to have very close to this ideal condition, so long as it doesn't have to operate across a wide range of altitudes. Interestingly, even an atmosphere-optimized nozzle works better in lower pressures, but it is the best you can get for atmospheric operation. I have a good diagram somewhere showing this - I'll post it if I can find it.

As for the aerospike nozzle that A.T. posted, that's not so much an attempt to optimize a rocket for atmospheric use as it is an attempt to optimize a rocket for use across a wide range of pressures. To greatly oversimplify, an aerospike basically uses the ambient pressure to affect how much the flow spreads after exiting the nozzle throat, which effectively varies the expansion ratio with ambient pressure. By varying this expansion ratio, the engine tries to keep P_e as close to P_amb as possible, across a wide range of P_amb. Ordinary nozzles have a fixed P_e determined by nozzle geometry and mass flow rate, so this could be very beneficial for a rocket that goes from the ground all the way to space while using the same set of engines. Most current rockets use multiple stages though, so the benefits of an aerospike are questionable at best (each stage simply uses engines optimized for its expected operating environment).

I think this thread has migrated a long way from the OP, which I don't think was really addressed at its basic level and leaped into higher things. (Rocketed away from us, in fact.)

I have to agree with you here (but rockets sure are interesting, aren't they?)