# Does a Rocket Engine Exert Same Force Regardless of Velocity?

1. Jul 15, 2013

### 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 space ship travelling 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 travelling 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

2. Jul 15, 2013

### 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

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.

Last edited: Jul 15, 2013
3. Jul 15, 2013

### 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.

4. Jul 16, 2013

### 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?

5. Jul 16, 2013

### rcgldr

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.

Last edited: Jul 16, 2013
6. Jul 16, 2013

### 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.

7. Jul 16, 2013

### rcgldr

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.

8. Jul 16, 2013

### 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

9. Jul 16, 2013

### rcgldr

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.

Last edited: Jul 16, 2013
10. Jul 19, 2013

### 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.

11. Jul 19, 2013

### sophiecentaur

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.

12. Jul 19, 2013

### sophiecentaur

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.

13. Jul 19, 2013

### bahamagreen

sophiecentaur,

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

*except the turtles, I know about that.

Last edited: Jul 19, 2013
14. Jul 19, 2013

### sophiecentaur

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.

15. Jul 19, 2013

### voko

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".

Which is does. Kinetic energy does that, too.

16. Jul 19, 2013

### 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.

17. Jul 19, 2013

### cjl

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).

18. Jul 22, 2013

### sophiecentaur

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

19. Jul 22, 2013

### D H

Staff Emeritus
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.

20. Jul 22, 2013

### sophiecentaur

I realise 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 realise 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.)