# Conservation of energy

1. Feb 23, 2005

### Felix83

consider 2 identical rockets, same mass and size, same engine which gives a constant thrust.

the rockets are both moving upward. the thrust of the engine is equal to the weight of the rocket (assume the mass of the fuel lost is negligable). therefore the forces are balanced so if one of the rockets is moving at v, it will continue to move at v until it runs out of fuel.

consider them at some arbitrary height and set this point as potential energy = 0. now consider the 2 rockets as they both move from this height to some greater height h. let the speed of rocket A be v, and the speed of rocket B be 10v.

since the rocket thrust equals the weight, the velocities will remain at v and 10v respectivly. therefore rocket A will take much longer to reach the height h than rocket B. if the engines of both rockets provide the same thrust, the would burn fuel at the same rate. therefore since rocket B reaches height h faster, it will burn less fuel, but gain more potential energy than rocket A.

how can this be????

mathmatically the work done by each engine will be the same since the force is the same and the distance is the same. The power would also work out as expected, the power provided by rocket B is greater since it does the same amount of work in less time.

however it seems then that rocket b will provide more power will burning less fuel...????

unless rocket B actually burns MORE fuel.....but why should a rocket engine burn more fuel just because it is going faster?

think of it like this, you have a large block with a small block attached to it. if you throw the small block away from the large block, accelerating it, you apply a force on it for a certain period of time. the force is also applied over a certian distance relative to the large block. an equal and opposite force acts on the large block for the same amount of time. this is the same regardless of how fast the system is going at the beginning.

2. Feb 24, 2005

### ramollari

The forces provided by the rocket engines will be the same, but the powers they deliver will be different $$P_b > P_a$$ as you said. So the engine in b will necessarily burn more fuel. Though the push force b provides is the same, it will be applied over a given distance in less time.
Again, in the second example you gave (though unrelated) the impulses will be equal (same force over same amount of time), not the power (depending on the distance over which the force is applied, i.e. initial speed).

Last edited: Feb 24, 2005
3. Feb 24, 2005

Looking at your rocket question, the Power provided by rocket B will be greater than the power provided by rocket A. But this is considering that you are flying them in ideal, no air friction conditions. Rocket B would NOT burn more fuel than rocket A. Also, to maintain the proper thrust for constant speed, you need to assume that the fuel has no mass, which you have done.

Regards,

Ned

4. Feb 24, 2005

### ramollari

Then where does the extra power result from?

5. Feb 24, 2005

### reilly

Think upon two elevators, rather than rockets. Play the same game, but have serious steriod-enhanced lifters, pulling the elevators up at constant v with a good stout rope-and-pully system. Now the problem is more realistic(right!). The elevators and lifters, one of each, form a closed system, rather than a sorta-adiabatic one. Who is going to run out of gas first? Charlie, the 10v guy, or Susan, the v young lady? (Both were well and comparably fed prior to execution.) Why?

Note that the burn rate for a constant speed rocket under gravity is proportional to the speed.

It's all in the wrists.

Regards,
Reilly Atkinson

6. Feb 24, 2005

### Felix83

Note that the burn rate for a constant speed rocket under gravity is proportional to the speed.

ok so your saying that if rocket a is moving at v, and rocket B has the same engine and starts at 10V, B will decellerate until he is moving at v ?

ok i think this makes sense now,
rocket As engine provides a certain thrust for a certain time period, while working against gravity. the gravity does work for a certain distance. if rocket B is moving faster, under the same time period, gravity does more work because the rocket travels a longer distance, therefore the engine has to do more work in the same time period, hence burn more fuel.

7. Feb 24, 2005

### reilly

Sounds good to me. (But, I don't get the 2nd paragraph. Why decellerate?)
A very clever question, indeed.

Regards,
Reilly Atkinson

8. Feb 24, 2005

### Romperstomper

Think of driving a car down the highway, you have to push the gas pedal down more to travel at 65 mph than you do to travel at 25 mph.

I think what he means about the decellerating back to v is if engine b starts buring the same amount of fuel as engine a, rocket b will decellerate back to v.

9. Feb 24, 2005

### kanato

I think you're right, rocket B would burn less fuel to reach a given height, travelling at constant velocity, in the absence of any velocity-dependent frictional force (ie. air resistance). We assume that the force from the thrust is proportional to the rate at which fuel is burnt, and since both are providing the same force ($$+mg\hat{z}$$), both are burning fuel at the same rate, so in the same amount of time, the same amount of fuel is burnt, and rocket B travels higher. But it's somewhat artificial; if we're going to have a constant gravitational field such as that near the surface of the earth, then we should probably think about air resistance, which will have a much greater impact on the faster travelling rocket. And we also haven't considered the extra amount of fuel rocket B would require to push its velocity up to 10 times the velocity that rocket A has.

10. Feb 25, 2005

### Felix83

yea the decellerating part, i meant if they were both burning fuel at the same rate, sorry i left that out

after what kanato said, i thought about it some more and now im not so sure. consider the decelerating situation. both rockets burn fuel at the same rate. rocket A starts at v and rocket B starts at 10V. assume rocket A provides enough thrust to continue at constant velocity. in order for B to decellerate to v, the forces must be unbalanced - the force of gravity must be greater than the force of thrust of the rocket engine. with rocket A, the thrust is equal to the weight. both rockets have the same mass so therefore, if the rocket engine in B is weaker than gravity, rocket A must provide more thrust than B. However, they are both still identical rocket engines burning fuel at the exact same rate.

How can a rocket engine produce less thrust force just because it is travelling faster?

You take a certain rocket engine and have it bolted down and tested while it is stationary, say the thrust is 1000N. You take the same rocket and put it on a wheeled cart with negligable friction with an accelerometer, fire it, and calculate the thrust at 100mph. You find that it is 700N. This would have to be possible in order for Rocket B to burn more fuel in the original situation. How can this be?

11. Feb 25, 2005

### reilly

All you need to know is Newton's 2nd law applied to a constant speed rocket -- it's stipulations are quite unambiguous.

Regards,
Reilly Atkinson

12. Feb 26, 2005

### Duarh

This is a reference frame problem. Energy is not the same in all reference frames. Both rockets will indeed burn fuel at the same rate. The fuel (burning gas) exiting the two rockets will have the same speed _relative to the particular rocket_ - but, in the Earth's reference frame, the speed will be much greater for the 10v rocket. So, from the Earth's point of view, the 10v rocket will be losing more energy per unit of time than the 1v rocket - even though the two rockets are burning the same amount of fuel. In both the rocket's own frames of reference, though, the value of the rate of energy loss is the same.

When we say energy is conserved, we mean it's conserved in a particular reference frame. The total energy of a system certainly isn't the same value if you decide to go to another reference frame (think a moving car - in its own reference frame, it has no kinetic energy).

13. Feb 26, 2005

### kanato

But the forces would be the same in different reference frames, provided we are not anywhere near relativistic speeds. In the earth's reference frame, there is a force of mg pulling down, and each rocket has to provide the same thrust pushing it up.

If both rockets have the same engines and masses, then in order for rocket B to decellerate it seems to me that the driver would have to reduce the engine thrust, or turn it off entirely, so that the forces would be unbalanced on rocket B.

If we're looking at the decellerating situation, the engine has to put out less thrust to slow down.. ie, the pilot has to try to slow down. It has nothing to do with the velocity the rocket is travelling, but just with the fact that the pilot wants to slow it down.

I think the thrust will still be 1000N, unless we are taking air resistance into account, which could be pretty significant at 100mph.

14. Feb 26, 2005

### Duarh

Absolutely, forces remain the same. That doesn't mean that the rate of energy loss remains the same. A force of 10 N will add more Joules of energy per second to an object moving at 100 m/s than to one at 10/s, for instance. (101^2-100^2>11^2-10^2). The rate of energy loss is the same _with respect to distance_ - that's the whole reason why the work-energy theorem is useful. The amount of energy lost per distance is the same for both rockets from the Earth's POV - but one loses it faster (since it's going faster), the other slower.

15. Feb 26, 2005

### RandallB

Why are you assuming they will both reach the same height ??? How could that be done??

Consider adding a third rocket to the mix with NO starting V at all.
Still sitting on the launch pad. Same deal thrust = weight so pad no longer has to hold the weight of the the third rocket. All three engines do the same thing - maintain the same acceleration on all three rockets that happens to be the same as the G acceleration they all feel from gravity. It would be as if you had turned gravity off for a while and they will move as if in space without the engines every being fired.
Fuel releases the same energy, in the same time. Third roket never moved. - SO More than change the height must be involved in accounting for all of it.

Also: You spend the day working at lifting and carring heavy boxes from one end the building to the other. Boss figures since the floor is the hieght you didn't really do any work so why should you be paid??
Maybe OK not to pay you if you didn't use any energy.

16. Feb 26, 2005

### reilly

This is NOT a "reference frame" issue. This is a relatively standard freshman physics (advanced wih calculus) problem -- as I mentioned before, just use Newton's 2nd Law. If one insists upon identical rockets, then the problem is not well-set, as mathematicians say- unless there's a throttle of some sort to control the rocket dynamics. With no adjustment mechanism, the problem is much like, "How much dirt can Phil dig out in 15 minutes from a hole 6'x6'x10'

RandallB got it.

Regards,
Reilly Atkinson

17. Feb 26, 2005

### Duarh

Hmm, I think I see where I got confused. . .the forces exerted by the rocket on the fuel and vica versa are just internal forces to the rocket-fuel system - the center of mass of the whole system will still fall or decelerate in the gravitational field the same as if there was no thrust present - and, of course, if the center of mass has a higher initial velocity, the system will acquire more potential energy per second than one going at lower velocity (it would be the same if you had two balls shot up in the air - if at the same instant one of them was going at 1v, the other at 10v, the 10v one would be losing more energy per second due to gravity). In this sense it's still the kind of reference frame issue I was referring to (that energy is not linearly dependent on velocity), though, of course, what I said in my previous posts didn't address the problem correctly. (To be clear, I was never implying that changing the reference frame changed the physics, but only that the change in kinetic energy you observe in the motion of any object when it is accelerated depends on your velocity with respect to that object).

So when the 10v rocket's center of mass slows down to 0v, it will be way higher than the 1v rocket's CM (so the formerly-10-v system will have way more potential energy) because of the energy contained in its initial velocity, not the burning of the fuel - it will have burned fuel at the same rate, and that energy (rate * time elapsed) will be reflected in the kinetic energy of the system's parts (gas + ship) at that point.

Is this what you were driving at?

Last edited: Feb 26, 2005