# Where'd the energy come from?

1. Nov 7, 2009

### trivia1

A rocket in space, of mass 1kg, accelerates at 2m/s squared. Between t=0 and t=1 it's change in KE is 0.5j, between t=999 and t=1000 it's change in KE is 999.5j. Yet the rocket motor power output hadn't changed. What explains the massive difference in KE transferred to the rocket?

2. Nov 7, 2009

### Danger

Welcome to PF, Trivia1.
I know nothing of math, so I'm just going to assume that your numbers are accurate. Consider that the speed, and thus KE, of the rocket is not increasing linearly. Speed builds up to some pretty ferocious levels given enough constant acceleration.

3. Nov 7, 2009

### Staff: Mentor

Hi trivia1, welcome to PF

D H's excellent rocket tutorial page explains it in detail: https://www.physicsforums.com/showthread.php?t=199087

The broad outline is that you need to consider the KE of the exhaust also. When you do that you find that the KE of the system (rocket + exhaust) changes at a constant rate equal to the power of the rocket motor.

4. Nov 7, 2009

### Sillyboy

from your data, I calculated it. It's reasonable, we can know that it is an accelaration process. You know, as the rocket accelarates, the traction decreased, so the resistence should decrease too if the mass of the rocket is constant. but as we know when the velocity increases, the resistance should increase too.so I think the mass of th rocket has decreased! It is just the KE of the decreased exghaust that contributes to the increased KE of the rocket!

I am sorry for my poor English!

Last edited: Nov 7, 2009
5. Nov 8, 2009

### trivia1

Thanks for the welcome Danger and DaleSpam
My difficulty with comprehending why there is such a large difference in KE growth over the two periods lies with the relationship between KE and Velocity squared. I've done a google search on this dynamic and the rocket model is the one generally used to try and explain it. I still don't get it. Here's the same problem, put in a different way. A rocket has a mass of 1kg and a velocity of 999m/s to observer A, and has a velocity of 0 to observer B. It accelerates at 2m/s squared. In the following second it gains 999.5j of KE in ref to A, and 0.5j in reference to observer B. It's the same rocket, same power. Obviously I'm missing something basic here.

6. Nov 8, 2009

### Danger

I see what you mean now. This is a problem of relativity. For the conditions that you've given, observer B must be moving in the same direction and at initially the same speed as the rocket.
Think of it as a gun problem as opposed to a rocket one (although the acceleration is eliminated). If you are standing still in relation to a shooter, and he shoots you, that bullet will impact you at 1,000 ft/sec (as an example; there's a vast range of ammo). If, on the other hand, you are running away from the shooter at 995 ft/sec, the bullet will eventually catch you and impact at 5 ft/sec. It wouldn't even leave a bruise.

7. Nov 8, 2009

### trivia1

In the examples above the acceleration should be 1m/s squared.
Danger, I understand that KE is relative to an observer, but what I don’t get is how the magnitude of the change in KE is related to the initial velocity. At extreme initial velocities the gain in energy for even slight increases in velocity is huge. A lkg rocket with initial V of 100000m/s has an engine applying a force of 1N for 1s. It gains 100000j of KE. That’s in addition to the KE the rocket already possessed. If it’s initial V is 0, change in KE is 0.5j. Yet the rocket engine converted the same amount of chemical energy in both cases. That’s the bit that confuses me. In one frame of reference huge gain, another frame of reference a tiny gain.

8. Nov 8, 2009

### rcgldr

You seemed to have missed DaleSpams point. The rocket engine is increasing the KE of both fuel and rocket, and the rate of increase in KE is constant if the rocket engine is producing constant thrust at the same mass flow rate of spent fuel.

Rocket propulsion relies on ejecting a part of its own internal mass (spend fuel) for propulsion. If no external forces are invovled, then note that the center of mass of the rocket and it's spent fuel never moves (regardless of the frame of reference).

If the frame of reference is the rockets initial velocity, then all of the starting increase in KE is going into the fuel. As the rockets speed increases, the KE of both the rocket and it's remaining fuel are increased, as well as the spent fuel. Eventually the rocket can reach a speed where it's moving faster than the terminal exhaust veolicity of the spent fuel, in which case the KE of the fuel being ejected is being decreased by the engine, relative to that original frame of reference where the rocket wasn't moving.

9. Nov 8, 2009

### qraal

You need to account for the kinetic energy of the exhaust required to produce the thrust. You'll find the kinetic energy all balances then. For example, say your rocket engine has an exhaust velocity, u. For a given thrust, T, the mass-flow rate μ = T/u. What's the kinetic energy of the exhaust? Starting at rest it's obvious, 1/2.μ.u2. But what about when you're at a speed v? The rocket, mass m, moves forward at v+T/m, while the exhaust jet goes backwards at (v - u) because it's pointed in the opposite direction to which the rocket is being propelled forward. Thus the exhaust's kinetic energy is 1/2.μ.(v-u)2 and the rocket's is 1/2.m.(v+T/m)2.

In sum: The mass ejected backwards loses kinetic energy while the mass moving forwards gains it. Jet power thus can rise even when the jet's exhaust velocity remains the same, relative to the rocket, the whole time.

Last edited: Nov 8, 2009
10. Nov 8, 2009

To get a constant acceleration you need a constant resultant force not a constant power.If the force remained constant then as the velocity increases the power (force times velocity) must increase also.

Last edited: Nov 8, 2009
11. Nov 8, 2009

### trivia1

Is the following correct?
At extreme initial velocities the gain in kinetic energy for even slight increases in velocity is huge. A l kg rocket with initial V of 100000m/s has an engine applying a force of 1N for 1s. It gains 100000j of KE. If it’s initial V is 0, and an engine applies a force of 1N for 1s the change in KE is 0.5j. Yet the rocket engine converted the same amount of chemical energy in both cases.

12. Nov 8, 2009

### qraal

I'm not so happy with my explanation, so I'll do a couple of expansions to illustrate what's going on a bit better.

Before an impulse the rocket + fuel's kinetic energy is 1/2(m+μ).v2, relative to a stationary observer. The potential energy of the fuel becomes kinetic energy and the tiny mass of fuel is propelled rearwards at speed (v-u), relative to the stationary observer.

But ask yourself: what is the rocket's speed relative to the rocket?

Prior to the impulse from the exhaust, by Galilean relativity, the speed is zero, then after the impulse, relative to that initial state, it gains by some small acceleration equal to the thrust/rocket-mass. And that's always true.

The confusion comes from comparing what a co-moving observer sees (constant jet-power) in the rocket's frame, and what a stationary observer sees the kinetic energy of the rocket to be. You just can't compare the two frames like that without confusing yourself.

So what does a stationary observer observe the jet-power to be when a rocket is in motion? Well the fuel packet starts with a kinetic energy of 1/2.μ.v2. It burns, expands in the combustion chamber and then exists at a speed (v-u) relative to the stationary observer. Thus the difference between before and after is 1/2.μ.(v-u)2 - 1/2.μ.v2 = 1/2.μ.u2 - μ.v.u.

Now 1/2.μ.v2 is obviously the initial kinetic energy of the propellant in the rocket's frame, but what is - μ.v.u? Oddly enough it's mirrored when we expand out the kinetic energy of the rocket, before and after...

the speed increment is μ.u/m, so after the impulse the rocket's KE is 1/2.m.(v + μ.u/m)2. Then the difference before and after is μ.v.u + (μ.u)2/2m. The second term (μ.u)2/2m is (Thrust)2/2m, which is the jet-power.

So what is μ.v.u? Well KE = 1/2.m.v2, thus d(KE)/dt is m.v.(dv/dt)... and in this case (dv/dt) is μ.u/m. That means μ.v.u is your "extra" kinetic energy and it was hiding in the maths the whole time.

13. Nov 8, 2009

### qraal

The propellant has gained kinetic energy along with the rocket that contains it. Its change in energy when it's burnt has to take that into account else you'll end up with this apparent paradox. When you do the maths it all adds up.

That being said it does tell you why ion rockets have such pitiful thrust levels for seemingly quite high power levels. For an exhaust velocity of 100,000 m/s you need 50 kW of power for every measly N of thrust, with perfectly efficient power conversion. Inefficiencies in power generation and powering the jet means an ion-rocket can't lift off from a planet with decent gravity. Sufficient power would melt the rocket from waste heat alone.

14. Nov 8, 2009

### Staff: Mentor

Yes, you are still neglecting the KE of the exhaust. Please read the tutorial and always do your analysis including the KE of the exhaust.

15. Nov 8, 2009

### D H

Staff Emeritus
The exhaust is important to understand real rockets. However, this apparent paradox is not limited to rockets. Cars can easily accelerate at 2 m/s2. The Bugatti Veyron, for example, accelerates from 0 to 100 km/h in 2.5 seconds (13.89 m/s2 average acceleration) and has a top speed of over 400 km/h. To a person standing on the ground, a Veyron starting from rest gains a specific kinetic energy of 385.8 joules/kg ($(100\,\text{km/s})^2/2$) in 2.5 seconds. From the perspective of another Veyron racing at a constant 400 km/h toward the accelerating Veyron, the accelerating Veyron gains a specific kinetic energy of 3472.2 joules/kg ($\left((500\,\text{km/s})^2-(400\,\text{km/s})^2\right)/2$) in that same 2.5 second interval.

So how does the exact same car gain 385.8 joules/kg in one frame and 3472.2 joules/kg in another? Where does that extra 3086.4 joules/kg come from? The car has to burn some fuel to accelerate. To the stationary observer, the energy of that fuel is (initially) purely potential energy. To the moving car, that same fuel has a lot of kinetic energy in addition to its potential energy. That extra 3086.4 joules/kg is, as qraal put it, "hiding in the maths the whole time."

16. Nov 8, 2009

### Bob_for_short

The work done by the engine is the force ma multiplied by the distance. That is the key to the answer. In the first case the distance is much smaller than in the second case. Than means, in order to keep the same acceleration the engine should do much more work in one second. So the engine output is in fact much larger in the second case. Surprise! The power is a frame-dependent thing. Fortunately a fast engine has some extra energy to spent.

Last edited: Nov 8, 2009
17. Nov 8, 2009

### D H

Staff Emeritus
This can lead you down a dangerous path, which is to conclude that a rocket's acceleration must decrease as it gains speed. This is after all exactly what happens with an automobile. This is not what happens with a rocket. In fact, the exact opposite is the case: The acceleration of a rocket with a constant thrust increases as fuel is burnt. This increased acceleration can be harmful to occupants of the rocket. For example, the Space Shuttle commences a "3-g throttle down" at about 7 minutes and 40 seconds into launch to compensate for this tendency of acceleration to increase as rocket mass decreases.

What is happening here is that you are ignoring the energy of the exhaust, Bob. If you take the energy transferred from the rocket proper to the exhaust it is clear that the engine's energy output can indeed be constant as posited in the original post.

18. Nov 8, 2009

Agreed and this is the point I was making in post number ten.Take a simple example where it is not necessary to consider efficiencies of engines,exhaust gases and so on,a mass being pulled across a smooth table by a string attached to a falling mass.The main energy conversion here is GPE to KE.The power input from the falling mass does not remain constant because it falls increasing distances in successive equal intervals of time.

Last edited: Nov 8, 2009
19. Nov 8, 2009