Why Does Kinetic Energy Increase So Much in a Constantly Accelerating Rocket?

[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?

 

[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.

 

[Dale]

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.

 

[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!

 

[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.

 

[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.

 

[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.

 

[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.

 

[qraal]

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.

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.

 

[Dadface]

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.

 

[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.

 

[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.

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.

 

[qraal]

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.

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.

 

[Dale]

Obviously I'm missing something basic here.

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.

 

[D H]

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."

 

[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.

 

[D H]

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!

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.

 

[Dadface]

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!

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.

 

[Dadface]

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.

I think some of us are talking at cross purposes here. I agree with your analysis of a rockets motion with a constant thrust but the OP was referring to a constant power not thrust.

 

[D H]

For a rocket, constant power means constant thrust.

 

[Bob_for_short]

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 intervals of time.

Absolutely correct! I added a couple of phrases to my post.

 

[Dadface]

For a rocket, constant power means constant thrust.

Clearly they are not the same thing,thrust is a force and is measured in Newtons and power is rate of doing work and is measured in Watts

 

[rcgldr]

So how does the exact same car gain 385.8 joules/kg in one frame and 3472.2 joules/kg in another?

When considering the power peformed by an engine of some type, the frame of reference should be related to the point of application of force to some external object. In the case of a car, the point of application of force is at the road surface, so the road the car drives on should be the frame of reference. If the car were to accelerate a short distance on a long flat bed, then the surface of that flat bed should be the frame of reference.

A rocket engine is a special case, because outside of the atmosphere, the rocket does not interact with any external objects, but instead relies on an internal interaction where part of it's own mass is accelerated and expelled at high speed.

 

[D H]

Clearly they are not the same thing,thrust is a force and is measured in Newtons and power is rate of doing work and is measured in Watts

Clearly so. However in the case of a rocket, a rocket with constant power output will indeed have constant thrust.

 

[D H]

When considering the power peformed by an engine of some type, the frame of reference should be related to the point of application of force to some external object.

Only because the math is easiest in this frame. The only thing that prevents me from modeling the behavior of a car accelerating down a road on the surface of the Earth from the perspective of a Neptune-centered, Neptune-fixed frame of reference is that the math becomes ridiculously convoluted from this perspective.

 

[Dadface]

Clearly so. However in the case of a rocket, a rocket with constant power output will indeed have constant thrust.

I am assuming this is a property characteristic of the rocket engine.Consider the main energy changes for two consecutive equal time intervals.For the first interval power times time equals gain of KE(rocket plus gases) and for the second interval the same power times the same time equals an even greater gain of KE.Where does the extra energy come from during the second interval?

 

[D H]

Consider the main energy changes for two consecutive equal time intervals.For the first interval power times time equals gain of KE(rocket plus gases) and for the second interval the same power times the same time equals an even greater gain of KE.Where does the extra energy come from during the second interval?

When one looks at the rocket+exhaust gases as a system, there is no extra energy during the second interval. The fourth post in my simple rocket tutorial, https://www.physicsforums.com/showthread.php?t=199087 investigates rocket behavior from the perspective of conservation of energy.

 

[Dadface]

When one looks at the rocket+exhaust gases as a system, there is no extra energy during the second interval. The fourth post in my simple rocket tutorial, https://www.physicsforums.com/showthread.php?t=199087 investigates rocket behavior from the perspective of conservation of energy.

Thanks for the exchanges I am enjoying it and I will look at your tutorial and hopefully learn something new.If I see something amiss I will be back and hopefully get some clarification.Unfortunately I've got to pop off soon, my wife wants to use the computer.

 

[Dale]

Where does the extra energy come from during the second interval?

Again, you are ignoring the exhaust. When you consider the KE of the exhaust you see that the KE of the rocket+exhaust increases at a constant rate equal to the power of the engine. See the tutorial I linked to earlier for details.

 

[Bob_for_short]

Where does the extra energy come from during the second interval?

From the kinetic energy of engine of any nature. In a moving RF it is higher.

 

[D H]

No, it isn't. You are insisting on ignoring the exhaust, Bob.

 

[Bob_for_short]

If an engine pulls the probe mass with a rope, then there is no exhaust. I am speaking of a general case.

 

[Dale]

By the way, to the OP, although it is certainly possible to analyze a rocket motion from the perspective of the conservation of energy it is too easy to neglect the energy of the exhaust as you and others have done. Therefore it is my recommendation to always work rocket motion problems from the perspective of the conservation of momentum. You will always get the same answer, but the conservation of momentum approach forces you to consider the exhaust.

 

[Bob_for_short]

Maybe but what if I want to consider a body of a constant mass? Why should I reduce the generality? The paradox of the problem is in neglecting the kinetic energy of the engine. It is the true source of the additional power. Power is a frame-dependent quantity.

 

[Dale]

Yes, Bob_for_short, I know, I am only talking about common mistakes made by beginning students when analyzing rockets and a simple way to avoid them. The OP's question is not about general engines, but specifically about rocket engines. I recommend that students use conservation of momentum for rocket problems whenever possible, simply because it avoids mistakes.

I am sure that you, Bob_for_short, will do fine and avoid mistakes using other approaches that will cause problems for students. So feel free to ignore my recommendation as it does not apply either to you nor to general engines.

 

[rcgldr]

If an engine pulls the probe mass with a rope, then there is no exhaust. I am speaking of a general case.

If there is a rope in involved then you have interaction with an external object, the point of appllication of force is upon the rope, and with constant power the rate of force will decrease with speed with respect to the rope according to the equation power = force x speed. This changes the situation from the rocket and its spent fuel, which is a closed internal interaction that increases kinetic energy of both rocket and fuel, but does not change the center of mass or the total momentum of the rocket and fuel system.

In the rocket case, the point of application is between the engine and the remaining on-board fuel. Regardless of the speed of the rocket and onboard fuel relative to some frame of reference, the engine and onboard fuel always have zero relative speed. For a given engine at a specific throttle setting the mass flow, thrust, power, and terminal velocity of the spent fuel relative to the rocket is constant, and the kinetic energy of the total system of rocket and spent fuel increases linearly with respect to time, regardless of the frame of reference.

 

[Bob_for_short]

Let me put it in a simple way; the power P is a frame- or velocity-dependent quantity:
P = Fv.

The force and acceleration are Galilean invariants but velocity is not. So in different RFs the powers are different.

Or in the same RF but at different time moments - when the body is moving slowly or quickly, the velocities are different, so are the powers.

 

[rcgldr]

Let me put it in a simple way; the power P is a frame- or velocity-dependent quantity: P = Fv.

Power can be calculated to avoid making it frame dependent. The rate of consumption of potential chemical energy of the fuel times an effeciency factor allows power to be calculated independent of the frame of reference. The work peformed (change in KE) per unit time on spent fuel and rocket is constant (for a given throttle setting) regardless of the frame of reference and would provide another means to calculate the power independent of the frame of reference.

Power is defined as a rate of work, from which the force x velocity form can be derived. In the case of a rocket, in order to use the force x velocity form, velocity should be defined based on the change in velocity of the fuel and rocket, or change in velocity of the fuel with respect to the rocket, which are frame independent.

DH's rocket tutorial thread covers this:

Power
The time derivative of the total mechanical energy simplifies to

\dot T_{r+e}(t) =<br /> \frac 1 2 \dot m_e(t) v_e(t)^2<br />

https://www.physicsforums.com/showthread.php?t=199087

 

[trivia1]

The following, regardless of the KE of the spent fuel being ignored in the calculation, is correct.

A rocket in space, of mass 1kg, accelerates at 1m/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. KE is relative to an observer stationary at t=0.

I’m assuming the engine applying the force is doing so by firing out minute particles at extremely high velocities, so as any loss in rocket mass –that is unspent fuel- is negligible. This is an ideal scenario in a Newtonian universe, so the particles can be as small as we wish them to be and the velocities as high as we decide. It just has to be consistent with Newton’s physics.
Gaining an understanding of the physics of rocket propulsion –interesting as it is- is not why I posted. It’s the unintuitive nature of the relationship between KE and V squared that I was trying to grasp. Bob for Short has explained that Power is relative to Velocity. That would mean that for an observer at rest in relation to the Rocket at t=0, at t=1000 that rocket’s engine is massively more powerful. That seems more than a little odd, but it does give me a new perspective. It means that an engine exerting a force on a mass, whether in deep space or on a road on Earth, is more powerful the higher it’s velocity. To a layperson in physics, that’s plain strange.
Is it correct to state that as an engine’s velocity increases, it’s power output in reference to an initial observer increases exponentially?

 

[A.T.]

It means that an engine exerting a force on a mass, whether in deep space or on a road on Earth, is more powerful the higher it’s velocity.

power = force * velocity

 

[qraal]

The following, regardless of the KE of the spent fuel being ignored in the calculation, is correct.

A rocket in space, of mass 1kg, accelerates at 1m/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. KE is relative to an observer stationary at t=0.

I’m assuming the engine applying the force is doing so by firing out minute particles at extremely high velocities, so as any loss in rocket mass –that is unspent fuel- is negligible. This is an ideal scenario in a Newtonian universe, so the particles can be as small as we wish them to be and the velocities as high as we decide. It just has to be consistent with Newton’s physics.
Gaining an understanding of the physics of rocket propulsion –interesting as it is- is not why I posted. It’s the unintuitive nature of the relationship between KE and V squared that I was trying to grasp. Bob for Short has explained that Power is relative to Velocity. That would mean that for an observer at rest in relation to the Rocket at t=0, at t=1000 that rocket’s engine is massively more powerful. That seems more than a little odd, but it does give me a new perspective. It means that an engine exerting a force on a mass, whether in deep space or on a road on Earth, is more powerful the higher it’s velocity. To a layperson in physics, that’s plain strange.
Is it correct to state that as an engine’s velocity increases, it’s power output in reference to an initial observer increases exponentially?

Power increases, from one point of view, but from the rocket's and the exhaust's perspective nothing changes - so long as the mass-flow is negligible. Power is the time derivative of energy. As a rocket gets faster so too its kinetic energy gets higher, but at the square of the velocity. So you'd expect the power, as time derivative (the instantaneous rate of change) of energy to increase quickly too, in step with the energy. There's nothing really strange about that is there?

 

[D H]

A rocket in space, of mass 1kg, accelerates at 1m/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. KE is relative to an observer stationary at t=0.

Gaining an understanding of the physics of rocket propulsion –interesting as it is- is not why I posted. It’s the unintuitive nature of the relationship between KE and V squared that I was trying to grasp. Bob for Short has explained that Power is relative to Velocity. That would mean that for an observer at rest in relation to the Rocket at t=0, at t=1000 that rocket’s engine is massively more powerful. That seems more than a little odd, but it does give me a new perspective. It means that an engine exerting a force on a mass, whether in deep space or on a road on Earth, is more powerful the higher it’s velocity. To a layperson in physics, that’s plain strange.
Is it correct to state that as an engine’s velocity increases, it’s power output in reference to an initial observer increases exponentially?

One explanation for this apparent paradox is that energy is a frame dependent concept. During that first second of operation, an observer initially at rest with respect to the rocket will see your rocket's energy change by 0.5 joules. Compare this to the change in energy as observed from the perspective of someone moving toward the accelerating rocket with a constant velocity of 999 m/s: A change of 999.5 joules.

An even better explanation, IMO, is that the paradox arises from looking at an open system. This view can make it look like rockets violate conservation of energy. One answer to this: There is of course no expectation that energy is conserved in an open system.

When you look at the rocket+exhaust cloud as a whole, the change in energy *is* frame-independent. A rocket in a pure vacuum and far removed from any gravitational bodies is not subject to any external forces. The rocket combined with the exhaust cloud the rocket leaves behind form an isolated system. The only change in the kinetic energy of the rocket+exhaust cloud system arises from converting the potential energy of the fuel to kinetic energy, and this change is frame-independent.

 

[Dadface]

trivia, forget rockets for a moment and look at another simpler case which might appeal to your intuition,an object in free fall in a vacuum.In this case the force on the object remains constant, its kinetic energy increase in the way you expect it to but then so does its power increase and all the sums add up.In this and other examples the force remains constant but the power(Fv) changes.
Rockets seem to be an exception to this and we have been told that with these that if the force remains constant the power remains constant also"the KE of the exhaust gases plus rocket increase at a constant rate".In my opinion this makes rockets really interesting and like yourself I am struggling a bit to get this concept in my head in an intuitive(and in my case non mathematical) way.I think I am beginning to gain an understanding but I need to look again at the excellent tutorials posted by DH.It's the hard stuff and the stuff which promotes discussion which makes physics such an interesting subject.

 

[Dale]

regardless of the KE of the spent fuel being ignored in the calculation

trivia1, it is really annoying, and frankly pretty rude, for you to come here and ask a question and then deliberately ignore the answer you received multiple times from multiple people. If you were not interested in hearing the responses and learning the resolution to your problem then why did you bother to post the question in the first place? All you have accomplished is to waste our time and make us feel foolish for bothering to try to help someone who didn't really want our help.

 

[trivia1]

trivia1, it is really annoying, and frankly pretty rude, for you to come here and ask a question and then deliberately ignore the answer you received multiple times from multiple people. If you were not interested in hearing the responses and learning the resolution to your problem then why did you bother to post the question in the first place? All you have accomplished is to waste our time and make us feel foolish for bothering to try to help someone who didn't really want our help.

By ignoring the KE of the spent fuel I hoped to simplify the model to gain some sort of insight. Bob_for_short explained the increase in KE by referring to the force acting over distance, which also ignores the spent fuel.
I'm not deliberately ignoring answers, I've read the posts, I simply don't get it intuitively. Mathematically it makes sense, but the concepts the maths support are difficult for me to grasp.
Why anyone would feel foolish for trying to answer a question is also a puzzle to me. If you've answered it, and I haven't got it, why feel foolish? People continue to post answers, that go along the same lines as the first, because they make the perfectly reasonable and correct assumption that it’s a genuine puzzle for me, and they know I’m not deliberately ignoring anything. If you don’t make that assumption, don’t post.

 

[Dale]

If you ignore the exhaust then your system is not closed and energy is not conserved.

 

[D H]

By ignoring the KE of the spent fuel I hoped to simplify the model to gain some sort of insight. Bob_for_short explained the increase in KE by referring to the force acting over distance, which also ignores the spent fuel.

Bob_for_short did a disservice. Rockets work by transferring momentum to space, and they do this by ejecting mass. Ignore that and you will come up with an apparent paradox. The paradox vanishes as soon as you realize

• The rocket by itself is an open system. Open systems do not in general conserve mass, linear momentum, angular momentum, or energy. Combining this with the fact that energy is frame-dependent leads to a big fat "So what!"
• OR you could look at the rocket+exhaust as a closed system. The paradox disappears: mass, momenta, and energy are all conserved.

However, a new apparent paradox appears in the form of Tsiolkovsky rocket equation. You want to accelerate a rocket from rest to three times the exhaust velocity? No problem. You just need the initial mass (vehicle+fuel) to be 95% fuel. Six times the exhaust velocity? Big problem. Now the initial mass needs to be 99.75% fuel.

 

[rcgldr]

The other issue you're ignoring is the point of application of that force. A car applies a force to the pavement it moves on, an airplane applies a force to the air it moves through, in these cases, the power is related to the force times speed at the point of application of that force. However a rocket doesn't interact with the space it moves through. Instead the force is generated internally, by expelling a bit of itself backwards at high speed. In this case the point of application of force is at the rocket nozzle, any remaining onboard fuel is accelerated along with the rocket, and the power generated is a function of how much and how fast the onboard fuel is accelerated, not related to the rockets speed relative to some other object or frame of reference in space.

 

[qraal]

The other issue you're ignoring is the point of application of that force. A car applies a force to the pavement it moves on, an airplane applies a force to the air it moves through, in these cases, the power is related to the force times speed at the point of application of that force. However a rocket doesn't interact with the space it moves through. Instead the force is generated internally, by expelling a bit of itself backwards at high speed. In this case the point of application of force is at the rocket nozzle, any remaining onboard fuel is accelerated along with the rocket, and the power generated is a function of how much and how fast the onboard fuel is accelerated, not related to the rockets speed relative to some other object or frame of reference in space.

Doesn't matter. The rules still rule regardless of the reference frame one picks. What doesn't work is comparing a measurement in one reference frame against a measurement in another and then wondering why they apparently disagree.

 

[FoxCommander]

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?

well consider the speed that the object is going at t=999, it is going 1998 m/s. which is pretty fast. the amount of work done by the rocket is the force of the rocket * the distance traveled. If it is going that fast for one second, not even counting acceleration, its distance will be 1998 meters (with acceleration it turns out to be 1999), that's almost 2Km in just 1 second! Now multiply that to the force, which is 2N and you get 3998Joules. which is also a very big amount of energy, for the size.

You can see why even a small force over a great distance can become a lot of energy. this is why the difference is so big.