A misunderstanding of the work equation?

[Mandlbaur]

If a rocket launches from earth, and lets say that it moves up 5m in the first second, then it will do work of force times distance - let's call the amount w. If we wait until it reaches a velocity of 100m/s, then we can calculate the work being done over that particular one second as 20w but the energy expenditure is the same.

 

How is this possible ?

[swansont]

If a rocket launches from earth, and lets say that it moves up 5m in the first second, then it will do work of force times distance - let's call the amount w. If we wait until it reaches a velocity of 100m/s, then we can calculate the work being done over that particular one second as 20w but the energy expenditure is the same.

 

How is this possible ?

 

 

You haven't actually shown that your scenario is correct, but in general, rocket engines are not 100% efficient at converting stored energy into the rocket's kinetic energy. There is also KE in the exhaust, and this efficiency is lowest at low speeds (and zero when the rocket is at rest)

 

https://en.wikipedia.org/wiki/Rocket#Energy_efficiency

[Mandlbaur]

 

 

You haven't actually shown that your scenario is correct, but in general, rocket engines are not 100% efficient at converting stored energy into the rocket's kinetic energy. There is also KE in the exhaust, and this efficiency is lowest at low speeds (and zero when the rocket is at rest)

 

https://en.wikipedia.org/wiki/Rocket#Energy_efficiency

 

 

Thank you very much for the information, but it doesn't really answer my question. Lets say for the purposes of illustration that the hypothetical rocket engine has a constant thrust and constant efficiency from the moment you light it up till it's expended in Mars orbit.

 

We have 20 times more work being done in the second section of the example for the same energy expenditure as in the first section.

 

What am I misunderstanding/miscalculating here because that does not make any sense?

[studiot]

Your title is a good one, because I think this is a straightforward misunderstanding.

 

 

 

..........................constant thrust ......................20 times more work

 

Therein lies your difficulty.

 

 

Yes the thrust (a force) is constant.

 

And if we take gravity as constant over the flight

 

The same acceleration numerically happens in each second of the flight since force is constant.

 

So each second the rocket gains say 5 m/s (or whatever) velocity.

 

So the velocity gain is additive not multiplicative as you have in your '20 times'

 

But of course if the acceleration is a

 

(a + a)² = (2a)² = 4a² and so on for higher powers.

 

It is the difference between an arithmetic progression and a geometric progression.

 

BTW

I also wonder if you actually realise what force is doing the work.

It is not the thrust alone.

[Mandlbaur]

Your title is a good one, because I think this is a straightforward misunderstanding.

 

 

Therein lies your difficulty.

 

 

Yes the thrust (a force) is constant.

 

And if we take gravity as constant over the flight

 

The same acceleration numerically happens in each second of the flight since force is constant.

 

So each second the rocket gains say 5 m/s (or whatever) velocity.

 

So the velocity gain is additive not multiplicative as you have in your '20 times'

 

But of course if the acceleration is a

 

(a + a)² = (2a)² = 4a² and so on for higher powers.

 

It is the difference between an arithmetic progression and a geometric progression.

 

BTW

I also wonder if you actually realise what force is doing the work.

It is not the thrust alone.

 

 

Thank you for your response.

I am not quite understanding how it applies though, maybe I haven't explained my question clearly enough:

 

Work = force x distance.

In instance 1) the work is the force x 5m. In instance 2) the work is force x 100m. The difference is 100/5=20 times more work being done in the second instance as compared to the first instance.

In both instances, the engine burns for the same period (1 second) so the same energy is spent.

 

If I am doing this calculation correctly, then it doesn't make sense when you consider the work/energy principle ?

 

BTW - I have studied physics at university level.

[studiot]

 

 

BTW - I have studied physics at university level.

 

Then you will realise that the mass of the rocket is not constant.

[swansont]

 

 

Thank you very much for the information, but it doesn't really answer my question. Lets say for the purposes of illustration that the hypothetical rocket engine has a constant thrust and constant efficiency from the moment you light it up till it's expended in Mars orbit.

 

We have 20 times more work being done in the second section of the example for the same energy expenditure as in the first section.

 

What am I misunderstanding/miscalculating here because that does not make any sense?

 

 

It doesn't have constant thrust efficiency. You can't just throw that constraint into the problem, it's unphysical. A rocket where thrust=weight has 0% thrust efficiency. All of the energy is going into heat and KE of the exhaust. No work is being done, yet whatever amount of energy conversion from combustion is happening. When the rocket velocity is the same as the exhaust velocity, it has maximum efficiency. None of the energy is ending up as KE of the exhaust. You get the maximum change in KE of the rocket for a given amount of fuel combusted.

[Mandlbaur]

 

Then you will realise that the mass of the rocket is not constant.

 

 

Absolutely, but how does that present any relevance to the problem I have presented?

 

 

It doesn't have constant thrust efficiency. You can't just throw that constraint into the problem, it's unphysical. A rocket where thrust=weight has 0% thrust efficiency. All of the energy is going into heat and KE of the exhaust. No work is being done, yet whatever amount of energy conversion from combustion is happening. When the rocket velocity is the same as the exhaust velocity, it has maximum efficiency. None of the energy is ending up as KE of the exhaust. You get the maximum change in KE of the rocket for a given amount of fuel combusted.

 

You are absolutely correct.

 

Let's make it an EmDrive thruster which has constant thrust and efficiency and does not use any propellant.

[studiot]

 

 

Let's make it an EmDrive thruster which has constant thrust and efficiency and does not use any propellant.

 

You can't change the rules half way.

 

A rocket is basically a small payload on top of a large fuel tank.

The fuel is often 90% of the mass of the rocket.

 

At any one time the payload plus remaining fuel is being propelled forwards thus the mass of this combination diminishes rapidly with progress.

 

It is true that

 

[math]KE = \frac{1}{2}m{v^2}[/math]

But m is a function of v or the other way round.

So you need to develop a mathematical statement of this relationship to assess the KE of the part travelling forwards.

[Mandlbaur]

 

 

You can't change the rules half way.

 

A rocket is basically a small payload on top of a large fuel tank.

The fuel is often 90% of the mass of the rocket.

 

At any one time the payload plus remaining fuel is being propelled forwards thus the mass of this combination diminishes rapidly with progress.

 

It is true that

 

[math]KE = \frac{1}{2}m{v^2}[/math]

But m is a function of v or the other way round.

So you need to develop a mathematical statement of this relationship to assess the KE of the part travelling forwards.

 

 

 

The equation for work is work = force x distance.

There is no way to incorporate v, m or KE into that simple equation.

Are you proposing that the equation is incorrect ?

Edited March 21, 2017 by Mandlbaur

[Strange]

The equation for work is work = force x distance.

There is no way to incorporate v, m or KE into that simple equation.

 

 

It also doesn't include acceleration but you are using that in your scenario.

 

If you are going to include acceleration then you have to also include all the other "real world" factors. Like, what is the actual force at different times.

 

You have raised a potentially interesting question but seem to be refusing to explore possible resolutions to the apparent problem.

 

 

Are you proposing that the equation is incorrect ?

 

Are you?

[swansont]

Mass being a function of time is another confounding effect but I think is not directly involved with the misconception of the OP, which is (in part) simple mis-accounting: The energy available goes into KE of more than just the rocket. Any energy balance is going to be wrong when you leave out one of the items that needs to be included. (It's also thermodynamics rather than mechanics, so energy is going into more than just CoM motion, but I'm not sure how big a role that is playing here)

[studiot]

Mass being a function of time is another confounding effect but I think is not directly involved with the misconception of the OP, which is (in part) simple mis-accounting: The energy available goes into KE of more than just the rocket. Any energy balance is going to be wrong when you leave out one of the items that needs to be included. (It's also thermodynamics rather than mechanics, so energy is going into more than just CoM motion, but I'm not sure how big a role that is playing here)

 

Yes but it depends what you (and the OP) mean by "The Rocket".

 

I am simply trying to point out that this is not a fixed object but is continually changing throughout the flight, unless you include the exhaust.

 

Which I suppose is another way of say the same thing.

Edited March 21, 2017 by studiot

[Bender]

I'll try to answer the OP as simply as I can. Therefore, I'm going to assume the rocket flies horizontal, has perfect efficiency and constant mass (in other words: it might as well be a car driving on a frictionless road in a vacuum). I have the impression that the rocket is overshadowing the essence of the question (although I could be wrong).

 

If the force is constant, energy expenditure is not. It takes more work to accelerate from 10 m/s to 20 m/s then it takes to accelerate from 0 m/s to 10 m/s.

 

Simply look at the formula for power:

P=v F

So, with constant force, the power (and thus work) increases with velocity.

Edited March 21, 2017 by Bender

[swansont]

I'll try to answer the OP as simply as I can. Therefore, I'm going to assume the rocket flies horizontal, has perfect efficiency and constant mass (in other words: it might as well be a car driving on a frictionless road in a vacuum). I have the impression that the rocket is overshadowing the essence of the question (although I could be wrong).

 

If the force is constant, energy expenditure is not. It takes more work to accelerate from 10 m/s to 20 m/s then it takes to accelerate from 0 m/s to 10 m/s.

 

Simply look at the formula for power:

P=v F

So, with constant force, the power (and thus work) increases with velocity.

 

 

The issue was how this could be, if the combustion power is constant.

[Bender]

The issue was how this could be, if the combustion power is constant.

The OP also assumes the force is constant. That's why he got 20 times the work. Obviously, both cannot be constant, but in the case of power being constant:

F=P/v

Edited March 21, 2017 by Bender

[swansont]

The OP also assumes the force is constant. That's why he got 20 times the work. Obviously, both cannot be constant, but in the case of power being constant:

F=P/v

 

 

Why can't they both be constant? Power propelling the rocket ≠ total power You appear to be making the same mistake; the rocket isn't the only thing that is being given kinetic energy.

[Bender]

Why can't they both be constant? Power propelling the rocket ≠ total power You appear to be making the same mistake; the rocket isn't the only thing that is being given kinetic energy.

Because I'm making a bunch of assumptions to get away from the rocket, which I interpret from the OP to be merely an example and detracts, as far as I can tell, from the real question. It is pretty obvious from the other replies in this thread that the OP is not interested in all the additional complications that come with the rocket.

 

Look at the title of the thread: it is about the work equation, not about rocket propulsion.

 

(also, I'm using output power, not input power)

Edited March 21, 2017 by Bender

[swansont]

Because I'm making a bunch of assumptions to get away from the rocket, which I interpret from the OP to be merely an example and detracts, as far as I can tell, from the real question. It is pretty obvious from the other replies in this thread that the OP is not interested in all the additional complications that come with the rocket.

 

Look at the title of the thread: it is about the work equation, not about rocket propulsion.

 

(also, I'm using output power, not input power)

 

 

But the fact that it's a rocket is a key difference and a source of the misconception: KE is given to the exhaust, and not just the rocket. (That's why rocket propulsion becomes more efficient as the rocket speeds up, up to the exhaust velocity), which is a difference from e.g. an engine in a car.

 

Otherwise, I agree, and previously pointed out that you can't arbitrarily place constraints on the problem, since it creates an unphysical situation.

The OP also assumes the force is constant. That's why he got 20 times the work. Obviously, both cannot be constant, but in the case of power being constant:

F=P/v

 

Your analysis assumes force is constant. That's not the issue.

[Mandlbaur]

I'll try to answer the OP as simply as I can. Therefore, I'm going to assume the rocket flies horizontal, has perfect efficiency and constant mass (in other words: it might as well be a car driving on a frictionless road in a vacuum). I have the impression that the rocket is overshadowing the essence of the question (although I could be wrong).

 

If the force is constant, energy expenditure is not. It takes more work to accelerate from 10 m/s to 20 m/s then it takes to accelerate from 0 m/s to 10 m/s.

 

Simply look at the formula for power:

P=v F

So, with constant force, the power (and thus work) increases with velocity.

 

 

Thank you for understanding the question.

If P=vF then F=P/v.

So what you are saying then is that the force/thrust that the rocket provides is dependent on the velocity ?

Also, the Force is infinitely large when the velocity is zero ?

Edited March 21, 2017 by Mandlbaur

[swansont]

 

 

Thank you for understanding the question.

If P=vF then F=P/v.

So what you are saying then is that the force/thrust that the rocket provides is dependent on the velocity ?

Also, the Force is infinitely large when the velocity is zero ?

 

 

No, the power delivered to the rocket is zero when v is zero. The power delivered depends on v. Some of the power goes into the exhaust.

[Bender]

Thank you for understanding the question.

If P=vF then F=P/v.

So what you are saying then is that the force/thrust that the rocket provides is dependent on the velocity ?

In the case of the rocket, the answer is provided by Swansont and others. There are several dependencies, one of which is the velocity.

Also, the Force is infinitely large when the velocity is zero ?

In practical systems, the power is usually not constant, especially the output power (output power = input power x efficiency). It is sometimes kept constant by a controller, e.g. in some electrical drives, but for practical reasons, the force (or torque) is limited at low speeds.

For a universal motor, which can often be found in mixers and vacuum cleaners, the torque seemingly goes to infinity when the speed goes to zero. Of course, the electrical wires burn before that can happen (among other things; resistance of the wire becomes an issue).

Edited March 21, 2017 by Bender

[Mandlbaur]

 

 

No, the power delivered to the rocket is zero when v is zero. The power delivered depends on v. Some of the power goes into the exhaust.

 

 

If the power is zero when the velocity is zero how does accelerate in the first place?

[Bender]

If the power is zero when the velocity is zero how does accelerate in the first place?

P=F.v

 

zero (output) power does not mean zero force. At the instant the thrust starts and v=0, the power will be zero. The "next instant", the rocket will have accelerated somewhat and the power is no longer zero.

 

In other words, there is no length of time, however short, for which the average power is zero, but at the start itself, the momentary power is zero.

Edited March 22, 2017 by Bender

[KipIngram]

Because kinetic energy goes as v^2, the energy being delivered by the first increment of force (just as the rocket begins to move), is zero "approximately." The velocity can experience a "first order change" while the energy experiences a "second order" change. In the limit of the "very first instant" your ratio of velocity change to energy change approaches infinity.

 

That said, the rocket is delivering a lot of energy to the exhaust - that mass is flying at exhaust speed out the back, and so has significant kinetic energy. That gets you back to inefficiency at low speed - significant energy is being given to the exhaust, while insignificant energy is being given to the rocket.

 

Over any small but not infinitesimal window of time beginning at t=0 the energy delivered to the rocket is small but not zero, so it will get itself underway. And after that the faster the rocket is going the more rocket v the force has to work with to produce delivered power. I remember watching Apollo launches when I was a kid, and always marveled at how gradually the rocket first lifted away from the pad. Then later it seemed to speed up faster and faster. But it all makes great sense given the math.