[Origin]

I want rigorous proof that the ke that the rocket achieves never exceeds the work done by the constant thrust, without invoking the coe.

Why?  Using the conservation of energy is by far the easiest way to solve the problem.  If your friend says the conservation of energy does not apply it is his responsibility to supply the math that disproves current theories.  Since he can't do that, the discussion will be over.
I suspect this all boils down to the confusion some people have around velocity and KE.  If the velocity of a mass increases linearly then the KE will increase exponentially.  This seems to really upset some people.

[paul cotter (OP)]

yes indeed, origin I agree totally.the "coe" is absolute in my world. however the op in question' for whom I have great respect, argues otherwise. I want a rigorous proof from first principles that the ke can never exceed the work done by the constant thrust, without invoking the "coe". the op says the ke rises on a parabolic function and will exceed the sum of energy added by the constant thrust as the energy expended by a constant combustion process is a linear function.

[Eternal Student]

Hi again.

    I still think the reply given in post #14 is the fastest or easiest way to demonstrate what you wanted to show.   However, I'm aware that sometimes we need to see where the errors exist in our own work to consider that another approach would have been better.

   I'm going to go through what you've written and how you had derived an expression for the k.e. of the rocket.

this is my take: F=MA, A=F/M, V=integral of Adt=integral F/M.  F is constant so V=Ftimes the integral1/Mdt.

    That bit was ok.   

You have    velocity of the rocket,  v    =    F  .   

Then you said:

M is a function of t(decreasing as fuel is burnt).

   Which is correct.   m = m(t).                     
However, you forgot that you were integraing with respect to time, t, and not with respect to m.   
So the indefinite integral is not   Ln(m),   it's slightly more complicated than that.       

Standard derivations of the "Tsiolkovsky rocket equation" are availble from many places.   The Wikipedia entry is reasonable:  https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation.
   Under the usual assumption that the exhaust gas always has a constant velocity relative to the rocket and the rocket was stationary initially (has v =0  at  t=0)  the Tsiolkovsky rocket equation already tells you exactly what you wanted to know   -   it will be precisely the same as the result of calculating your integral:

    v(t)   =   

[LATE EDITING:  Note that the expression is really showing v as a fuction of m_final,  v = v(m_final).  However, m_final is a function of t, which you can determine explicitly.]

It's similar to the expression you thought you obtained from your integral.  The main adjustments are that is has more constants appearing in it:
(i) It has a constant,  v_e,  which is the exhaust velocity relative to the rocket, appearing at the front of the equation.
(ii) it involves a ratio of masses under the natural Logarithm.   (You could re-arrange this to show it has a constant that is added if you prefer).

   The nice thing about this expression is that it didn't depend on the thrust being constant, it can be constant (as you stated) but it doesn't matter at all.  All that was required is that the exhaust is sent away from the rocket at the same speed all the time,  i.e.   that the engines work in much the same way all the time.

   Anyway, you can use that expression for v,  the velocity of the rocket if you want   and  then  ½ . m . v²   is the kinetic energy of the rocket you were seeking.

   You still have to compare this  k.e. against work done by the thrust on the rocket to reach your ultimate goal   (showing the  k.e.  ≤   the work done by the thrust).

   Overall, I think there's enough work here that I wouldn't bother to continue or present this method to your friend.   The method in post #14 seems to be a faster and easier method to reach your goal and it uses only basic principles of Newtonian mechanics and not the Conservation of Energy.   However, you are obviously free to continue and do as you wish.   I hope that going through your own work has been of at least some small help.

Best Wishes.

[alancalverd]

Or most simply take the initial and final conditions of an ideal rocket.

The system consists of a payload of mass m and a fuel load M with specific energy s per unit mass.

Initial potential chemical energy = Ms. Therefore however you arrange the fuel burn profile, final kinetic energy ≤ Ms, there being no other source of energy. 

[paul cotter (OP)]

thanks very much, "eternal student".its 50years since I did calculus at college and I see my error. in my first derivation I put the mass (as a function of time)=initial mass of fuel +mass of casing-Mt/T, where t is the independent variable and T is the time for fuel exhaustion.

[Petrochemicals]

in the rocket's frame of reference, constant thrust produces constant acceleration....

   I'm concerned about what you've said there.   
   "The rocket's frame of reference"  is usually taken to mean the frame of reference in which the rocket is stationary.   So the rocket actually has 0 acceleration in that frame at all times.  I mean that is a constant acceleration but I'm not sure it's what you wanted. 
   

Accelleration but no velocity.

[Eternal Student]

Hi.

Accelleration but no velocity.

   Yes.   If you use an instantaneous inertial frame   (sometimes called "the instantaneous rest frame") for the rocket then it can have an acceleration but 0 velocity.   This probably was what the original poster was intending and I probably just misunderstood which reference frame(s) they were using.   Technically, you keep changing the frame of reference in every instant to make sure the object always has 0 velocity.  So this is a whole class of frames not one fixed frame for all time.
    The main alternative is to use a genuinely accelerated frame of reference - but that probably wasn't what the original poster had in mind and there's little point in me spending any more time to discuss how an accelerated frame would differ.

Best Wishes.

[Petrochemicals]

Hi.

Accelleration but no velocity.

   Yes.   If you use an instantaneous inertial frame   (sometimes called "the instantaneous rest frame") for the rocket then it can have an acceleration but 0 velocity.   This probably was what the original poster was intending and I probably just misunderstood which reference frame(s) they were using.   Technically, you keep changing the frame of reference in every instant to make sure the object always has 0 velocity.  So this is a whole class of frames not one fixed frame for all time.
    The main alternative is to use a genuinely accelerated frame of reference - but that probably wasn't what the original poster had in mind and there's little point in me spending any more time to discuss how an accelerated frame would differ.

Best Wishes.

But what does this all mean?

[Eternal Student]

Hi.

But what does this all mean?

   
   i.d.k.    Which bits and do you need to start a new thread to discuss it?
   The main point was that I probably did mis-understand what the OP was trying to talk about.

Best Wishes.

[wolfekeeper]

FWIW here's a graph of energy efficiency versus final velocity (in red):



To generate the graph you use the rocket equation to calculate the remaining rocket mass after reaching a particular velocity. You can then calculate the kinetic energy of the vehicle at that speed. The loss of mass (i.e. the fuel burnt) will happen at a particular exhaust velocity, so you can calculate how much energy was used and divide one by the other to get the red line.

The scale along the bottom is in multiples of the exhaust velocity and the scale up the left hand side is in multiples of the engine efficiency. It obviously never reaches 100% and in fact carries on going down.

[Eternal Student]

Hi.

Nice diagram  @wolfekeeper .    Out of interest, what was the blue line?

Best Wishes.

[Eternal Student]

Hi.

   I made a mistake in post #14,  sorry.   I do not have time to fix that at the moment but I've just marked where the mistake appears.  It should actually make the situation a bit more interesting.
  It's good Job I wasn't actually at mission control but was just chatting on a forum.   I'll send a mail message to @paul cotter  to notify them of the mistake.

Best Wishes.

[paul cotter (OP)]

not sure if that will work, ie email. this is going to be longwinded but it's the only way to explain :I have worked in two main areas (1) broadcast transmission and (2) power generation and transmission. but in addition I have repaired electronic equipment for over 55 years and being retired now I want a break. it would not be uncommon for a metal box covered in cow sh1t or cement with twenty wires hanging out to be delivered to me with the only info "it doesn't work".i can't do this anymore. as such I try to keep a low profile and do not have email, I used my wife's email to register. there was a box to enable other members to send email and I did not tick it. hope you understand and thank you very much for your input.

[Eternal Student]

Hi.
   When I said "mail",  I meant whatever they call the built-in messages system in this website.   There is one, occasionally I get a message from someone.   Anyway, when you log in you should see something flagged as a message along the top of the screen if you have a message.  If you've already disabled that, don't worry you haven't missed anything.
   The message didn't say anything different to what was in the previous post,  I just wasn't sure you would ever need or want to check this forum thread again but you might have seen the message.
    Privacy and not wanting any proper email is something most of us can fully understand.
    Anyway, I might write some more stuff about rockets and possibly fix the mistake I made -  but the interest in the topic might have gone, so I wasn't going to rush.   

Best Wishes.   

[paul cotter (OP)]

I am sorry, I misread your post and saw e-mail when I should have seen MAIL. and I did get your message and would love to see further analysis. thank you again and apologies for my off-topic rant.

[Eternal Student]

Hi again.

  ... ( I ) would love to see further analysis

   
  Well, actually it's very difficult.   In no small part it's because there just isn't a universally accepted definition of what the force acting a body which changes mass is supposed to be.

   Here's an extract from Wikipedia to illustrate the point:

Variable-mass systems, like a rocket burning fuel and ejecting spent gases, are not closed and cannot be directly treated by making mass a function of time in the second law  (Newton's 2nd law)

   To make this clear:   Newton's law states that force on a body is proportional to (equal to if you use the right units) the time rate of change of the momentum of the body     
F  = 

   However you CANNOT just make mass a function of time and end up with something like  F = m    +  v

   It is an old idea that Newton's 2nd law would always hold perfectly well if you used in its proper form as a rate of change of momentum when an object changes its mass.   It actually doesn't hold at all well in most situations.  To make it work, some terminology has to be used very carefully. 
   It's actually better if we just change the wording of Newton's 2nd law to something like  "Force ON A CLOSED SYSTEM  is proportional to the rate of change of momentum of that system"  but this just wasn't done mainly for historical reasons - Newton's laws were too heavily entrenched in all the main works of physics by the time this was realised.   Instead, if Newton's 2nd law gets any revision at all it's just that some other terminology is introduced and used very carefully.

What terminology needs to be used carefully?
   The definition of "a body" needs to be kept consistent.
   Newtons 2nd law,  in the form   F =     will hold if the body you are considering is consistently identified as the same set of particles.   
   That's where we have a problem when we consider rockets.  We call a body  "the rocket" but it actually isn't one consistently identified set of particles, i.e. it isn't one consistently identified body at all.     "The rocket"  is always understood to mean  "the rocket at a particular time t" and it is a physically different collection of particles then  "the rocket at time t=0"   or   "the rocket at time t = 1000".   
    After some fuel has burnt and some exhaust has been discarded then the original object    "the rocket at t = initial time"  is now scattered over a region of space.   There were only ever internal forces acting on "the rocket at the initial time", there were no net external forces on it at all. 
    Let's just state this clearly because it takes a moment to fully appreciate:  At any time t,  the net force on "the object" = 0;  the work done on "the object" = 0    and the acceleration of "the object" = 0.   This all makes perfect sense if you replace the word   "the object"   with  "the rocket at the time t".   It's just unfortunate that we human beings tend to shorten our description of "the object" so that  "the rocket at time t"   is just said to be  "the rocket"  and forget that it's actually never the same object as time progresses (when exhaust is being expelled).
    - - - - - - - - - - - - - - - - -

   OK, that was long but, I think quite essential to understand.
Anyway, as tempting as it might be to assume  F = m    +  v     will hold for a rocket, it just won't.    However, using Newtons 2nd law on the entire closed system of the rocket + exhaust gases we CAN obtain a suitable equation of motion for a rocket (and then we can seriously consider what we might identify as the Force term).

   Here's another quote from Wikipedia:

.... the equation of motion for a body whose mass m varies with time by either ejecting or accreting mass is obtained by applying the second law to the entire, constant-mass system consisting of the body and its ejected or accreted mass. The result is

F + u   =  m

where u is the exhaust velocity of the escaping or incoming mass relative to the body.

Under some conventions, the quantity u on the left-hand side, which represents the advection of momentum, is defined as a force (the force exerted on the body by the changing mass, such as rocket exhaust) and is included in the quantity F. Then, by substituting the definition of acceleration, the equation becomes F = ma.

   Anyway, that's the first major problem.  Exactly as stated above  under SOME conventions  the term u   is included as a Force acting on the rocket and it is simply merged with  F (the net external force),   under other conventions it isn't.   Without agreement about what is a force acting on a rocket we can't determine what the work done by the thrust (which is the integral of force over distance) is supposed to be.

[This post is too long and I am breaking it here].     All Wikipedia quotes are from this page:  https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Variable-mass_systems

Best Wishes.

[paul cotter (OP)]

this is starting to look insoluble if only constant thrust is specified without speed of exhaust. my derivation for m(t) must then be incorrect but I cannot see why. without math symbols it is very hard to present,  m(t)=M+Mc-Mt/T, where M=initial mass of fuel, Mc= mass of casing, and T=time for fuel exhaustion. since most of the mass of a rocket is fuel I simplified by removing Mc . so m(t)=M-Mt/T, M and T being constants  integrating the reciprocal of this expression gives a -ln  expression .

[Eternal Student]

Hi again.

this is starting to look insoluble....

    Well, there is some good news.   If you do use the convention mentioned earlier where u is considered as a force, then the method and proof presented in post #14  is perfectly fine, it will go through without needing any alteration.   In my opinion that is the most sensible convention and it does match up most easily with what you can measure in practice if you tried to install some force measuring devices (force gauges) into the structure of the rocket just above the jet propulsion engines.  You would measure a force (a thrust) on that instrument.   It is what most people would reasonably be imagining when they talk about the thrust on the rocket due to the expulsion of exhaust gas.

    If you use the other convention, then there is never any force on a rocket (unless it is actually near some gravitational source like a planet and then there is a genuine external force).  I know that seems a little arbitrary but that's just how it is.  Sadly, I don't set the conventions.   People who use that convention would argue that you can't measure the force on the rocket by installing a force gauge only on the structure above the engine.  The gas which is being expelled now (or is about to be expelled) is also part of the rocket and you need to try and install a force gauge in that as well.  Obviously in practice you can't actually do that.  Anyway, in their approach the ejection of gas is just a consequence of some internal forces and does not produce any net force on the object at all.
   You can't "solve" this with some clever mathematics, it's just a matter of which convention you choose to use.

my derivation for m(t) must then be incorrect but I cannot see why. without math symbols it is very hard to present,  m(t)=M+Mc-Mt/T, where M=initial mass of fuel, Mc= mass of casing, and T=time for fuel exhaustion. since most of the mass of a rocket is fuel I simplified by removing Mc . so m(t)=M-Mt/T, M and T being constants

   I can see what you've done.  It's perfectly fine, you are assuming a constant burn rate.    = constant independent of time t.   This does produce a constant force (a constant thrust) on the rocket   BUT  only if you're using the (I would say better) convention where the term  u is considered as a force on the rocket.

Best Wishes.

[paul cotter (OP)]

hi again. other people on different fora have been debating the question. member "uatu" on the german allmystery.de forum has provided what I consider the definitive solution with a graph illustrating the key parameters. my primitive derivation indicated an initial parabolic rise in ke, followed by an asymptote at around 3/4 fuel consumption. the rigorous expression provided by "uatu" does the same followed by a sharp downturn. this derivation seems to be rock-solid, in my limited mathematical abilities. bottom line: the coe is safe and emmy noether can rest peacefully.

[wolfekeeper]

Hi.

Nice diagram  @wolfekeeper .    Out of interest, what was the blue line?

Blue line is the instantaneous energy efficiency (force times vehicle speed/half the exhaust velocity squared) of the rocket engine expressed as a percentage of the internal engine efficiency at turning available chemical heat energy into fast moving exhaust.