[paul cotter (OP)]

an acquaintance in an other forum has suggested that a rocket breaks the conservation of energy because the ke follows a parabolic rise and will outpace the energy input(from a constant thrust) at some point.i say that because of propellant mass loss the ke rise will hit an asymptote and not cross the energy input(a linear function).I have the highest possible respect for this individual but we just can' agree. I say the ke at any point is fm/2 (lin m)squared,where f=constant thrust, and m is the remnant mass after a burn.

[paul cotter (OP)]

error; ke=m(flin m)squared/2

[Bored chemist]

He is not taking account of the fact that, to produce a constant thrust doesn't mean a constant power.
As the rocket gets faster  the power needed to maintain a constant thrust rises parabolically too.

Power is energy / time
And the energy is force times distance.
So the power needed is force times distance over time.
But distance / time is velocity so, as the rocket goes faster it becomes harder and harder to produce a constant thrust.
That cancels out the face that the energy is a parabolic function of velocity.

[paul cotter (OP)]

in the rocket's frame of reference, constant thrust produces constant acceleration.if applied from a static frame of reference, the power required would increase with speed.just to be clear this is in vacuo and devoid of gravitational effects.

[alancalverd]

Constant thrust does not produce constant acceleration because the mass is decreasing.

In vacuo and devoid of gravitation it will fly in a straight line, not a parabola.

[paul cotter (OP)]

the parabola reference was the function that represents the ke of the rocket.off course I was wrong to say "constant thrust produces constant acceleration".constant thrust produces an increasing acceleration as the mass of propellant reduces.the question is :can the ke exceed the work done by the thrust. the coe prohibits this.i am looking for a proof without invoking the coe. I say the ke=m(f linm)sq and this is an asymptotic function an gives a limit to the max ke achievable and that this is less than or equal to the work done by the thrust.

[Eternal Student]

Hi.

   To be honest, I'm not entirely sure I understood the phrasing in the original question.   The formatting of your equations also didn't seem to be very clear...
What is this:

m(flin m)squared/2

"flin"   might be    f . Lin  and    Lin  might actually be what is often written as  Ln  or the natural Log of something  -  but I just don't know, I'm guessing what you meant to write there.

    The forum does support LaTex mark-up language for equations although it's a bit of nuisance to use.  An alternative might be to use any software you want to create the equation and just attach a picture of the equation in the forum post.

   Anyway, I'm really sorry that I can't comment more.  I just can't be sure what it was you were asking.
-  - - - - - -

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. 
    This frame of reference isn't an inertial frame of reference anyway and most principles of physics won't hold in that frame the way you may want or expect.  For example, energy and momentum do not need to be conserved in that frame of reference.
   There's a fair chance you really wanted the rocket to have a constant thrust as measured in some inertial reference frame  -   but I don't know, I'm just guessing what you wanted.
 
Best Wishes.   

[vhfpmr]

The rocket equation is: m.dV/dt = Ve.dm/dt

m is the mass
dV/dt is acceleration
Ve is exhaust gas velocity
dm/dt is rate of fuel burn

For a constant exhaust gas velocity and fuel burn rate, acceleration increases with reducing mass.

[Bored chemist]

In vacuo and devoid of gravitation it will fly in a straight line, not a parabola.

Nobody said otherwise.

[Bored chemist]

The problem with the rocket's frame of reference is that it always has zero velocity.

[paul cotter (OP)]

I have been unclear and muddled.what I want is proof, from first principles, without invoking the coe, that a rocket's ke cannot exceed the work done by a constant thrust.i say the ke =the mass m, multiplied by the square of the thrust times the nat log m, where m= the residual mass after a burn.i say that this is an asymptotic function and cannot be greater than the work done by the thrust. god, I wish I had math symbols on this keyboard.

[Eternal Student]

Hi.

i say the ke =the mass m, multiplied by the square of the thrust times the nat log m, where m= the residual mass after a burn.

   If it helps, I can see there is something wrong with that formula.  You may want to check how you derived that.

   You have written   k.e.  =   m. t² . Ln(m)        with  t = thrust,  m = mass of the rocket remaining.

As the rocket burns fuel, the mass m goes down.    Sadly, as soon as  m < 1 unit,     Ln (m) becomes negative and then the whole expression for  the kinetic energy becomes negative.   Obviously that can't be right,  kinetic energy is never going to be negative.

Best Wishes.

[Halc]

what I want is proof, from first principles ... that a rocket's ke cannot exceed the work done by a constant thrust.

For a rocket, that's pretty trivial.
In the inertial frame in which the rocket is initially stationary, the energy is mostly wasted. The rocket gains a little KE by going from nothing to not-much, and most of the energy is wasted making the exhaust move fast. But the KE of the rocket goes up a little with this tremendous expenditure of chemical energy.  As it goes faster and faster, the rocket's KE levels off and starts dropping again (from reduction of mass). A KE that is dropping is not going to have any trouble staying below the final energy expenditure value. With most rockets, the final KE is probably less than 1% of the energy with which it was fueled because it masses almost nothing at the end.

A rocket is always less efficient than some system where all the energy is exerted by some mass (Earth say) that is stationary in the frame in question. In such a scenario, Earth picks up negligible energy and the projectile gets it all, and the final KE of the projectile still cannot exceed the work done by whatever is driving this. So since at all times the rocket is less efficient than that (by a lot), its final KE cannot be greater than the work done by the fuel it carries.

[evan_au]

a rocket's ke cannot exceed the work done by a constant thrust

The answer is that:
(1) The work done by a constant thrust approaches infinity as time approaches infinity. This will exceed the energy in any rocket (eventually).
(2) A rocket carries a finite amount of fuel, so the constant thrust cannot continue for an infinite time. So this is not a paradox.

So the resolution is that the fuel runs out before a rocket's ke can exceed the work done by a constant thrust (...maybe moments before...)

This applies even in deep space, where it is not fighting the gravitational well of a nearby planet or star...

[Eternal Student]

LATE EDITING:   This post did not end up needing any adjustment.  However, the definition of the force on a rocket used here follows only one of two common conventions.  This is discussed in later posts.
  - - - - - - - - - - - - - - - -

Hi again.

   

what I want is proof, from first principles, without invoking the coe, that a rocket's ke cannot exceed the work done by a constant thrust.

    Let's just do that first:

Here's a simple rocket:

rocket.png (16.57 kB . 1171x633 - viewed 3211 times)

   That's the rocket initially.   At any later time, some of the fuel has been used and some of the lower stages have been discared.

   So at any fixed later time "the rocket" is usually less than shown in the diagram.  We don't care about the exhaust gas or any bits of rocket that have been discarded.   "The rocket" is just whatever we have left.

   Let's consider the rocket at some (arbitrary) fixed time, T, when it was just the the grey bit on the diagram which I have called "the payload".   Note that the name "payload" is just for convenience, I don't care if the rocket still has more fuel and could still make another burn and possibly discard a few more bits.  It doesn't matter,  the grey bit ("the payload") is what "the rocket" will be at the last time, T, that we will be examining the situation.   At any earlier time "the rocket" will be the payload PLUS at least some of the red stuff on the diagram (the rest of the rocket).

   Let's call the force on the rocket,   F_rocket.   You state this is to be kept constant (we will do that - but that isn't actually going to be necessary anyway).
Now there is also a force just on the payload (the grey bit) which we can call   F_payload.
At any time before T,  the payload must stay with the rest of the rocket:   The payload cannot accelerate faster than the rocket or else it will become separated.  Similarly it can't accelerate slower than the rest of rocket or else it will be crushed.   The acceleration of "the rocket" and of just "the payload" is identical at all times between t= 0 and t = T and we will just call this acceleration, a.
   Using Newton's Laws we have:
F_payload    =   m_payload . a
F_rocket      =   m_rocket . a      <---- LATE EDITING  Another convention would disagree (see later posts if interested).

However, the mass of rocket,  m_rocket  ≥  m_payload  (the mass of the payload)  for all times before T.
Therefore,  F_rocket  ≥  F_payload  .
Let the rocket move along the x-axis.  Set x=0 as the location of the rocket when t=0 and we're assuming it was stationary at that time.  Set x= X as the location of rocket when t = T.
The kinetic energy of the payload is then precisely         ≤     
   
   The expression on the right is just exactly the definition of work done by the thrust on the rocket.   The left hand side requires you to know how the kinetic energy of a body is defined in Newtonian mechanics (it's ½mv²) and that the change in that quantity is equal to the integral shown.  This is not a difficult result to show, it's in most books but if you want it we could show it here in a few minutes.   This isn't the conservation of energy principle, which you specifically asked us to avoid using.  Instead it follows just from the basic principles or definition of what kinetic energy must be in Newtonian mechanics.  For a body which retains fixed mass and is accelerated with the force shown then that integral holds:  Note that the payload really did retain fixed mass throughout all times before t=T so we can just find the force acting on the payload and throw that in the formula.   Meanwhile, "the rocket" as a whole did not retain a fixed mass and so the force on the rocket could not be used in that integral formula to determine the kinetic energy of the rocket.

    Anyway, the expression on the right is the work done by the thrust on the rocket and we have just shown the kinetic energy of the rocket (the payload or the bit of rocket we have left) at any arbitrary time T is always less than or equal to the work done by the thrust on the rocket up to that time.   It's actually a lot of words and mathematical expressions to show a simple idea:   The mass of the rocket at an earlier time is always greater than the mass of the rocket at a later time.  So a lot of the work done by the thrust gives kinetic energy to bits of the rocket that will be thrown away either as exhaust gas or discarded rocket stages.  Very little of the work done becomes kinetic energy of the final payload.

Best Wishes.

[paul cotter (OP)]

eternal student, I am crippled by not having math symbols:i said the SQUARE of (the thrust times the nat logm).halc, I suspect you understand my take on the matter.leaving out the efficiency of combustion and nozzle enthalpy use, I want rigorous proof that the ke that the rocket achieves never exceeds the work done by the constant thrust, without invoking the coe.

[paul cotter (OP)]

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.   M is a function of t(decreasing as fuel is burnt).   V=Ftimes nat log M.    ke= Vsquared/2 times M=M/2 times the square of nat log M. F=constant thrust, M=residual mass at any time.           

[alancalverd]

Evan has alluded to the implicit fallacy in the OP. If you accelerate at a constant rate indefinitely, you will each an infinite speed, but you don't have an infinite amount of fuel.

vhfmpr has given the instantaneous rocket equation.

The initial and final masses are known, so you need to integrate from m₀ to m_p, the initial and payload masses respectively. Which is what ES has done.

[paul cotter (OP)]

maybe i'm not explaining things properly(highly likely). forget about payload and just consider a rocket in free flight with  constant thrust. my acquaintance says the rocket breaks the "coe" by virtue of a parabolic increase in ke versus a constant thrust(and hence a linear increase in work done by the thrust). I say he's wrong as the ke levels off due to mass loss and reaches an asymptote, equal or less than the work done by the thrust.i believe "halc" understands my point of view.

[paul cotter (OP)]

I made another error in the last post: I meant to say 'a parabolic increase in ke versus a linear increase in energy from a constant combustion process. I sometimes seem to have an edit function and sometimes not. any clues?