Post by: paul cotter on 13/04/2022 15:57:42
Post by: paul cotter on 13/04/2022 16:02:34
Post by: Bored chemist on 13/04/2022 17:55:58
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.
Post by: paul cotter on 13/04/2022 18:26:22
Post by: alancalverd on 13/04/2022 18:45:16
In vacuo and devoid of gravitation it will fly in a straight line, not a parabola.
Post by: paul cotter on 13/04/2022 19:02:17
Post by: Eternal Student on 13/04/2022 19:33:53
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.
Post by: vhfpmr on 13/04/2022 19:59:38
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.
Post by: Bored chemist on 13/04/2022 20:01:46
In vacuo and devoid of gravitation it will fly in a straight line, not a parabola.Nobody said otherwise.
Post by: Bored chemist on 13/04/2022 20:03:05
Post by: paul cotter on 13/04/2022 20:31:12
Post by: Eternal Student on 13/04/2022 22:51:03
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.
Post by: Halc on 13/04/2022 22:51:54
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.
Post by: evan_au on 13/04/2022 22:58:59
Quote from: paul cotter
a rocket's ke cannot exceed the work done by a constant thrustThe 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...
Post by: Eternal Student on 14/04/2022 03:18:08
- - - - - - - - - - - - - - - -
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 3089 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, Frocket. 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 Fpayload.
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:
Fpayload = mpayload . a
Frocket = mrocket . a <---- LATE EDITING Another convention would disagree (see later posts if interested).
However, the mass of rocket, mrocket ≥ mpayload (the mass of the payload) for all times before T.
Therefore, Frocket ≥ Fpayload .
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.
Post by: paul cotter on 14/04/2022 09:15:20
Post by: paul cotter on 14/04/2022 09:51:44
Post by: alancalverd on 14/04/2022 11:03:26
vhfmpr has given the instantaneous rocket equation.
The initial and final masses are known, so you need to integrate from m0 to mp, the initial and payload masses respectively. Which is what ES has done.
Post by: paul cotter on 14/04/2022 12:16:01
Post by: paul cotter on 14/04/2022 12:37:25
Post by: Origin on 14/04/2022 12:43:44
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.
Post by: paul cotter on 14/04/2022 13:32:50
Post by: Eternal Student on 14/04/2022 14:12:56
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 mfinal, v = v(mfinal). However, mfinal 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, ve, 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.
Post by: alancalverd on 14/04/2022 14:33:16
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.
Post by: paul cotter on 14/04/2022 17:41:07
Post by: Petrochemicals on 14/04/2022 20:02:39
Accelleration but no velocity.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.
Post by: Eternal Student on 14/04/2022 23:53:23
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.
Post by: Petrochemicals on 15/04/2022 00:06:28
Hi.But what does this all mean?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.
Post by: Eternal Student on 15/04/2022 03:26:35
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.
Post by: wolfekeeper on 15/04/2022 03:42:17
(https://upload.wikimedia.org/wikipedia/commons/5/50/Average_propulsive_efficiency_of_rockets.png)
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.
Post by: Eternal Student on 16/04/2022 01:31:25
Nice diagram @wolfekeeper . Out of interest, what was the blue line?
Best Wishes.
Post by: Eternal Student on 16/04/2022 13:12:40
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.
Post by: paul cotter on 16/04/2022 16:51:07
Post by: Eternal Student on 16/04/2022 21:12:12
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.
Post by: paul cotter on 16/04/2022 21:59:26
Post by: Eternal Student on 17/04/2022 03:36:20
... ( 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:
Quote
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:
Quote
.... 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
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.
Post by: paul cotter on 17/04/2022 10:48:21
Post by: Eternal Student on 17/04/2022 15:04:37
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 constantsI 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.
Post by: paul cotter on 18/04/2022 14:37:23
Post by: wolfekeeper on 22/04/2022 02:12:47
Hi.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.
Nice diagram @wolfekeeper . Out of interest, what was the blue line?
Post by: wolfekeeper on 22/04/2022 21:26:00
coe
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.FWIW this is somewhat subtle stuff. There was actually some disagreement about the relationships between rockets and energy but it was laid to rest by Hermann Oberth.
He reported that at one point in his investigations of rocketry he believed he had completely disproven conservation of energy too, but he later was able to show that he hadn't.
See:
https://en.wikipedia.org/wiki/Oberth_effect
This links to 'Ways to spaceflight' which describes his mathematics which covers this and other topics.