Physics in troubles: the real equation of force is F = ma and not F = dp/dt

[sethoflagos]

Just now, martillo said:

You can't. The momentum of the total system of the rocket and the exhausted fuel includes all the mass exhausted from the initial state to the end. You cannot just discard all that mass already exhausted time ago in the conservation of momentum. If you don't agree with this I cannot proceed anymore.

You want me to yield to your opinion? It isn't me that's having a problem with established science.

I suspect pride is clouding your judgment. Good night.

[martillo]

13 minutes ago, sethoflagos said:

You want me to yield to your opinion? It isn't me that's having a problem with established science.

I suspect pride is clouding your judgment. Good night.

No, I'm just saying I cannot proceed any more. It is not to yield to opinions. It is not about opinions, is to go on from what is right. We both cannot go on if we don't reach an agreement at this point. Good night.

Edited February 23, 2023 by martillo

[joigus]

Dear @martillo, there's only one person here who's just been adhering to his own opinion, gritting his teeth against all evidence and reasoning. The evidence being that these are the concepts that engineers apply on a regular basis.

All of this is established science. Nothing more to say on my part, as you totally ignored my arguments. Other members seem to have come to similar conclusions.

Edited February 24, 2023 by joigus
minor correction

[studiot]

10 hours ago, martillo said:

10 hours ago, studiot said:

Think carefully about what you just proposed.

is p_r constant ?

If so why?

If not why not ?

No, p_r is not constant, is just  p_r = p₀ - p_e . Both increasing in time with the energy of the combustion of the fuel.

I'm glad you understant that.

But what do you mean by 'both'  ?

I only asked about p_r 

But since you mention it, what about p_o and p_e  ?

It is necessary to make some assumption about p_e ;  it is usual to assume a constant burn so p_e is then constant.

But p_o is more tricky since p_o = psystem  = p_rocket at the start of the burn.
But as the burn proceeds the rocket not only gains velocity, it looses mass.

 

Edited February 24, 2023 by studiot

[martillo]

23 minutes ago, joigus said:

Nothing more to say on my part, as you totally ignored my arguments.

I didn't ignored you at all. I answered you several times as you can see at page 8 of this thread. What arguments do you want me to take into consideration? I don't understand...

20 minutes ago, studiot said:

But what do you mean by 'both'  ?

I mean both p_r and p_e augments their magnitude (in absolutes values) with time while the rockets move.

23 minutes ago, studiot said:

It is necessary to make some assumption about p_e ;  it is usual to assume a constant burn so p_e is then constant.

Right, a constant burn exhausting burnt fuel at constant velocity v_e relative to the rocket and constant mass being expelled all the time (dm_e/dt = constant).

27 minutes ago, studiot said:

But p_o is more tricky since p_o = psystem  = p_rocket at the start of the burn.
But as the burn proceeds the rocket not only gains velocity, it looses mass.

I was making calculations. Please consider p = p₀ at the start of the burn but p₁ at the launching when the rocket starts to move. There's an interval of time between the start of burning and the start of the rocket's landing which is important to take into consideration. In this interval fuel mass is being expelled so p_e augments in magnitude as the momentum of the total system p augments in magnitude from 0 to some p₁ = p_e1 . In this interval the momentum of the rocket remains p_r = 0 since the rocket does not move (v = 0). At time t = 0 momentum p = p₀ = 0. 

[joigus]

16 minutes ago, martillo said:

I didn't ignored you at all. I answered you several times as you can see at page 8 of this thread. What arguments do you want me to take into consideration? I don't understand...

That's fairly clear to me. You don't understand...

Here (from page 8, where you apparently answered 'several times':

16 hours ago, martillo said:

You are right. They must be conserved for closed systems only without external forces or energies being applied to or delivered from it. It is just that it was the case we were analyzing, a closed system.

15 hours ago, joigus said:

Which is included as a particular case in my analysis, of which you said nothing.

And indeed,

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow\frac{d}{dt}\left(m\dot{x}\right)=-\frac{dV}{dx}\Rightarrow m\ddot{x}+\dot{m}\dot{x}=F_{\textrm{ext}}=-V'\left(x\right) \]

Take \( F_{\textrm{ext}}=0 \), and there's your super-special instance. I told you, didn't I?

More from unbearable page 8...

15 hours ago, martillo said:

That was said for the case presented by Genady of a melting ice moving at some constant velocity v. We were analyzing that how th principle of conservation of momentum applies.

I told you conservation of momentum applies for overall rocket + exhaust in free space, and I told you how.

16 hours ago, joigus said:

Rocket:

Energy is not conserved

Momentum is not conserved

Exhaust:

Energy is not conserved

Momentum is not conserved 

Rocket + exhaust:

Mechanical energy is not conserved

Mechanical energy + chemical energy stored in fuel is conserved

Momentum is not conserved if rocket + exhaust are in a gravitational field

Momentum is conserved if fuel + rocket are in free space

You see, when you burn fuel, some kind of potential energy that's stored in the molecular bonds gets converted into heat + kinetic energy for both rocket and exhaust. We call this energy Gibbs free energy (of fuel).

So no, mechanical energy is not conserved in this case. If the rocket is stationary in free space at initial time, K.E = P.E. = 0 initially. Then it starts burning fuel and starts moving. Now P.E. is not cero, but K.E. is K.E. of exhaust + K.E. of rocket. So where did this energy come from?

I've highlighted it. Maybe I'm lucky enough that you happen to read it this time.

Etc.

 

 

PS: By fuel + rocket I mean 'exhaust + rocket.'

Good day, sir.

[martillo]

18 minutes ago, joigus said:

That's fairly clear to me. You don't understand...

Here (from page 8, where you apparently answered 'several times':

And indeed,

 

ddt∂L∂x˙−∂L∂x=0⇒ddt(mx˙)=−dVdx⇒mx¨+m˙x˙=Fext=−V′(x)

 

Take Fext=0 , and there's your super-special instance. I told you, didn't I?

More from unbearable page 8...

I told you conservation of momentum applies for overall rocket + exhaust in free space, and I told you how.

I've highlighted it. Maybe I'm lucky enough that you happen to read it this time.

Etc.

 

 

PS: By fuel + rocket I mean 'exhaust + rocket.'

Good day, sir.

I apologize. I read it to fast. I think I can agree in everything there. That's a problem I have while discussing with several ones at the same time, to read to fast. By the way we have discussed about some things may be time ago and I find you very knowledgeable and reasonable and have all my respect. 

I see now your approach is very similar to that we are having with studiot now. Your comments will be very important to me as always have been.

18 minutes ago, joigus said:

So where did this energy come from?

From the combustion of the fuel. Energy is stored in the fuel.

26 minutes ago, joigus said:

And indeed,

 

ddt∂L∂x˙−∂L∂x=0⇒ddt(mx˙)=−dVdx⇒mx¨+m˙x˙=Fext=−V′(x)

 

Take Fext=0 , and there's your super-special instance. I told you, didn't I?

I totally agree with this I think...

Edited February 24, 2023 by martillo

[studiot]

1 hour ago, martillo said:

I apologize. I read it to fast. I think I can agree in everything there. That's a problem I have while discussing with several ones at the same time, to read to fast. By the way we have discussed about some things may be time ago and I find you very knowledgeable and reasonable and have all my respect. 

I see now your approach is very similar to that we are having with studiot now. Your comments will be very important to me as always have been.

 

1 hour ago, joigus said:

PS: By fuel + rocket I mean 'exhaust + rocket.'

There is much to be learned from the study of the rocket system including just how much is hidden in assumptions and definitions.

For instance what is the coordinate system we are placing our equations in ?

The centres of mass of both the rocket and the exhaust are not only moving, relative to a some fixed 'ground' observer, but also moving within the rocket and exhaust subsystems respectively.

This makes the application of calculus more difficult, particularly if we stick to the traditional d/dx notation for space.

Ther is not such issue with d/dt however.

But where is the origin of x located ?

 

Then the question of definitions

By 'rocket' I mean the rocket plus any unburnt fuel, at all times.

By exhaust or firing or burning I mean burnt and expelled fuel.

There is no delay between burning and expulsion that needs to be considered.

 

But a further interesting question that comes out of my earlier one.

Can the rocket system drive forward under power at constant velocity?

If so under what conditions ?

 

I feel sure that overhasty appreciation (or not) of these fundamentals lies behind all the misunderstands shown in this thread.

[martillo]

2 hours ago, joigus said:

And indeed,

 

ddt∂L∂x˙−∂L∂x=0⇒ddt(mx˙)=−dVdx⇒mx¨+m˙x˙=Fext=−V′(x)

 

Take Fext=0 , and there's your super-special instance. I told you, didn't I?

I just disagree with your comment below of the equation because if the rocket is treated as an open system as you said at page 6 of the thread then F_ext ≠ 0. F_ext is equal to the force that the exhaust produce on the rocket F_ext = -v_edm/dt which is the Thrust Force.

Edited February 24, 2023 by martillo

[Lorentz Jr]

Explaining relativity: 8 pages.

Explaining Newtonian mechanics: 9 pages and counting. 😂

[studiot]

11 minutes ago, martillo said:

I just disagree with your comment below of the equation because if the rocket is treated is as an open system as you said at page 6 of the thread then F_ext ≠ 0. F_ext is equal to the force that the exhaust produce on the rocket F_ext = -v_edm/dt which is the Thrust Force.

Thank you for bringing that up,

you have reminded me that I wanted to refer to your use of opn / closed systems.

By a closed system you should actually mean an isolated system.

Open systems allow mass and energy exchange with the outside world.

Closed systems allow energy but not mass exchange with the outside world.

Isolated systems do not allow either energy or mass exchange with the outside world.

https://mechaengineerings.wordpress.com/tag/closed-system/

 

This is really as used in Physics and Thermodynamics, ansd does not take forces into account.

Mechanics adds some additional concepts for Forces

The free body diagram

Force isolation

Neither of which mean that no forces act on the body of interest.

[martillo]

53 minutes ago, studiot said:

There is much to be learned from the study of the rocket system including just how much is hidden in assumptions and definitions.

For instance what is the coordinate system we are placing our equations in ?

Seems straightforward to consider horizontal coordinate x and vertical coordinate z. Then it can be considered two equations, one in the x axis without gravity force involved and the other in the z axis with gravity included. The equation in the x axis is which we were considering in the entire thread and it has all Mathematical and Physical strictness I think.

53 minutes ago, studiot said:

The centres of mass of both the rocket and the exhaust are not only moving, relative to a some fixed 'ground' observer, but also moving within the rocket and exhaust subsystems respectively.

This makes the application of calculus more difficult, particularly if we stick to the traditional d/dx notation for space.

Ther is not such issue with d/dt however.

I agree although I don't think there's a problem with the x notation.

 

53 minutes ago, studiot said:

But where is the origin of x located ?

Depends on the considered problem,. If the rocket is ready to launch then the origin is at the base of the rocket. If we consider just the rocket flying in the air from some instant then the origin would be the place of the rocket at that instant.

 

53 minutes ago, studiot said:

Then the question of definitions

By 'rocket' I mean the rocket plus any unburnt fuel, at all times.

By exhaust or firing or burning I mean burnt and expelled fuel.

I agree.

53 minutes ago, studiot said:

There is no delay between burning and expulsion that needs to be considered.

I think it would be much better but I will try to follow your approach. I will point out if I see some problem with this at some time.

53 minutes ago, studiot said:

But a further interesting question that comes out of my earlier one.

Can the rocket system drive forward under power at constant velocity?

If so under what conditions ?

Well, it will need a mobile exhausting system to control the direction of the flux of the exhaust and/or some mobile wings as some missiles have.

I'm just looking forward for your announced treatment of the problem of the rocket...

Edited February 24, 2023 by martillo

[studiot]

3 minutes ago, martillo said:

18 minutes ago, studiot said:

But a further interesting question that comes out of my earlier one.

Can the rocket system drive forward under power at constant velocity?

If so under what conditions ?

I didn't get the point...

What exactly did you not understand ?

If the rocket is moving forwards whilst burning and expelling exhaust is it possible to adjust the rocket drive so that its velocity, V, is constant  ?

If it is possible how would you achieve this ?

 

Also please please read the rest of my post properly before making wrong guesses about coordinate systems.

[martillo]

18 minutes ago, Lorentz Jr said:

Explaining Newtonian mechanics: 9 pages and counting. 😂

Is much more than that I think...

9 minutes ago, studiot said:

What exactly did you not understand ?

If the rocket is moving forwards whilst burning and expelling exhaust is it possible to adjust the rocket drive so that its velocity, V, is constant  ?

If it is possible how would you achieve this ?

 

Also please please read the rest of my post properly before making wrong guesses about coordinate systems.

A control in the rocket's motor on the quantity of exhausting (control of v_e ) would be needed.

 

Edited February 24, 2023 by martillo

[studiot]

35 minutes ago, martillo said:

Is much more than that I think...

A control in the rocket's motor on the quantity of exhausting (control of v_e ) would be needed.

 

To do what exactly ?

I'm not being funny, this is a fundamental property of a rocket drive we are talking about.

You have not directly answered my question which started    -  Is it possible ...?

You have made nearly 10 pages of complaint about conventional analysis of the rocket drive.

So exactly what value of v_e would be required to maintain the condition V = a constant ?

What is your mathematical relationship between v_e and V that would make this happen ?

[martillo]

16 minutes ago, studiot said:

To do what exactly ?

I'm not being funny, this is a fundamental property of a rocket drive we are talking about.

You have not directly answered my question which started    -  Is it possible ...?

You have made nearly 10 pages of complaint about conventional analysis of the rocket drive.

So exactly what value of v_e would be required to maintain the condition V = a constant ?

What is your mathematical relationship between v_e and V that would make this happen ?

In the problem we are analyzing we have constant exhaust in velocity and mass, v_e = constant and dm_e/dt = -dm/dt = constant. We have then a constant force F on the rocket which is the Thrust Force, F = -v_edm/dt and the rocket have a constant acceleration continuously increasing its velocity v.

To have constant velocity v no force would be applied except for the case of air resistance...

 

Edited February 24, 2023 by martillo

[joigus]

18 minutes ago, martillo said:

F_ext is equal to the force that the exhaust produce on the rocket F_ext = -v_edm/dt which is the Thrust Force.

No. F_ext stands for 'total external force' for both the rocket --with unburnt fuel-- and the exhaust. You didn't understand me. I will need some more time, as comments are piling up. I see no end to this...

 

[martillo]

4 hours ago, joigus said:

And indeed,

 

ddt∂L∂x˙−∂L∂x=0⇒ddt(mx˙)=−dVdx⇒mx¨+m˙x˙=Fext=−V′(x)

 

Take Fext=0 , and there's your super-special instance. I told you, didn't I?

Just to clarify, what I agree here is that this is the equation currently being applied in the rocket motion. Here is used the Force Equation dp/dt = F_ext. Is not my proposed equation, I claim that actually the equation should be: ma = mdv/dt = F_ext and is not zero in the problem we analyze.

Edited February 24, 2023 by martillo

[joigus]

2 minutes ago, martillo said:

Just to clarify, what I agree here is that this the equation currently being applied in the rocket motion. Here is used the Force Equation dp/dt = F_ext. Is not my proposed equation, I claim that actually the equation should be: ma = mdp/dt = F_ext and is not zero in the problem we analyze.

This is not even dimensionally correct. Besides, m in both sides of the equation would simplify. 

Please, review your maths before you commit another badly wrong comment.

[martillo]

3 minutes ago, joigus said:

No. F_ext stands for 'total external force' for both the rocket --with unburnt fuel-- and the exhaust. You didn't understand me. I will need some more time, as comments are piling up. I see no end to this...

But you said to be considering the rocket as an open system with an external force applied to it... 

1 minute ago, joigus said:

This is not even dimensionally correct. Besides, m in both sides of the equation would simplify. 

Please, review your maths before you commit another badly wrong comment.

Right I edited it:

5 minutes ago, martillo said:

Just to clarify, what I agree here is that this the equation currently being applied in the rocket motion. Here is used the Force Equation dp/dt = F_ext. Is not my proposed equation, I claim that actually the equation should be: ma = mdv/dt = F_ext and is not zero in the problem we analyze.

 

[joigus]

1 minute ago, martillo said:

But you said to be considering the rocket as an open system with an external force applied to it... 

Yes. The rocket --never mind it being an open system-- has its own Lagrange equation. The beauty of Lagrange is you don't even have to think about forces. It's all in describing so-called configurations of the system. That is: How many variables do I need to know where it's at in its evolution?

You don't even need to consider whether you are in an inertial system or not. Coordinates could be curvilinear for all you care. Lagrange's method takes care of everything else, including inertial forces.

Edited February 24, 2023 by joigus
minor addition

[martillo]

13 minutes ago, joigus said:

Yes. The rocket --never mind it being an open system-- has its own Lagrange equation. The beauty of Lagrange is you don't even have to think about forces. It's all in describing so-called configurations of the system. That is: How many variables do I need to know where it's at in its evolution?

You don't even need to consider whether you are in an inertial system or not. Coordinates could be curvilinear for all you care. Lagrange's method takes care of everything else, including inertial forces.

Unfortunately I'm not able to analyze in terms of Lagrangians. That is Variational Calculus and as I said several times I have no expertise at all in this area. That's why I'm waiting for studiot's treatment of the problem now...

Edited February 24, 2023 by martillo

[joigus]

1 minute ago, martillo said:

Unfortunately I'm not able to analyze in terms of Lagrangians. That is Variational Calculus and as I said several times I have no expertise at all in this area.

OK. I'm analising everything in terms of Newtonian mechanics. I need some time to react to @studiot and @sethoflagos, and see if we agree on reference systems and everything else.

You can either ignore external fields --or consider the system in free space-- or not. I'll think about a situation that's as general as possible without the whole thing being a mess.

Real rockets, of course, require many other considerations.

[martillo]

17 minutes ago, joigus said:

OK. I'm analising everything in terms of Newtonian mechanics. I need some time to react to @studiot and @sethoflagos, and see if we agree on reference systems and everything else.

You can either ignore external fields --or consider the system in free space-- or not. I'll think about a situation that's as general as possible without the whole thing being a mess.

Real rockets, of course, require many other considerations.

Thanks! I would appreciate your reasonings on the problem very much.

 

 

Just for the case, I post here my derivation of the thrust based on the Force Equation as F = ma valid even for variable mass. I presented it at the OP but with a couple of stupid silly mistakes I have corrected in it:

 

Rocket thrust force derivation

The thrust equation of the rocket is derived here considering the approximation that the mass of the expelled fuel of the rocket is negligible compared with the mass of the mass of the rocket plus the mass of its contained fuel:

Momentum and Force equations:

p = mv, F = ma = mdv/dt

dp/dt = mdv/dt + vdm/dt = F + vdm/dt

 

Considerations:

a) masses equations:

m = mass of the rocket with its contained fuel

m_e = mass of the expelled fuel

m’ = m + me = constant = total mass of the system rocket with total fuel

dm’/dt = dm/dt + dm_e/dt = 0

Then: dm_e/dt = -dm/dt

b) velocities equations (one dimension):

v = absolute velocity of the rocket

v_e = velocity of the expelled fuel in relation to the rocket assumed constant

u = absolute velocity of the expelled fuel

v_e = v – u = constant

dv_e/dt = dv/dt – du//dt = 0

Then: du/dt = dv/dt

 

Derivation:

Total momentum of the system rocket with total fuel: p’ = mv + m_eu

dp’/dt = d(mv + m_eu)/dt = mdv/dt + v/dm/dt + m_edu/dt + u/dm_e/dt

Considering: du/dt = dv/dt and dm_e/dt = -dm/dt

dp’/dt = (m + m_e)dv/dt + (v-u)dm/dt

As v – u = v_e and considering m_e << m (m + m_e ≈ m) then:

p’ ≈ mdv/dt + v_edm/dt and as dp’/dt = 0 then:

mdv/dt ≈ -v_edm/dt

Finally F = ma = mdv/dt ≈ -v_edm/dt under the approximation m_e << m

Then the rocket thrust force is: F ≈ -v_edm/dt

Edited February 24, 2023 by martillo

[studiot]

1 hour ago, martillo said:

To have constant velocity v no force would be applied except for the case of air resistance...

At last.

We are getting somewhere.

No just air resistance of course.

 

The conclusion is that

By itself in free space the rocket cannot 'run the rocket engine' and travel at constant velocity.

Alternatively the a working rocket muct be under acceleration unless it is also being acted upon by an external force for example gravity or air resistance.

If it did not do so it would be breaking all 3 of Newton's laws.