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

[sethoflagos]

6 minutes ago, martillo said:

If you consider F = dp/dt then F = mdv/dt + vdm/dt which would be equal to mdv/dt only if dm/dt = 0 what would mean constant mass.

Is not for me, for me is very right to apply F = ma always even for variable mass. For me is the right equation of force. That is the title of the thread.

dp/dt gives you BOTH sides of the equation

dp/dt = d(mv)/dt = mdv/dt + vdm/dt =0

Hence:     mdv/dt = -vdm/dt

[martillo]

7 minutes ago, sethoflagos said:

dp/dt gives you BOTH sides of the equation

dp/dt = d(mv)/dt = mdv/dt + vdm/dt =0

Hence:     mdv/dt = -vdm/dt

That is valid for the total system of the rocket with the total fuel (contained and expelled) only.

[sethoflagos]

5 minutes ago, martillo said:

That is valid for the total system of the rocket with the total fuel (contained and expelled) only.

So? What is omitted that is pertinent to your OP?

 

[martillo]

15 minutes ago, sethoflagos said:

So? What is omitted that is pertinent to your OP?

 

You posted:

36 minutes ago, sethoflagos said:

dp/dt gives you BOTH sides of the equation

dp/dt = d(mv)/dt = mdv/dt + vdm/dt =0

Hence:     mdv/dt = -vdm/dt

In case of that applied to the total system m is constant and everything is zero:

dpdt = 0 and also dm/dt and dv/dt are zero. The mass is constant, the acceleration would be constant and zero.

Just to mention, just for the case, you arrived at mdv/dt = -vdm/dt but this is not the thrust equation. You must decompose the system into rocket and fuel and elaborate. The thrust equation is mdv/dt = -v_edm/dt where v_e is the constant velocity of the expelled fuel relative to the rocket and m is the mass of the rocket with its contained fuel only.

Edited February 21, 2023 by martillo

[studiot]

38 minutes ago, martillo said:

In current Physics is widely considered F = dp/dt and that in the case, and only in the case of constant mass, the equation F = ma applies. Any Physics' textbook says that. So it would be wrong, in current Physics, to apply it on something with variable mass as it is done for rockets, or not? I do not understand how you don't see this.

If you consider F = dp/dt then F = mdv/dt + vdm/dt which would be equal to mdv/dt only if dm/dt = 0 what would mean constant mass.

Is not for me, for me is very right to apply F = ma always even for variable mass. For me is the right equation of force. That is in the title of the thread.

 

Surely you dont need a course in basic calculus ?

 

Definition momentum = mass x velocity

p = mv

Therefore if both m and v are variable

[math]\frac{{dp}}{{dt}} = m\frac{{\partial v}}{{\partial t}} + v\frac{{\partial m}}{{\partial t}}[/math]

Where both m and v are functions of t.

To work this out you would need to have equations for both m and v as functions of t, unless you could devise an equation connecting m and v.

Do you have such equations ?

Some treatments specify that the mass flows out at a constant rate, that is

[math]\frac{{\partial m}}{{\partial t}} = C[/math]

 

Alternatively you can calculate the changing mass flow required to produce constant acceleration.

 

 

[Genady]

4 minutes ago, martillo said:

constant velocity of the expelled fuel relative to the rocket

Constant velocity relative to the rocket, but it changes relative to an inertial frame together with a changing velocity v of the rocket. Check in your source, how that v_e is defined. 

[sethoflagos]

1 minute ago, martillo said:

You posted:

In case of that applied to the total system m is constant and everything is zero:

Someone gives you $5. Someone takes $5 off you. You are left with $0. 

4 minutes ago, martillo said:

dpdt = 0 and also dm/dt and dv/dt are zero. The mass is constant, the acceleration would be constant and zero.

Someone gives you $5. Someone takes $5 off you. You are left with $0.

6 minutes ago, martillo said:

Just to mention, just for the case, you arrived at mdv/dt = -vdm/dt but this is not the thrust equation. You must decompose the system into rocket and fuel and elaborate. The thrust equation is mdv/dt = -v_edm/dt where v_e is the constant velocity of the expelled fuel.

Neither did I mention that integration of the equation yieds the Tsiolkovsky rocket equation. The OP didn't ask for it.

But you can mug up on it at https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

[martillo]

16 minutes ago, Genady said:

Constant velocity relative to the rocket, but it changes relative to an inertial frame together with a changing velocity v of the rocket. Check in your source, how that v_e is defined. 

In the inertial frame considered for the rocket... I will check it out but seems to be u = v - v_e.

Edited February 21, 2023 by martillo

[Genady]

4 minutes ago, martillo said:

In the inertial frame considered for the rocket... I will check it out but seems to be u = v - v_e.

The rocket is not an inertial frame because it accelerates. Consider relative to the ground.

[martillo]

38 minutes ago, studiot said:

Surely you dont need a course in basic calculus ?

Where you think I went wrong?

38 minutes ago, studiot said:

To work this out you would need to have equations for both m and v as functions of t, unless you could devise an equation connecting m and v.

Do you have such equations ?

m and v as a function of t would give the trajectory of the rocket. That is not necessary to calculate here. The equation connecting v and m is the Thrust Equation we are already considering since the beginning: mdv/dt = -v_edm/dt.

Solving this equation would give v as a function of m for the considered value of v_e if you want that.

34 minutes ago, sethoflagos said:

Someone gives you $5. Someone takes $5 off you. You are left with $0. 

Someone gives you $5. Someone takes $5 off you. You are left with $0.

Neither did I mention that integration of the equation yieds the Tsiolkovsky rocket equation. The OP didn't ask for it.

But you can mug up on it at https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

I don't understand your point of discussion. Please explain.

16 minutes ago, Genady said:

The rocket is not an inertial frame because it accelerates. Consider relative to the ground.

I said the inertial frame considered for the rocket not on the rocket. That is ground.

Edited February 21, 2023 by martillo

[Boltzmannbrain]

16 hours ago, martillo said:

Yes, equation 1.5. Right below they say:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

nm

Edited February 21, 2023 by Boltzmannbrain
The website is not working well with my cell phone.

[sethoflagos]

41 minutes ago, martillo said:

I don't understand your point of discussion. Please explain.

Yes, mdv/dt = -vdm/dt is indeed the overall force balance. So let's separate rocket from exhaust components.

Consider an observer travelling with a constant velocity v

He sees the rocket of mass m gain a velocity dv.

He sees a small amount of mass dm ejected at a velocity of v_e in the opposite direction

No nett motion of the total becomes mdv = -v_edm

Hence we recover your thrust equation mdv/dt = -v_edm/dt

It really isn't rocket science ..... er 🤨

[martillo]

6 minutes ago, sethoflagos said:

Yes, mdv/dt = -vdm/dt is indeed the overall force balance. So let's separate rocket from exhaust components.

Consider an observer travelling with a constant velocity v

He sees the rocket of mass m gain a velocity dv.

He sees a small amount of mass dm ejected at a velocity of v_e in the opposite direction

No nett motion of the total becomes mdv = -v_edm

Hence we recover your thrust equation mdv/dt = -v_edm/dt

It really isn't rocket science ..... er 🤨

Well may be you found a different way to reach at the Thrust equation. Actually I don't know if it is strictly right but there's no problem with that equation. The problem is how you derive the Thrust Force from it. The force on the rocket.  How would you proceed?

Edited February 21, 2023 by martillo

[sethoflagos]

23 minutes ago, martillo said:

The problem is how you derive the Thrust Force from it. The force on the rocket.  How would you proceed?

I've just done it. According to your earlier definition:

2 hours ago, martillo said:

The thrust equation is mdv/dt = -v_edm/dt where v_e is the constant velocity of the expelled fuel relative to the rocket and m is the mass of the rocket with its contained fuel only.

A more interesting question would be to ask "How was the exhaust decelerated from v to v - v_e?"

One for another day perhaps. 

Edited February 21, 2023 by sethoflagos

[martillo]

6 minutes ago, sethoflagos said:

A more interesting question would be to ask "How was the exhaust decelerated from v to v - v_e?"

The motor of the rocket does it. It depends on how the motor was designed and built. Lot of engineering behind...

[Genady]

1 hour ago, martillo said:

I said the inertial frame considered for the rocket not on the rocket. That is ground.

I see. When you said, 'considered for' you meant, 'relative to'. OK. Lost in translation.

[martillo]

14 minutes ago, Genady said:

I see. When you said, 'considered for' you meant, 'relative to'. OK. Lost in translation.

You mean misunderstanding or translation to other language? If translation, just curious about your language... 😄 May be Spanish too... Working in English makes a foreign one think very much in the right meaning of each word and phrase. Good to gain formality while writing the things.

Edited February 21, 2023 by martillo

[Genady]

2 minutes ago, martillo said:

You mean misunderstanding or translation to other language?

I mean misunderstanding.

 

2 minutes ago, martillo said:

just curious about your language

I have four in total. But everyday language is English.

[martillo]

Fortunately we have automatic orthographic correction. It helps a lot.

[studiot]

43 minutes ago, sethoflagos said:

A more interesting question would be to ask "How was the exhaust decelerated from v to v - v_e?"

Perhaps @MigL knows the answer to this but I would be very suprised if any useful rocket was designed for constant mass exhaust.

Different accelerations and velocities will also be required at different stages of the flight.

This implies that the force applied to the rocket will vary over the flight time.

 

These are parameters that can be worked out for a desired flight history.

[swansont]

2 hours ago, martillo said:

It is used F = ma on something with variable mass. Don't you see this at least as suspicious?

The only place where F=ma appears is right after equation 1.2, where they note that it comes from F=dp/dt

There’s nothing suspicious that I see. 

4 hours ago, martillo said:

Any Physics' textbook says that. So it would be wrong, in current Physics, to apply it on something with variable mass as it is done for rockets, or not?

It’s not being applied. Conservation of momentum is, and then dp/dt is used to get the force.

 

3 hours ago, martillo said:

That is valid for the total system of the rocket with the total fuel (contained and expelled) only.

And that’s how it is applied.

[martillo]

24 minutes ago, swansont said:

The only place where F=ma appears is right after equation 1.2, where they note that it comes from F=dp/dt

There’s nothing suspicious that I see. 

What about right below the Equation 1.5. The Thrust Equation. They do not derive the force as F = dp/dt. In spite of this they just say:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

This is to apply F = mdv/dt what is F = ma where m is the (variable) mass of the rocket, very dissimulated, I know. It is wrong in current Physics to apply F = ma on something with variable mass. In current Physics F = ma only applies for constant mass m. But for rockets F = ma works very well (as they function very well, isn't it?) so, it follows that the right equation for variable mass is F = ma. This is the key point in the subject.

1 hour ago, sethoflagos said:

A more interesting question would be to ask "How was the exhaust decelerated from v to v - v_e?"

The link I provided (http://www.braeunig.us/space/propuls.htm) also treats: Combustion and Exhaust Velocity, Specific Impulse, Rocket Engines, Power Cycles, Engine cooling, Solid Rockets Motors, Monopropellant Engines and Staging. You should take a look on it as you would be interested in those things.

Edited February 21, 2023 by martillo

[Genady]

20 minutes ago, martillo said:

In current Physics F = ma only applies for constant mass m.

It applies to a momentary mass, a mass at a moment in time.

[swansont]

26 minutes ago, martillo said:

What about right below the Equation 1.5. The Thrust Equation. They do not derive the force as F = dp/dt. In spite of this they just say:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

This is to apply F = mdv/dt what is F = ma where m is the (variable) mass of the rocket, very dissimulated, I know. It is wrong in current Physics to apply F = ma on something with variable mass. In current Physics F = ma only applies for constant mass m. But for rockets F = ma works very well (as they function very well, isn't it?) so, it follows that the right equation for variable mass is F = ma. This is the key point in the subject.

They refer to them as forces because they are. They took momentum and applied dp/dt to it. dp/dt is a force, despite your insistence to the contrary.

This is quite clearly spelled out in the derivation.

What they did not do is apply F=ma to anything with variable mass. I don’t know how you can claim that they are.

 

 

 

 

 

[martillo]

1 minute ago, Genady said:

It applies to a momentary mass, a mass at a moment in time.

No, in current Physics it applies for constant m only. This is clear in any basic Physics' text.