Real equation of Force F = ma not dp/dt

[martillo]

This is an extract of "Appendix A" in my manuscript on a new theory which can be accessed through the link in my profile if someone would be interested.

The real Equation of Force is F = ma 
 
Today's Physics is stating that the Equation of Force is F = dp/dt. 
We will analyze the equation of motion of rockets to see that the real Equation of Force is: 

F = ma
 
A rocket has variable mass in its trajectory and it's important to see its motion equation. 
Let m be its variable mass at any instant in its movement composed by the mass of the rocket plus the mass of its contained fuel. 
I have made a search in the internet about rocket motion equations and all the sites agree in the equation: 
F = m(dv/dt) = –ve(dm/dt) where ve is the speed of the fuel expelled relative to a frame at rest. 
 
One web site:
http://www.braeunig.us/space/propuls.htm
 
They all agree that the force acting on the rocket is due to the expelled mass and is F = –ve(dm/dt) and that the equation of motion is F = m(dv/dt) = ma. 
I assume the equation have been completely verified experimentally with enough precision from a long time ago. 
It is evident that it is used the equation: 
F = ma
 for the force and not: 
F = dp/dt 

... ... ...

This indicates that today's Physics is wrong stating the Equation of Force as F = dp/dt. 
 
The right equation for force is F = ma even when mass varies. 
 
Note that the natural derivation of the famous equation E = mc2 by Relativity Theory has no sense since it is based in the wrong relation F = dp/dt. 
Relativity Theory becomes a wrong theory since it is based on a wrong law. 
 

[Strange]

15 minutes ago, martillo said:

The real Equation of Force is F = ma 

You are only 300 years too late with this breakthrough:

Quote

Second law:

In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma.

https://en.wikipedia.org/wiki/Newton's_laws_of_motion

17 minutes ago, martillo said:

Equation of Force is F = dp/dt. 

That is the same thing.

[martillo]

Not the same thing, not at all.

I think you didn´t get to the point...

[Strange]

12 minutes ago, martillo said:

Not the same thing, not at all.

It is trivial to see that it is identical. 

ma = m(dv/dt) = mdv/dt = dp/dt

41 minutes ago, martillo said:

Note that the natural derivation of the famous equation E = mc2 by Relativity Theory has no sense since it is based in the wrong relation F = dp/dt. 

How is F=dp/dt used in the derivation of that equation?

 

[swansont]

19 minutes ago, martillo said:

Not the same thing, not at all.

I think you didn´t get to the point...

A conveyor belt moves at speed v. You drop masses on it. Solve for the motion of the masses using F=ma

3 minutes ago, Strange said:

It is trivial to see that it is identical. 

ma = m(dv/dt) = mdv/dt = dp/dt

 

m (dv/dt) + v (dm/dt)

 

 

[martillo]

27 minutes ago, Strange said:

It is trivial to see that it is identical. 

ma = m(dv/dt) = mdv/dt = dp/dt

How is F=dp/dt used in the derivation of that equation?

 

 As swansont posted:

p = mv

dp/dt = mdv/dt + vdm/dt = ma + vdm/dt

 

Here is a link with the theoretical derivation of the equation E = mc2 based on F = dp/dt = d(mv)/dt:

http://www.emc2-explained.info/Emc2/Deriving.htm#.XUdykvZFyM8

 

 

[Strange]

7 hours ago, martillo said:

I think you didn´t get to the point...

What is the point?

Quote

The second law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.

[math]\mathbf{F} = \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = \frac{\mathrm{d}(m\mathbf{v})}{\mathrm{d}t}[/math]

The second law can also be stated in terms of an object's acceleration. Since Newton's second law is valid only for constant-mass systems,^[20][21][22] m can be taken outside the differentiation operator by the constant factor rule in differentiation. Thus,

[math]\mathbf{F} = m\,\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = m\mathbf{a}[/math]

https://en.wikipedia.org/wiki/Newton's_laws_of_motion

[martillo]

The point is that in both cases in consideration, Rocket Dynamics and Relativity Theory, the mass varies so dp/dt is not the same that ma. They are different: 

dp/dt = d(mv)/dt = mdv/dt + vdm/dt

dp/dt = ma + vdm/dt

There´s a problem then because Rocket Dynamics implies force defined as F = ma while Relativity Theory needs F = dp/dt.

 

[Strange]

6 minutes ago, martillo said:

They are different: 

He says and then proceeds to show that they are equivalent. (Not that, as your equation shows, F=ma is only valid for constant mass.)

[martillo]

4 minutes ago, Strange said:

He says and then proceeds to show that they are equivalent. (Not that, as your equation shows, F=ma is only valid for constant mass.)

But Rocket Dynamics shows F = ma necessarily valid even for variable mass.

Rockets' experiments validate this.

[Eise]

Martillo,

Imagine a simple rocket, having one engine, giving a constant thrust. As it is climbing higher, it of course loses mass: the fuel it uses. For ease also assume the gravity field constant (e.g. assume s small travelling distance).

What happens to the acceleration of the rocket during the engine works? Does it stay the same, does it increase, or even decrease? How do you calculate this?

Just now, martillo said:

But Rocket Dynamics shows F = ma necessarily valid even for variable mass.

Rockets' experiments validate this.

So m stays the same, even when the rocket loses weight due to its using fuel?

[martillo]

13 minutes ago, Eise said:

Martillo,

Imagine a simple rocket, having one engine, giving a constant thrust. As it is climbing higher, it of course loses mass: the fuel it uses. For ease also assume the gravity field constant (e.g. assume s small travelling distance).

What happens to the acceleration of the rocket during the engine works? Does it stay the same, does it increase, or even decrease? How do you calculate this?

So m stays the same, even when the rocket loses weight due to its using fuel?

No, m not "stays the same". m is variable and the valid relation is F = ma with variable m.

The well known "Thrust Equation" for rockets is: F = ma = -v_edm/dt

You can easily Google for it.

Edited August 5, 2019 by martillo

[Strange]

30 minutes ago, martillo said:

the mass varies so dp/dt is not the same that ma. They are different: 

That makes no sense.

 dp/dt accounts for varying mass. If you want to take varying mass into account in F=ma then you need the time derivative of mass dm/dt and so you end up with dp/dt.

 

 

8 hours ago, martillo said:

Today's Physics is stating that the Equation of Force is F = dp/dt. 

Actually, that is what Newton said. It is usually taught (initially) as F=ma because that is simpler. 

[martillo]

5 minutes ago, Strange said:

That makes no sense.

 dp/dt accounts for varying mass. If you want to take varying mass into account in F=ma then you need the time derivative of mass dm/dt and so you end up with dp/dt.

 

 

Something does not make sense. The definition F = dp/dt makes no sense. The true is that F = ma even for varying mass m. Rocket Dynamics demonstrates it and Relativity Theory fails.

Edited August 5, 2019 by martillo

[Strange]

Just now, martillo said:

Something does not make sense. The definition F = dp/dt makes no sense.

I assume that is why this form is not taught first. It requires the student to have gained a small amount of math and physics understanding. 

[Eise]

44 minutes ago, martillo said:

m is variable and the valid relation is F = ma with variable m.

Now apply this to the question you did not answer in my previous posting:

49 minutes ago, Eise said:

What happens to the acceleration of the rocket during the engine works? Does it stay the same, does it increase, or even decrease? How do you calculate this?

Please, show your calculation, using F = ma.

[martillo]

1 hour ago, Eise said:

Now apply this to the question you did not answer in my previous posting:

Please, show your calculation, using F = ma.

Ok, here my calculation:

momentum: p = mv

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

First consider the total system: Rocket plus total fuel, contained and expelled one.

Total mass: m' = m + m_e where m is the mass of the rocket and its contained fuel at velocity v and m_e is the mass of the expelled fuel at velocity u relative to a frame at rest.  

m' constant then dm'/dt = dm/dt + dm_e/dt = 0

Then:

dm_e/dt = -dm/dt

Now:

p' = mv + m_eu

dp'/dt = mdv/dt + vdm/dt + m_edu/dt + udm_e/dt = 0 (net force F' = 0 and dm'/dt = 0)

u constant then du/dt = 0

mdv/dt + vdm/dt - udm/dt = 0

mdv/dt = -(v - u)dm/dt

(v - u) = v_e velocity of expelled fuel relative to the rocket.

Then the thrust equation:

mdv/dt = -v_edm/dt

a = dv/dt 

and as is considered F = ma

F = -v_edm/dt is the force exerted on the rocket

 

 

 

 

 

 

Edited August 5, 2019 by martillo

[Eise]

6 minutes ago, martillo said:

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

So you are using dp/dt.

And you did not quite answer my question, which was:

1 hour ago, Eise said:

What happens to the acceleration of the rocket during the engine works? Does it stay the same, does it increase, or even decrease? How do you calculate this?

As said, for simplicity, assume a constant thrust, a homogeneous gravitational field, no air resistance etc etc.

[swansont]

3 hours ago, martillo said:

 There´s a problem then because Rocket Dynamics implies force defined as F = ma while Relativity Theory needs F = dp/dt.

No, rocket dynamics uses dp/dt and uses the changing mass in the analysis

 

1 hour ago, martillo said:

 (v - u) = v_e velocity of expelled fuel relative to the rocket.

Then the thrust equation:

mdv/dt = -v_edm/dt

a = dv/dt 

and as is considered F = ma

F = -v_edm/dt is the force exerted on the rocket

Your link doesn't say that's the net force. It says it is a force (and it has units of force), called the thrust. But you have to realize that they are treating the rocket and the exhaust as one system in the derivation. IOW, they aren't talking about the force on the rocket. The net force is zero in their derivation.

Asking what is the force on the rocket is a different question than what the derivation in your link addresses. The end goal is solving for v(t), which is expressed as Tsiolkovsky's rocket equation. But even so, they address this when discussing the action/reaction force pair. There's an action force of the mass leaving the rocket, and a reaction force, acting on the rocket, in accordance with Newton's third law. That force is given as ma.

[Strange]

1 hour ago, martillo said:

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

So:

  dp/dt = F + vdm/dt

Therefore

  F = dp/dt - v.dm/dt

If the mass is constant, dm/dt = 0 and therefore

  F = dp/dt - 0

  F = dp/dt

Q.E.D.

This is schoolboy level mathematics. (That is why even I can do it!)

11 hours ago, martillo said:

They all agree that the force acting on the rocket is due to the expelled mass and is F = –ve(dm/dt) and that the equation of motion is F = m(dv/dt) = ma. 
I assume the equation have been completely verified experimentally with enough precision from a long time ago. 
It is evident that it is used the equation: 
F = ma
 for the force and not: 
F = dp/dt 

Read what you wrote:

the force acting on the rocket is due to the expelled mass and is F = –ve(dm/dt)

In other words, F = -dp/dt

the equation of motion is F = m(dv/dt)

In other words, fo = dp/dt

You contradict yourself at every step.

[martillo]

I apologize. There´s something wrong in the calculation I tried for Eise.

It's wrong to state p' = mv + m_ev_e for the rocket and the expelled fuel.

The calculation is wrong twice giving a right result...

But the initial post is right...

I'm working on that.

 

 

5 hours ago, swansont said:

Your link doesn't say that's the net force. It says it is a force (and it has units of force), called the thrust. But you have to realize that they are treating the rocket and the exhaust as one system in the derivation. IOW, they aren't talking about the force on the rocket. The net force is zero in their derivation.

So do I. The net external force on the total system is zero. 

 

5 hours ago, swansont said:

Asking what is the force on the rocket is a different question than what the derivation in your link addresses. The end goal is solving for v(t), which is expressed as Tsiolkovsky's rocket equation. But even so, they address this when discussing the action/reaction force pair. There's an action force of the mass leaving the rocket, and a reaction force, acting on the rocket, in accordance with Newton's third law. That force is given as ma.

I treat that the same way. The reaction force on the rocket is given by the well known thrust equation: F = ma = v_edm/dt

4 hours ago, Strange said:

So:

  dp/dt = F + vdm/dt

Therefore

  F = dp/dt - v.dm/dt

If the mass is constant, dm/dt = 0 and therefore

  F = dp/dt - 0

  F = dp/dt

Q.E.D.

This is schoolboy level mathematics. (That is why even I can do it!)

Read what you wrote:

the force acting on the rocket is due to the expelled mass and is F = –ve(dm/dt)

In other words, F = -dp/dt

the equation of motion is F = m(dv/dt)

In other words, fo = dp/dt

You contradict yourself at every step.

Strange, can't you understand that, in what is considered, the MASS m IS NOT CONSTANT? Not for rockets, not for Relativity!

Edited August 5, 2019 by martillo

[swansont]

39 minutes ago, martillo said:

 So do I. The net external force on the total system is zero. 

 

I treat that the same way. The reaction force on the rocket is given by the well known thrust equation: F = ma = v_edm/dt

OK, then.  ma = v_edm/dt is not a statement of Newton's second law; it's the third law.  

It is derived from the second law, using F = dp/dt

 

 

[Strange]

29 minutes ago, martillo said:

Strange, can't you understand that in what is considered the MASS m IS NOT CONSTANT? Not for rockets, not for Relativity!

If it is not constant, then you can't use F=ma.

Quote

The second law can also be stated in terms of an object's acceleration. Since Newton's second law is valid only for constant-mass systems,^[20][21][22] m can be taken outside the differentiation operator by the constant factor rule in differentiation. 

...

Any mass that is gained or lost by the system will cause a change in momentum that is not the result of an external force. A different equation is necessary for variable-mass systems 

https://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton's_second_law

 

16 hours ago, martillo said:

One web site:
http://www.braeunig.us/space/propuls.htm

Equation 1.2 F = dp/dt

Even your own sources prove you wrong.

[martillo]

2 minutes ago, swansont said:

OK, then.  ma = v_edm/dt is not a statement of Newton's second law; it's the third law.  

It is derived from the second law, using F = dp/dt

That´s right. The rockets dynamics works fine with cassical Mechanics and the Newton's Laws if the second is expressed as F = ma. They are fine for me. What I argue is that F = ma has general aplication and is valid even when mass varies. The problem is that this contradicts the statement F = dp/dt which is largely used in Relativity.

8 minutes ago, Strange said:

If it is not constant, then you can't use F=ma.

https://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton's_second_law

 

Equation 1.2 F = dp/dt

Even your own sources prove you wrong.

That equation is wrong. Newton's original formulation of the Second Law is wrong. They actually use the relation F = ma as can be read right after. Is not a source o f mine. Is what can be found in the web. I just analize it.

[swansont]

3 minutes ago, martillo said:

That´s right. The rockets dynamics works fine with cassical Mechanics and the Newton's Laws if the second is expressed as F = ma. They are fine for me. What I argue is that F = ma has general aplication and is valid even when mass varies. The problem is that this contradicts the statement F = dp/dt which is largely used in Relativity.

No. It's probably important to note that (AFAICT) nobody asked or stated what the net force on the moving rocket was in the link. It's not needed to solve for v(t), and usually the whole point of such analysis is to get equations of motion.

Knowing the net force on the moving rocket isn't really useful information, precisely because you can't just equate it to ma, since F≠ma for a varying mass.

But saying that "rockets dynamics works fine with cassical (sic) Mechanics and the Newton's Laws if the second is expressed as F = ma" is ludicrous if you look at the derivation, since it's obvious that they did not do that. They used F = dp/dt. You are the only one here making the case for F=ma being the proper form, and you have not provided any evidence to support your position.