martillo

Real equation of Force F = ma not dp/dt

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Posted (edited)
11 minutes ago, swansont said:

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.

They use F = dp/dt for the total system rocket plus total fuel and apply conservation of momentum with net force F = 0 and dp/dt = 0. After, they derive another force, the force on the rocket alone as F = ma = vedm/dt.

They are two different forces!

Edited by martillo

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1 minute ago, martillo said:

They use F = dp/dt for the total system rocket plus total fuel and apply conservation of momentum with net force F = 0. After, they derive another force, the force on the rocket aloneas F = ma = vedm/dt.

They are different forces!

You just agreed that this was derived from the reaction forces. It's not a statement of the second law

NOBODY has applied F = ma while claiming you can do so from the second law. (Other than you)

And yes, they are different forces, acting on different objects. But since they must have an equal magnitude, it's very useful to use them, since we want v(t) and now we have an equation that has dv/dt in it, so we can integrate, and get the result without (mis)using Fnet = ma. Pretty neat, huh? 

 

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Posted (edited)
22 minutes ago, swansont said:

You just agreed that this was derived from the reaction forces. It's not a statement of the second law

NOBODY has applied F = ma while claiming you can do so from the second law. (Other than you)

And yes, they are different forces, acting on different objects. But since they must have an equal magnitude, it's very useful to use them, since we want v(t) and now we have an equation that has dv/dt in it, so we can integrate, and get the result without (mis)using Fnet = ma. Pretty neat, huh? 

 

No, they must not have equal magnitude. They don`t have. One is zero. The other is vedm/dt not zero. The net force on the total system is zero. The force on the rocket is not zero.

 

Remember this:

p = mv

F = ma

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

F = dp/dt only when dm/dt = 0. Particular case.

This is what I agree.

 

Edited by martillo

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59 minutes ago, martillo said:

No, they must not have equal magnitude. They don`t have. One is zero. The other is vedm/dt not zero. The net force on the total system is zero. The force on the rocket is not zero.

The rocket feels no force and is not accelerating? I don't think so. That's nonsense, and also a violation of Newton's third law.

 

59 minutes ago, martillo said:

 

Remember this:

p = mv

F = ma

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

F = dp/dt only when dm/dt = 0. Particular case.

F = dp/dt always. Your disagreement with this is something you have to justify. With something more than assertion.

 

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Posted (edited)
18 minutes ago, swansont said:

The rocket feels no force and is not accelerating? I don't think so. That's nonsense, and also a violation of Newton's third law.

 

F = dp/dt always. Your disagreement with this is something you have to justify. With something more than assertion.

 

What do you think I posted here?

Rockets dynamics not only justifies but demonstrates I'm right.

Edited by martillo

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Posted (edited)

Maybe it will help if you realize that it is net force not individual forces

Try this form

[latex]F_{net}=\frac{\Delta p}{\Delta t}[/latex]

[latex]\Delta p=m\Delta v=m(a\Delta t)=F_{net}\Delta t[/latex]

 

Edited by Mordred

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Posted (edited)
31 minutes ago, Mordred said:

Try this form

Fnet=ΔpΔt

Δp=mΔv=maΔ=FnetΔt

Missing a "t" after [math]ma \Delta[/math]?

[math]\Delta p=m \Delta v=ma \Delta t=Fnet \Delta t [/math]

Edited by Ghideon

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Posted (edited)

Good catch thanks was distracted by wife lmao corrected above

Edited by Mordred

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Posted (edited)
50 minutes ago, Mordred said:

Maybe it will help if you realize that it is net force not individual forces

Try this form

Fnet=ΔpΔt

Δp=mΔv=m(aΔt)=FnetΔt

 

Same mistake again. That is valid for constant mass m only which is valid for the total system of the rocket and the total fuel (contained plus expelled) but not for the rocket alone. The force on the rocket is a different one.

Edited by martillo

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Posted (edited)

How is that a mistake with regards to the rocket ship scenario ?

The original equation is for constant mass how is that different describing Newton's laws of inertia ?

You adapt the original equation for two varying values.

Edited by Mordred

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Posted (edited)
9 minutes ago, Mordred said:

How is that a mistake with regards to the rocket ship scenario ?

The net force is the sum of the force on the rocket plus the force on the expelled fuel which are equal but opposite following the action-reaction Newton's third law. The net force on the total system is zero. The force on the rocket the same as the force on the expelled fuel are not zero, they are given by the thrust equation F = vedm/dt which depends on the varying mass m.

Edited by martillo

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Posted (edited)
3 hours ago, martillo said:

They use F = dp/dt for the total system rocket plus total fuel and apply conservation of momentum with net force F = 0 and dp/dt = 0. After, they derive another force, the force on the rocket alone as F = ma = vedm/dt.

 

Just trying to catch up; need some help. To me it looks like:  [math]F = ma = v_{e} \frac{dm}{dt}[/math] means "force F on the rocket comes from the thrust given by the mass of the fuel exiting the rocket each second". OK I get that*.

But what is [math]ma[/math] in the middle actually physically describing in this case? Mass m and acceleration a is not input to the equation, free to have any value, but a result?

To me [math]ma[/math] in [math]F = ma = v_{e} \frac{dm}{dt}[/math] looks like they say: given an exhaust velocity [math]v_{e}[/math] and the time derivate of mass exiting the rocket  [math]\frac{dm}{dt}[/math] we can calculate what acceleration [math]a[/math] that thrust would give to a rocket with a fixed mass m. OR we could calculate the mass m that a rocket has to have, to be able to accelerate at acceleration=a given the thrust [math]v_{e} \frac{dm}{dt}[/math].

I'm fully aware of Newton's equations and a few basic formulas for rockets but I try to find what you are stating in this specific case, to be able to understand the claimed issue with relativity and take part in the discussion.

 

*) I hope, otherwise my question is invalid.

Edited by Ghideon

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Posted (edited)
35 minutes ago, Ghideon said:

Just trying to catch up; need some help. To me it looks like:  F=ma=vedmdt means "force F on the rocket comes from the thrust given by the mass of the fuel exiting the rocket each second". OK I get that*.

But what is ma in the middle actually physically describing in this case? Mass m and acceleration a is not input to the equation, free to have any value, but a result?

To me ma in F=ma=vedmdt  looks like they say: given an exhaust velocity ve  and the time derivate of mass exiting the rocket  dmdt  we can calculate what acceleration a  that thrust would give to a rocket with a fixed mass m. OR we could calculate the mass m that a rocket has to have, to be able to accelerate at acceleration=a given the thrust vedmdt .

I'm fully aware of Newton's equations and a few basic formulas for rockets but I try to find what you are stating in this specific case, to be able to understand the claimed issue with relativity and take part in the discussion.

 

*) I hope, otherwise my question is invalid.

I agree with everything here.

The issue is that F = ma is not the same as F = dp/dt.

Experimentally, in rockets dynamics, is proven that the relation F = ma = mdv/dt is valid even for a mass varying systems. But Relativity largely uses and depends on the relation F = dp/dt. They are different relations. The issue is which definition would be given for force in the most general case, F = ma or F = dp/dt?

Just mathematically can be derived the following:

By definition

p = mv

Then:

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

Now, for me, the true relation for force is F = ma and the equation for momentum and force can be written as:

dp/dt = F + vdm/dt

It can be seen that F = dp/dt is valid only in the cases where dm/dt = 0 which means cases or systems with constant mass but it is not the case for the rockets nor for relativistic systems!

I conclude then that Relativity Theory fails assuming generally F = dp/dt.

I conclude Relativity Theory a wrong theory.

 

Edited by martillo

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He is trying to account for two simultaneously varying values under constant thrust.

That's not what these equations are designed for. If you have constant thrust and fuel consumption varying the mass the acceleration also changes.

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44 minutes ago, martillo said:

The issue is that F = ma is not the same as F = dp/dt.

Not exactly the same, no.

As noted, you can't use F=ma when the mass is changing. Trying to calculate F(t) = m(t)a doesn't work.

F = dp/dt is more general. It works for constant mass (where only the velocity changes, in which case it is equivalent to F=ma) and it works when m varies.

I have seen people basing their attempt to disprove SR on some dodgy grounds, but this is like claiming that  5x2 is not the same as 2x5.

 

(And what it is about crackpots and relativity, anyway. We have known this is the way the universe works for well over 100 years. I grew up with this. How can it be surprising or unexpected? And even if people come to science late so it is unexpected, what is so unacceptable about it anyway. It's the way the world is. C is invariant. Get over it.)

21 hours ago, martillo said:

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.

I should warn anyone of a rational disposition that one of the starting points is that some of the results of modern physics are "hard to accept by common sense". Clearly, anything starting like that has zero scientific merit.

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Posted (edited)
9 minutes ago, Strange said:

Not exactly the same, no.

As noted, you can't use F=ma when the mass is changing. Trying to calculate F(t) = m(t)a doesn't work.

F = dp/dt is more general. It works for constant mass (where only the velocity changes, in which case it is equivalent to F=ma) and it works when m varies.

I have seen people basing their attempt to disprove SR on some dodgy grounds, but this is like claiming that  5x2 is not the same as 2x5.

 

(And what it is about crackpots and relativity, anyway. We have known this is the way the universe works for well over 100 years. I grew up with this. How can it be surprising or unexpected? And even if people come to science late so it is unexpected, what is so unacceptable about it anyway. It's the way the world is. C is invariant. Get over it.)

Well I think that actually the case is exactly the inverse: 

F = ma valid for the general case even when mass varies.

F = dp/dt valid for constant mass only.

Rockets' dynamics shows me right.

Relativity Theory wrong.

 

 

Edited by martillo

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7 minutes ago, martillo said:

Well I think that actually the case is exactly the inverse

I think it has been extensively demonstrated that you are mistaken.

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Posted (edited)
1 hour ago, martillo said:

The issue is that F = ma is not the same as F = dp/dt.

Experimentally, in rockets dynamics, is proven that the relation F = ma = mdv/dt is valid even for a mass varying systems. But Relativity largely uses and depends on the relation F = dp/dt. They are different relations. The issue is which definition would be given for force in the most general case, F = ma or F = dp/dt?

I do not see [math]F=ma[/math] in [math]F=ma=v_{e} \frac{dm}{dt}[/math].I see [math]ma=v_{e} \frac{dm}{dt}[/math].
Am I missing something? If mass m is not constant, what does it depend on? If mass decreases with time what function is [math]m(t)[/math]
(and [math]a(t)[/math]). 

 

*) and/or acceleration a

Edited by Ghideon
format. Missing word. Disclaimer added. "of same size." added. More clarifications.

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Posted (edited)
25 minutes ago, Ghideon said:

I do not see F=ma in F=ma=vedmdt .I see ma=vedmdt .
Am I missing something? If mass m is not constant, what does it depend on? If mass decreases with time what function is m(t)
(and a(t) ). 

 

*) and/or acceleration a

You don't see but it is wrote and not by me but the entire world of rockets dynamics???

How is that?

In vain to explain if you actually don't want to see.

Edited by martillo

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22 minutes ago, martillo said:

You don't see but it is wrote and not by me but the entire world of rockets dynamics???

How is that?

In vain to explain if you actually don't want to see.

I would like to see the issue you claim exists, that is why I am asking questions.

Your reference http://www.braeunig.us/space/propuls.htm says:

Quote

I do not find the construction [math]F=ma=v_{e} \frac{dm}{dt}[/math] in that reference. 

 

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Here is your initial error.

 

23 hours ago, martillo said:

where ve is the speed of the fuel expelled relative to a frame at rest. 

 

This incorrect ve is relative to the rocket, as correctly stated in your own reference.

 

Quote

 

 Did you understand the derivation of 'The rocket equation' given in your reference and in particular the significance of setting the condition that delta t tneds to zero?

This is necessary to achieve an inertial frame (which is your observer "at rest" ) by a Galilean tansformation, given by

Quote

in your reference.

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Posted (edited)
45 minutes ago, studiot said:

Here is your initial error.

 

 

This incorrect ve is relative to the rocket, as correctly stated in your own reference.

 

 

 Did you understand the derivation of 'The rocket equation' given in your reference and in particular the significance of setting the condition that delta t tneds to zero?

This is necessary to achieve an inertial frame (which is your observer "at rest" ) by a Galilean tansformation, given by

in your reference.

Right, you are right. I must work better on this, I made a mistake there. ve is the velocity of the expelled fuel relative to the rocket. I agree. Working on the derivation of the thrust equation for rockets I cannot make such mistake.

But the rest including the initial post is right.

The real general equation for force is F = ma (valid even for variable mass) and not F = dp/dt (which becomes valid for constant mass only).

 

Edited by martillo

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Posted (edited)

Thanks for your contribution studiot. I'm correcting the manuscript for my book and my site now.

That's what the new theory needs now, constructive critics.

 

Edited by martillo

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13 hours ago, martillo said:

p = mv

Then:

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

Now, for me, the true relation for force is F = ma and the equation for momentum and force can be written as:

dp/dt = F + vdm/dt

It can be seen that F = dp/dt is valid only in the cases where dm/dt = 0 which means cases or systems with constant mass but it is not the case for the rockets nor for relativistic systems!

That is 'begging the question': you put F = ma as only valid equality, and oh miracle, you also get it out! Further, dm/dt = 0 is not valid for a rocket, because it loses mass. Say a rocket burns constantly 1 kg fuel per second, so dm/dt = -1 kg/s.

Further you say for rockets

F = m(dv/dt) = –ve(dm/dt)

But the factor dm/dt shows that m is not constant, it is a function of time. For a constant mass dm/dt would be zero, so F would be too. Your rocket keeps standing on the ground...

I am still waiting for an answer on my questions:

On 8/5/2019 at 9:25 AM, Eise said:

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?

And of course, do not use F = dp/dt, not even implicitly.

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14 hours ago, Ghideon said:

 

To me ma in F=ma=vedmdt  looks like they say: given an exhaust velocity ve  and the time derivate of mass exiting the rocket  dmdt  we can calculate what acceleration a  that thrust would give to a rocket with a fixed mass m. OR we could calculate the mass m that a rocket has to have, to be able to accelerate at acceleration=a given the thrust vedmdt .

 

The F in that equation is not the net force (which is what martillo is missing, apparently). It's not written that way in the link provided in the OP; they just equate the two because that's what falls out of the derivation. In fact, they solve it by conservation of momentum, not by applying Newton's second law (though, of course, that's equivalent when you do the math in the time domain.)

10 hours ago, martillo said:

 The real general equation for force is F = ma (valid even for variable mass) and not F = dp/dt (which becomes valid for constant mass only).

Repeating this does not make it true.

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