martillo

You see, you are using F = ma and F = dp/dt. Here again. It is always used F = ma for the rocket,always.

I understand that. Could seem silly until analyzing appropriately the presented problem. Everybody (Science, developers, fabrics...) is using F = ma for a rocket which is something with variable mass. It must be generalized then for the general case. Well, fine, the initial total system is the rocket with its fuel... You want to analyze the problem from the beginning again, fine, I will try for you.

Well, let me summarize the problem here for you although I would recommend to see the problem as presented in the already posted link: http://www.braeunig.us/space/propuls.htm. The total system is the rocket with its expelled fuel presenting a total constant mass. It is applied dP/dt = 0 for this system to derive the force on the rocket alone, I mean to derive what force the expelled fuel exerts on the rocket. Then at a second place it is considered the rocket alone accelerated by a force which is the Thrust Force on the rocket. The problem is that what is applied in this derivation is F = a and not F = dp/dt.

I don't know if I get your point. The Euler Force is a transversal force in non uniform motion when transversal acceleration applies. This is not the case. May be it applies if gravity is considered on the rocket for instance, but this is not what is considered to obtain the Rocket Thrust equation. Because you said: " Start at the beginning and try to see where all these websites you refer to have hidden the analysis. You have a complete rocket standing there on the ground ready for launch. " I'm analyzing what other ones, everybody, calculate to find the force on the rocket. I just observe, I'm not calculating anything on my own. Come on, now seems you don't know what are we talking about in all this discussion, how can it be after all that was already posted? I suggest you to read with attention the first post of the thread describing with much care the problem discussed here.

It's not me applying the equation F = ma. It is what is actually applied in Science and in rockets development and fabrication and what works when they move, right? Everybody applies F = ma and not F = dp/dt. That's the point, F = ma is the valid one. No, the system is not the rocket ready to launch. Is the rocket already in the air at velocity v and expelling fuel at constant velocity -ve. Please take a look at the link in the first post to have a better view with good simple graphics (http://www.braeunig.us/space/propuls.htm). I don't want to calculate anything, I just observe which equations are used by everybody for the rockets. F = ma is used.

Well, my response is that or you don't get my point or I don't get your point. Well, right. We have two equations to apply. One is the Thrust Equation (mdv/dt = vrel(dm/dt)), the second is the problem, which equation to apply: F = ma or F = dp/dt. F = ma is applied and gives right results for rockets (they work isn't it?). Then follows that the right equation of motion is F = ma and not F = dp/dt, even when mass varies. Don't you get it?

Well, I did Electrical Engineering and I didn't see Variational Calculus (UNICAMP - Brasil). Not at college at all. I have a book given me as present because I was good in mathematics but never read it (Variational Calculus Krasnov, Makarenko, Kiseliov). I just took a look. Is something not for me.

Well, I studied Mechanics but not Variational Calculus. Do you think it is really needed to treat this problem? If so actually I would not be able to discuss anymore, of course. But here there is just one simple linear equation of motion. The problem is which formula is applied on it. F = ma or F = dp/dt? That gives the right equation to apply for objects with variable mass. F = ma is applied.

That's a problem here. This is not a static system. Seems you are suggesting I didn't studied Mechanics, I did. This thread is all about Mechanics. You can develop what you think it must apply.

I just said I'm not able to discuss here because goes beyond my expertise and cannot help. I'm just saying I cannot discuss here anymore.

No, you cannot say that. The evidence I'm presenting is the rockets' motion working properly and that what is found to be applied for them is F = ma and not dp/dt. This can be well verified on the entire web. In Wikipedia for instance it is found the same approach using F = ma and not dp/dt. May be you want the thread to be discarded and sent to trash for it to not be discussed anymore, that would not be fair...

What I'm saying is that the formula that really applies is F = ma and not dp/dt. This implies that even the classical formulation is wrong. Newton second law is formulated as F = dp/dt. I'm saying that is wrong and that the right formula is F = ma even when mass varies. Newton didn't have the opportunity to take missiles in consideration to verify the formula. They didn't exist in that epoch, right? That's the problem. I do care.

It must be remembered the point of discussion is that in the current solution for the twins' paradox the travelling one will meet the stationary one with less beard and it can be only speculated now a possible different solution with both meeting with same beard and so same age. This cannot be demonstrated so easily. It would involve apply General Relativity on the time of the acceleration of the travelling twin in the turning back considering the relativity of simultaneity in the same way we applied for our case of twins travelling just in inertial frames without acceleration. The point is that if current solution has applied this properly or not. This is area of General Relativity only where I have no expertise at all to enter and so I cannot help here. I agree with Boltzmannbrain point of view but I cannot help.

A rocket is may be the best example for variable mass and the equation F = ma is what is applied and not dp/dt. This show that the real equation for force is F = ma and not dp/dt. There are ones with high variation in their mass, missiles for instance. They all work very fine. So they are the proof that the equation F = ma applies and not dp/dt even for variable mass. This changes it all in Physics.

Do you really understand what you say? You say "using F = ma...". That's the point! If you use F = ma, you are not using F = dp/dt. You know: dp/dt = mdv/dt + vdm/dt. They are different things particularly when mass varies (dm/dt not zero). What I claim is precisely that they use F = ma and not F = dp/dt for the rocket.

I said at the OP that was something that passed unperceived in Physics, I don't expect it to be perceived as fast now.

Applying DP/dt they arrive at the equation: m(dv/dt) = vrel(dm/dt) where m is the mass of the rocket at any instant. How they calculate the thrust force now? Just saying: "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." Where is the consideration of dp/dt = d(mv)/dt with the mass of the rocket? Nowhere. They just apply F = m(dv/dt).

In this case it is. There's no p = mv for the rocket anywhere. They only apply dP/dt for the total system.

Just below equation 1.5 they just say that because of units matching the left side is the force. So, this is to apply F = ma.

You don't see it yet. They apply dp/dt to the total system, not to the rocket. For the rocket they apply: "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 = ma here.

They apply dp/dt to the total system rocket + expelled fuel but they do not apply it to the rocket alone, for the rocket they apply F = ma just saying: "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." Right for the total system but what about for the rocket? They don't apply the same for the rocket.

That is the momentum p of the total system rocket + expelled fuel. They do not apply that on the force on the rocket. In the equation 1.5 m is the mass of the rocket alone. To relate equation 1.5 to the thrust force 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." That is not applying F = dp/dt anyway. I'm saying that because of rockets work very well the equation of force is F = ma and not F = dp/dt.

Right, this is what is considered in the problem. No external forces. The force exerted on the rocket is the thrust force.

Yes, but they do not apply this in the derivation of the thrust force. 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." The net force on the rocket is just the thrust force. There's no other force considered.

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."