martillo

I can assume a = a(t) as known then, right? May be is a joke but fine. I will try to follow you... Assuming a = a(t) then you will have F = F(t) = ma(t). Now what? What does this mean for you?

You have one equation with two variables. You must have one determined to find the other one. Seems you are joking me...

For a constant mass m just F = ma. We are going in circles... I think I will take some time out now...

I don't know if I understand your point. The equation must be solved. a depends on what force is applied to the object. For instance, for a constant force Fc then a = Fc/m.

I hope you can understand... F is proportional to a if m is constant. F = ma then m is the constant of proportionality.

No, for any acceleration a.

Can you show then where and how they derive the Thrust Force from the Thrust Equation (Equation 1.5) applying F = dp/dt? What I read is just to say (simplified): "the left-hand term is the force as it matches the units". This is to apply F = mdv/dt = ma. Can't you see it? Well, no, I don't have a basic Physics text. I have read that in innumerable physics texts all my life. Starting at college. I only have a Blatt's Physics' text but is not so basic, and in Spanish... I can't believe you are asking me this.

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

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. 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.

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

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.

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

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?

Where you think I went wrong? 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 = -vedm/dt. Solving this equation would give v as a function of m for the considered value of ve if you want that. I don't understand your point of discussion. Please explain. I said the inertial frame considered for the rocket not on the rocket. That is ground.

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

You posted: 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 = -vedm/dt where ve 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.

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

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.

It is wrong in current Physics to consider F = ma on an object of variable mass. That's the point. And that is what is done in your book, in my provided link, in Wikipedia and everywhere else. Something wrong I have realized and wanted to share here in the forum.

Well, if you don't see wrong the applying of F = ma on an object of variable mass then I have nothing else to say...

It is used F = ma on something with variable mass. Don't you see this at least as suspicious? F = ma is used in the middle over an object with variable mass. Suspicious or not?

The initial dP/dt = 0 in the total system is always valid since the total mass is constant. What matters is after. As you said: The point is that now we have parts with variable masses and so here is where we must pay much attention which formula is used. It is found that it is used F = ma and not F = dp/dt. This is the point.

Of course it must be started on the equation dp/dt in the total system of constant mass and so dp/dt = 0. Yes I did:

Note dishonest, no. I meant I'm not able to analyze how the principle is applied to derive the equation of motion. Is a matter of VAriational Calculus and I have no expertise at all in this area. Difficult to read but I did. In the third page is written: This is to apply F = ma. Exactly the same it is said in at the link I provided: "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." Saying like "the left term is the force..." is to apply F = ma. They use the Thrust Force F = -vedm/dt of the rocket which is derived with F = ma. So they use F = ma at the end. Please read the answer above for @studiot to see better.

The "principle of east action" is a variational principle. As I said, I'm not able to discuss the problem in terms of Variational Calculus". I have no expertisse at all in that. Everybody is using F = ma for the rockets, Physics Science, developers, fabrics, and not use F = dp/dt. You can search for this in the web. They just use the Thrust Equation which is derived with F = ma. That's my claim. F = ma is the right equation then.