A New Light In Physics

[martillo]

A great challenge to open minds:

A New Light In Physics

"Classical Physics is coming back, RELOADED!"

[Boerseun]

Did you write that yourself, or did you find it on the 'net?

I just read the "New Evidence" section, and there are some glaring mistakes in it.

[martillo]

Boerseun,

Did you write that yourself, or did you find it on the 'net?

I wrote it.

I just read the "New Evidence" section, and there are some glaring mistakes in it.

Please tell me about.

[Erasmus00]

Consider your claim that the "real" equation of motion is F=ma instead of This is completely false. Lets look at your rocket equation.

Consider a rocket of mass M moving at speed v, expelling exhaust of mass dm at a velocity u. Note, u is a negative quantity if v is positive.





Now, there is no external force on this system (nothing external). Hence,


This leads to the traditional (and experimentally verified) rocket equation. Using F=ma does NOT result in the correct result.
-Will

[martillo]

Erasmus00,

This leads to the traditional (and experimentally verified) rocket equation. Using F=ma does NOT result in the correct result.

Yes it does and as is described at the end of the page.

By definition p = Mv and dp/dt = M(dv/dt) + v(dM/dt)
Now if F = Ma we have dp/dt = F + v(dM/dt)
Now if you apply this to the composed system of the rocket and the total fuel (the contained plus the expelled one you have:
dM/dt = 0 since the total fuel is constant and so:
dp/dt = F = 0 since there are no external forces applied.
Then we have the same initial equation.

The momentum must be conserved so the momentum of the rocket with its contained fuel must be equal and opposite to the momentum of the expelled fuel which is exactly the same conditon from which the equation of the rocket is currently derived! This leaves to the same equations in terms of the mass of the rocket and its contained fuel:
m(dv/dt) = -u(dm/dt)
the known equation of the rocket.

You must also note that in the pages cited in the text it is stated that the right term of the equation represents the force on the rocket. Then as the left part is m(dv/dt) = ma this represents ma = F and not dp/dt = F, no way.

[Erasmus00]

Originally Posted by martillo

You must also note that in the pages cited in the text it is stated that the right term of the equation represents the force on the rocket. Then as the left part is m(dv/dt) = ma this represents ma = F and not dp/dt = F, no way.

Then that particular page is wrong. The force on the rocket is a nebulous thing- considering that at any given instant much of the fuel is still part of the rocket. Thus, separating force on the rocket from force on the fuel is an impossible task.

Much better is considering the entire system, rocket+fuel and noting there are no external forces, so momentum is conserved. Further, without Newtonian mechanics would not always have conserved momentum in the absence of external forces.

Finally, its important to realize that the rocket equations you cite (while tested experimentally) where derived FROM Newtonian mechanics. They are derived in nearly every mechanics book, straight from F = \frac{dp}{dt}
-Will

[martillo]

Erasmus00,

Then that particular page is wrong.

No they are right.

The force on the rocket is a nebulous thing- considering that at any given instant much of the fuel is still part of the rocket. Thus, separating force on the rocket from force on the fuel is an impossible task.

Not at all.
The mass m in the rocket equation mdv/dt = u(dm/dt) represents the mass of the rocket plus the contained fuel and u(dm/dt) is the force on the rocket with its contained fuel.
You should take more care in the analisis of the problem.

Further, without F=dp/dt Newtonian mechanics would not always have conserved momentum in the absence of external forces.

Wrong. It is conserved as I have already shown.
The relation dp/dt = m(dv/dt) + v(dm/dt) = F + v(dm/dt) shows that when
F=0 and dm/dt=0 the momentum is conserved. The condition dm/dt=0 is the same as the condition that the conservation of momentum be valid for "closed systems" only (constant mass) as presented in current Physics. There are cases to show that when mass varies the momentum is not conserved. With the new aproach the condition surges naturally from the rellation which requires dm/dt=0.

Finally, its important to realize that the rocket equations you cite (while tested experimentally) where derived FROM Newtonian mechanics. They are derived in nearly every mechanics book, straight from F = \frac{dp}{dt}

The same can be derived with the new approach.

[Erasmus00]

Originally Posted by martillo

The condition dm/dt=0 is the same as the condition that the conservation of momentum be valid for "closed systems" only (constant mass) as presented in current Physics. There are cases to show that when mass varies the momentum is not conserved. With the new aproach the condition surges naturally from the rellation which requires dm/dt=0.

When mass varies, if there is no external force, momentum is STILL conserved. Consider the following: a raft of mass m is moving at a speed v in still water. A person (of mass M) falls into the boat with a downward velocity of V.

In this problem, the mass of the "raft" system is not conserved (it goes up by M). However, momentum IS conserved in the x direction (though not in the y).
-Will

[martillo]

The system of the rocket with its contained fuel can be an example of how momentum is not conserved when mass varies.
Consider the rocket in the steady state of travelling with constant velocity. The net force acting on it is zero since the force of the expelled fuel is opposite to the resistence of the air. F=0. The mass m of the system is continuosly decreasing (and thus dm/dt is not zero) while the velocity v is mantained.
It is obvious that the momentum p = mv is not conserved for this system. So the momentum is not always conserved when the mass of the considered system varies.
I think this consideration is in basic texts of Physics!

[Erasmus00]

Originally Posted by martillo

Consider the rocket in the steady state of travelling with constant velocity. The net force acting on it is zero since the force of the expelled fuel is opposite to the resistence of the air. F=0. The mass m of the system is continuosly decreasing (and thus dm/dt is not zero) while the velocity v is mantained.

Again, treating the rocket as a system separate from its fuel can only lead to confusion, as your system is dynamically changing as the rocket expels mass.

It is far simpler to treat rocket+fuel as one system, in which case you can note that there IS a net external force, the air resistance, so obviously momentum is not conserved.
-Will