How to interper thrust of the rocket

[amiras]

For example then talking about the rocket the trust is:

F_thrust = u*dm/dt

where u is speed of the gases relative to the rocket, and dm/dt rate of change of ejected mass of gases.

Now if the rocket is being launched vertically so force of gravity acts upon it and:

F_net = F_thrust - mg = u*dm/dt - mg

According to the 2nd Newton's law:

F_net = dp/dt = v*dm/dt + m*dv/dt = v*dm/dt + ma

If comparing these equations:

ma=-mg in this case:

should that be interpreted that the acceleration a is caused by the forces other then thrust.

And is it possible to write:

F_net = m*dv/dt + F_other = F_thrust + F_other

 

[bp_psy]

For example then talking about the rocket the trust is:

F_thrust = u*dm/dt

where u is speed of the gases relative to the rocket, and dm/dt rate of change of ejected mass of gases.

Now if the rocket is being launched vertically so force of gravity acts upon it and:

F_net = F_thrust - mg = u*dm/dt - mg

According to the 2nd Newton's law:

F_net = dp/dt = v*dm/dt + m*dv/dt = v*dm/dt + ma

If comparing these equations:

ma=-mg in this case:

should that be interpreted that the acceleration a is caused by the forces other then thrust.

And is it possible to write:

F_net = m*dv/dt + F_other = F_thrust + F_other

u*dm/dt =/= v*dm/dt
Since u is the speed of the ejected gas with respect to the rocket and v is the velocity of the rocket with respect to some fixed frame.This is clearly ma=-mg wrong.
The acceleration of the rocket depends on both the weight of the rocket and the thrust. The higher the weight the smaller the acceleration for some given thrust.

 

[K^2]

amiras, you are forgetting that the rocket is also losing amount of momentum proportional to lost mass. So dp/dt = F_net + v*dm/dt, so you get F_net = m dv/dt = ma. (This is one of the subtler parts of deriving rocket formula.)

So: ma = u dm/dt - mg.

 

[Delta Kilo]

First, as bp_psy said, u /= v

F_net = dp/dt = v*dm/dt + m*dv/dt

While this looks like a mathematically correct way to differentiate a product of two functions of t, you have to be carefult about physical significance of that.

The second term m*dv/dt is the force required to accelerate mass m from v to v+dv during time dt. This is ok. But the first term v*dm/dt means the force required to accelerate a small mass increment dm from 0 to v during time dt (or decelerate it from v to 0, depending on the sign of dm). In other words this equation quietly assumes that the extra mass acquired (or lost) by the moving body has initial (or final) velocity of 0.

The reason why it doesn't quite work is because mass (unlike other parameters, like velocity or temperature) does not just appear out of nowhere, it actually moves from one part of the system to another and carries its momentum with it. For that reason it is easier to treat the mass as a constant during differentiation and then explicitly account for the momentum brought in (or carried away) by the mass flow.

It's a subtle point, see here for some examples and discussion:
http://books.google.com.au/books?id=Ni6CD7K2X4MC&pg=PA690

PS the correct expression in your case if F_net = ma = u dm/dt - mg
PPS just noticed K^2 said the same thing already