Deduction of thrust force

[Misr]

hello people

 

This is an extract from my book about my problem

 

http://img693.imageshack.us/img693/3506/thrustforce.jpg

 

I have an explanation for this but not sure about it :

 

Pushing the rocket with a high speed

 

when a great amount of

 

burnt gases are pushed backwards

 

with a great momentum , the rocket

 

gains the impulse forward with

 

an equal momentum (According to

 

the law of conservation of linear momentum).

 

Also thrust exerts a force

 

upon the rocket , the rocket answers in the opposite direction with a force equal in magnitude (According to Newton's third law).

 

 

 

F net = d(mV)/dt

 

and since both mass and velocity are changing from time to another:

 

 

 

therefore :

 

d(mV)/dt = mdV/dt + Vdm/dt (According to the product rule where the formula of the product rule is(f.g)'=f '.g + f.g' or

 

d(u.x)/dy = xdu/dy + udx/dy )

 

 

 

And since this system (rocket and thrust) is an isolated system (which means that no external force acting on the system ) therefore the linear momentum is conserved and the the change in momentum is zero , consequently the net fore is zero .

 

 

 

therefore F net = mdV/dt + Vdm/dt = zero

 

where F net is the net force on the system .

 

 

 

therefore : mdV/dt = - Vdm/dt = Fimp

 

where Fimp is the thrust force which is a self generated force and is directly proportional to the rate of emission of exhaust or the rate of decrease of mass of the gases

 

 

 

m ----- indicates the mass of rocket plus the remainig fuel ,

 

dV/dt ------indicates the acceleration of rocket plus the acceleration of exhaust,

 

-V-------indicates the velocity of exhaust with respect to the rocket,

 

 

 

dm/dt-----indicates the rate of emission of thrust .

 

 

so is my explanation right???Thanks

[Amr Morsi]

I think that Fnet is not zero, it is equal to -dp/dt; where p is the momentum of gases.

Vdm/dt is an additional term due to the change of the total mass of the rocket. But, it is not the only effect on the acceleration.