I can't understand deduction of thrust force by this way?
This is an extract from my book
Can u explain this to me
Thanks

I can't understand deduction of thrust force by this way?
This is an extract from my book
Can u explain this to me
Thanks
Force is proportional to the change in momentum divided the change in time.Originally Posted by Misr
As both the mass and the velocity are changing within the momentum, the force depends on the amount both of these are changing seperately, added together.
Because this is the only force acting on the system, the two must be equal but opposite.
Imagine that instead of a steady stream of gas, you were shooting a series of bullets of mass delta_m, at velocity v, at the rate of one per unit time delta_t. If your space ship has a mass M, then you could use conservation of momentum to determine that during time delta_t your ship has gained velocity V=delta_m v/M. So the average acceleration is the change in velocity divided by the time, or delta_m*v/(M*delta_t). Then the force on your space ship is the mass M multiplied by acceleration, or delta_m*v*M/(M*delta_t) = v*delta_m/delta_t.
I'm still confused :?
Does the book mean by external force the net force acting on the rocket ???
The mass of rocket is constant while the velocity is changing but the mass of thrust is changing while the velocity is constant ??
Mdv/dt is the reaction of the rocket on thrust force because the mass of the rocket is constant
Vdm/dt is the force acting on the rocket and mass is variable bec. The mass of gasses varies from time to time.
Since these forces are equal in magnitude and opposite in direction
Therefore the sum of them is zero (The external force is zero according to newton's third law )
So is this explanation right?
Thanks for helping me
The mass of a rocket is NOT constant. If it is then that is called a failure.Originally Posted by Misr
The correct form of Newton's law is not F=ma but rather F=dp/dt where p is momentum.
The thrust on a rocket is determined by the velocity of the exhaust gasses and the mass rate of discharge, which is equivalent to the time rate of change of momentum.
So F = dm/dt X Vexhaust where m is the mass of the rocket and Vexhaust is the speed of the exhaust gasses. This applies in a vacuum. In the presence of atmospheric pressure there is a correction made for that effect.
The Vexhaust term is usually calles specific impulse, or Isp, again in a vacuum.
The rocket and its fuel comprise an isolated system. There is no external force, so the only force on the rocket is the reaction from the exhaust.Originally Posted by Misr
The rocket's mass and velocity are not constant, but at any given instant, they have some value.The mass of rocket is constant while the velocity is changing but the mass of thrust is changing while the velocity is constant ??
isn't the same thing as acceleration?? therefore gives newton's second law
of course is equivalent to as
may i ask what does represent?
That is the rate at which the fuel is being burned, and the rate at which mass is leaving via the rocket's exhaust.Originally Posted by Heinsbergrelatz
if it is so then the net force = zero right?or wrong?The rocket and its fuel comprise an isolated system. There is no external force, so the only force on the rocket is the reaction from the exhaust.
another problem is why d(mv)/dt = mdv/dt+vdm/dt??i can't really understand how this come can u help?
thanks so much
Correct. The center of mass of the ship plus exhaust gases does not move. The ship goes one way, the exhaust goes the other. This is basic conservation of momentum.Originally Posted by Misr
Its the product rule.another problem is why d(mv)/dt = mdv/dt+vdm/dt??i can't really understand how this come can u help?
thanks so much
http://en.wikipedia.org/wiki/Product_rule
What are you tring to do ?Originally Posted by Misr
Maybe this will help. It is called the "rocket equation" and it applies in a vacuum in the absence of outside forces, in particular in the absence of gravity.
Where is the speed of the exhaust gases relative to the rocket. nozzle. If is expressed in the inconsistent, but customary units of "seconds" then the rocket equation is
and the result is in feet per second.
thrust exerts a force on the rocket , the rocket answers by a force equal in magnitude and opposite in direction which makes the rocket accelerate upwards
since there is no external force affecting the system (because the system is isolated)
therefore the net force acting on the rocket is zero (the sum of thrust force and rocket's reaction on thrust = 0)
therefore F=d(mv)/dt=0
where Fis the net force acting on the system
by using the product rule we get :
F= mdV/d t+ Vdm/dt=0
Therefore mdV/d = Vdm/dt = Fimp
Where the negative sign indicates the direction
SO AM I RIGHT?
If yes then write "right" if no say why?
Thanks so much
Hello there ??
I want to ask also
if Fimp=Vdm/dt=mdV/dt
then what does Vdm/dt and mdV/dt indicate
You are wroing.Originally Posted by Misr
The net force on a rocket is not 0. If it were the rocket would not move.
The net force on the system consisting of the rocket and all the exhaust gasses is zero. That is true but not particularly useful.
Yeah yeah right.You are wroing.
The net force on a rocket is not 0. If it were the rocket would not move.
The net force on the system consisting of the rocket and all the exhaust gasses is zero. That is true but not particularly useful.
thrust exerts a force on the rocket , the rocket answers by a force equal in magnitude and opposite in direction which makes the rocket accelerate upwards
since there is no external force affecting the system (because the system is isolated)
therefore the net force affecting the system is zero (the sum of thrust force and rocket's reaction on thrust = 0)
therefore F=d(mv)/dt=0
where F is the net force acting on the system
by using the product rule we get :
F= mdV/d t+ Vdm/dt=0
Therefore mdV/dt = Vdm/dt = Fimp
Where the negative sign indicates the direction
huh, Right??
I have also another question
what does each of mdV/dt and Vdm/dt indicate???
does mdV/dt indicates the reaction of the rocket upwards and Vdm/dt indicate the thrust on rocket???
Thanks
m is the mass of the rocket plus the remaining fuel on board.Originally Posted by Misr
dv/dt is the acceleration of the rocket plus the remaining fuel on board.
v is the velocity of the exhaust gas out of the nozzle, with respect to the rocket.
dm/dt is the mass flow rate.
Ok ,Thanks very much i got it now
I want to make sure if this is right?
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
Thanks
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