Why is the thrust force equal to the change in momentum of the rocket?

[Misr]

Hello People actually i want to make sure of this explanation

The external force acting on the rocket is zero
Therefore : F = mdv/dt+vdm/dt=0

Therefore mdv/dt=-vdm/dt=Fimp(Impulse force or thrust force)

Well I have an explanation for this but I'm not sure about it

The external force is zero according to Newton's third law

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

So is this explanation right?

 

[tiny-tim]

Welcome to PF!

Hello Misr! Welcome to PF!

The external force is zero according to Newton's third law

Nooo …

Newton's third law says action is equal and opposite to reaction (and is true both for internal and external (pairs of) forces).

It has nothing to do with any force being zero.

The external force acting on the rocket is zero
Therefore : F = mdv/dt+vdm/dt=0

No, the force from the gas is an external force on the rocket.

Newton's second law (usually written F = ma) is really F = d(mv)/dt = mdv/dt + vdm/dt.

F in this case will be the force from the gas.

But a more direct method would be simply to use conservation of momentum.

 

[Misr]

The external force acting on the rocket is zero
Therefore : F = mdv/dt+vdm/dt=0

No, the force from the gas is an external force on the rocket.
I'm not the person who is saying that actually its a mysterious point in my book and I'm trying to explain it
here it is an extract from the book

[PLAIN]http://img693.imageshack.us/img693/3506/thrustforce.jpg
[/PLAIN]
i can't understand this point
i thought that the external force is the net force acting on the rocket

Newton's second law (usually written F = ma) is really F = mdv/dt + vdm/dt

why??

Thanks

 

[Misr]

Hello there?

 

[tiny-tim]

Hi Misr!

I'm sorry, I must have deleted the email about your post yesterday without ever opening it.

I'm not the person who is saying that actually its a mysterious point in my book and I'm trying to explain it
here it is an extract from the book

[PLAIN]http://img693.imageshack.us/img693/3506/thrustforce.jpg
[/PLAIN]
i can't understand this point
i thought that the external force is the net force acting on the rocket

ah, your book means there is no external force on the rocket-and-gas together.

So the momentum of the rocket-and-gas together doesn't change.

What it calls Fi is the external force on the rocket alone (or on the gas alone).

Newton's second law (usually written F = ma) is really F = mdv/dt + vdm/dt

why??

Because it is … that is Newton's second law …

F = d(mv)/dt.

(and F = mdv/dt + vdm/dt = ma + vdm/dt comes from the Chain Rule of calculus, and if m is constant, then it's just F = ma )

 

[Misr]

ah, your book means there is no external force on the rocket-and-gas together.
So the momentum of the rocket-and-gas together doesn't change.

Yeah and I'm asking why?

Thanks so much for helping me

 

[tiny-tim]

Because of Newton's second law … applied force = rate of change of momentum …

so if there's no external force, there's no change of momentum.

 

[Misr]

i'll try to define my prob exactly

First why is the external force = zero ?
second : i still don't understand why F = mdv/dt+vdm/dt

so can u explain this exactly
Thanks so much

 

[tiny-tim]

First why is the external force = zero ?

What else can it be (for the rocket-and-fuel combined)?

The force of the rocket on the fuel, and vice versa, is purely internal, and there's no other forces.

second : i still don't understand why F = mdv/dt+vdm/dt

Newton's second law (usually written F = ma) is really F = d(mv)/dt

F = ma is a short version, which only applies if the mass (m) is constant.

Here, the mass is not constant, so you have to use the full version, F = d(mv)/dt.

(and, using the Chain Rule, that's the same as mdv/dt + vdm/dt)

 

[Misr]

Ok , thanks so much i think i got it but still have one more question

What it calls Fi is the external force on the rocket alone (or on the gas alone).

ok if Fi= Fimp=-Vdm/dt=mdV/dt
then which one is the force on the rocket and which one is the gas force
Thanks again

 

[tiny-tim]

ok if Fi= Fimp=-Vdm/dt=mdV/dt
then which one is the force on the rocket and which one is the gas force

let's see …

assuming that F_imp is being measured in the same direction as V (eg both along the x-axis), and since dV/dt is positive (and dm/dt is negative), that means that F_imp is in the same direction as the rocket is going, so it must be the force on the rocket (and -F_imp must be the force on the gas).

 

[Misr]

(and dm/dt is negative)

but mass and time are both scalar
do u mean that Vdm/dt is negative and the velocity of the rocket upwards is positive?
and we don't know exactly if -Vdm/dt or mdV/dt is the force on the rocket or the gas because both mass and velocity are changing ?

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

 

[tiny-tim]

but mass and time are both scalar
do u mean that Vdm/dt is negative and the velocity of the rocket upwards is positive?

m is the mass of the rocket-plus-remaining-fuel, so m is decreasing.

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

You have a totally weird view of force …

force on the rocket makes the rocket accelerate,

force by the rocket (on something else) has nothing to do with it.

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)

No, the net force acting on the rocket-plus-all-the-fuel (including the fuel that's gone) is zero.

therefore F=d(mv)/dt=0
where Fis the net force acting on the system

No, F_imp = d(mv)/dt, which is not zero.

ah, now I see I took your word for it when you wrote …

ok if Fi= Fimp=-Vdm/dt=mdV/dt

… no, that's wrong, F_imp = d(mv)/dt = vdm/dt + mdv/dt.

 

[Misr]

Ok ,thanks so much for helping