Principle of linear momentum for a rocket

AI Thread Summary
The discussion focuses on calculating the relative velocity of gas expelled from a rocket using the principle of linear momentum. Key equations include the thrust force, fuel consumption rate, and the rocket's mass over time. The participant struggles to reconcile different solutions for the change in velocity, particularly regarding the inclusion of gravity in their calculations. Clarifications indicate that thrust does not depend on gravity, but the rocket's motion does. Ultimately, the participant realizes their mistake in separating thrust-induced acceleration from gravitational effects.
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Homework Statement


vertical rocket start:
F_thrust (simple F)
Δm (fuel consumed per Δt)
m_tot

the relative velocity of the gas w is demanded.

Homework Equations


vgas=v+Δv+w, w<0
m(t)=mtot-Δm*t

The Attempt at a Solution


rocket acceleration.
F-G=m(t)*a(t)
a(t)=F/m(t)-g

first:
v*m=Δm*vgas+(v+Δv(t))(m-Δm)
⇒w=-Δv(t)*m(t)/Δm
Δv(t)=a(t)*Δt
or second direct way:
w=F/Δm*Δt

so the first one should lead to the direct solution if I'm not mistaken, but I struggle to get the correct solution.
To be more specific, it seems in order to get the same result Δv=(a+g)Δt but this wouldn't make much sense for me.

thanks in advance
 
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Welcome to PF!

What is given in the problem?

oOoOoOo said:

Homework Statement


vertical rocket start:
F_thrust (simple F)
Δm (fuel consumed per Δt)
m_tot

the relative velocity of the gas w is demanded.

Homework Equations


vgas=v+Δv+w, w<0
m(t)=mtot-Δm*t

The Attempt at a Solution


rocket acceleration.
F-G=m(t)*a(t)
a(t)=F/m(t)-g

Without the external force (gravity) :
oOoOoOo said:
first:
v*m=Δm*vgas+(v+Δv(t))(m-Δm)
⇒w=-Δv(t)*m(t)/Δm
Δv(t)=a(t)*Δt

The thrust can be considered as force F=ma=mΔv(t)/Δt=-w Δm(t)/Δt.

oOoOoOo said:
or second direct way:
w=F/Δm*Δt

so the first one should lead to the direct solution if I'm not mistaken, but I struggle to get the correct solution.
To be more specific, it seems in order to get the same result Δv=(a+g)Δt but this wouldn't make much sense for me.

thanks in advance

If you know the thrust force F and the fuel consumed in unit time, Δm/Δt, .you can calculate the velocity of the gas with respect to the rocket.

##w=-\frac{F}{\Delta m/\Delta t} ## .
 
Maybe I wasn't clear enough, given are thrust, Δm and the total mass.

So to calculate the relative velocity w $$ w=-\frac{F}{\Delta m / \Delta t} $$ could be used, I know that, but I would prefer to calculate it with the conservation of linear momentum, since I try to explain the concept. The last time I had to solve an exercise like this one is a little bit in the past.. Now my problem is that I get two different solutions and I can't find my mistake.
Did I forget to include the gravity in the conservation somewhere?
 
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oOoOoOo said:
Maybe I wasn't clear enough, given are thrust, Δm and the total mass.

So to calculate the relative velocity w $$ w=-\frac{F}{\Delta m / \Delta t} $$ could be used, I know that, but I would prefer to calculate it with the conservation of linear momentum, since I try to explain the concept. The last time I had to solve an exercise like this one is a little bit in the past.. Now my problem is that I get two different solutions and I can't find my mistake.
Did I forget to include the gravity in the conservation somewhere?
It came from conservation of momentum and it is the same (but a sign ) what you wrote w=F/Δm*Δt, if you meant (F/Δm)*Δt. I do not see two different solutions.
The thrust does not depend on gravity. The motion of the rocket depends on it.
 
ehild said:
It came from conservation of momentum and it is the same (but a sign ) what you wrote w=F/Δm*Δt, if you meant (F/Δm)*Δt. I do not see two different solutions.
The thrust does not depend on gravity. The motion of the rocket depends on it.

So then my assumption $$\Delta v=a*\Delta t \text{ where } a=\frac{F-G}{m}$$ is wrong, because I am only interested in the acceleration and Δv which came from the thrust only?
 
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You said the thrust F was given and w was the question. You determined Δv from conservation of momentum, and made m (Δv/Δt) equal to the thrust force. It was correct.
The motion of the rocket is an other question. If gravity is involved, the acceleration depends on it. You can consider the whole motion as repeated "collisions" ( acceleration because of the ejected gas) followed by motion under gravity.
 
I finally got my mistake.
Thank you, very much.
 
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