Principle of linear momentum for a rocket

  • #1
4
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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
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909
Welcome to PF!

What is given in the problem?

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) :
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.

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} ## .
 
  • #3
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0
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?
 
Last edited:
  • #4
ehild
Homework Helper
15,543
1,909
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.
 
  • #5
4
0
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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  • #6
ehild
Homework Helper
15,543
1,909
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.
 
  • #7
4
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I finally got my mistake.
Thank you, very much.
 

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