Principle of linear momentum for a rocket

In summary, the problem involves a vertical rocket start with given values for F_thrust (thrust force), Δm (fuel consumed per unit time), and m_tot (total mass). The question is to calculate the relative velocity of the gas, w, in accordance with the equation vgas=v+Δv+w, where w<0. The attempt at a solution involved using the conservation of linear momentum to determine Δv, and then setting m(Δv/Δt) equal to the thrust force F, which resulted in the correct solution. The motion of the rocket also depends on gravity, which can be considered as repeated "collisions" followed by motion under gravity.
  • #1
oOoOoOo
4
0

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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  • #2
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} ## .
 
  • #3
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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  • #4
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.
 
  • #5
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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  • #6
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
I finally got my mistake.
Thank you, very much.
 

1. What is the principle of linear momentum for a rocket?

The principle of linear momentum for a rocket states that for every action, there is an equal and opposite reaction. This means that when a rocket expels gas out of its engine, the gas exerts a force on the rocket in the opposite direction, propelling it forward.

2. How does the principle of linear momentum apply to rocket propulsion?

The principle of linear momentum explains the basic concept of rocket propulsion. As the rocket expels gas out of its engine, the gas exerts a force on the rocket in the opposite direction, resulting in the rocket moving forward with an equal and opposite force.

3. What is the relationship between thrust and mass in a rocket?

The amount of thrust produced by a rocket is directly proportional to the mass of gas expelled from the engine. This means that the more mass of gas that is expelled, the greater the thrust and acceleration of the rocket will be.

4. How does the principle of linear momentum affect the design of a rocket?

The principle of linear momentum plays a crucial role in the design of a rocket. Engineers must consider the amount of thrust needed to overcome the weight of the rocket and payload, as well as the amount of mass that needs to be expelled to achieve the desired acceleration and velocity.

5. Can the principle of linear momentum be used to control the motion of a rocket?

Yes, the principle of linear momentum is used in rocket propulsion systems to control the motion and trajectory of the rocket. By adjusting the direction and amount of gas expelled from the engine, engineers can manipulate the thrust and ultimately control the rocket's motion.

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