Thrust force of a rocket ejecting mass

[Emspak]

Homework Statement

OK, this seems simple but I want to make sure I am not doing something totally wrong. The problem says: use the conservation of mass of a system of many particles to shoe that the thrust force of a rocket that ejects mass at rate [itex]\frac{dm}{dt}[/itex] is equal to [tex]F=-v_e \frac{dm}{dt}[/tex] where [itex]v_e[/itex] is the velocity of the mass ejected.

The Attempt at a Solution

I looked at it this way: momentum is conserved so if we start with mass m of the rocket, mv = k (where k is a constant).

SInce we have a simple differential equation [itex]F=-v_e \frac{dm}{dt}[/itex] it can be integrated as [itex]-m v_e = Ft[/itex]. Taking the derivative w/r/t time we get [itex]-m \frac {d v_e}{dt} = F[/itex]

That gets us the F=ma part of the equation, showing that that works. But I notice that if momentum is a k (constant) then [itex]-m v_e = k = Ft[/itex].

There's a step I am missing here I think. I feel like I am almost there. Any hep -- and anyone telling me I have approached this in entirely the wrong way -- would be appreciated.

 

[D H]

The momentum of the rocket is *not* a conserved quantity. What you are missing is that to look at things from the perspective of conservation of momentum you need to look at the rocket+exhaust cloud system.

Another issue is how you look at force. There is a direct connection with the conservation laws if you use ##\vec F=d(m\vec v)/dt##. However, this means force is a frame-dependent quantity if mass is not constant. Force becomes frame invariant if you use ##\vec F = m\vec a##, but now the immediate connection with the conservation laws is lost.

It might help if you look at things from the perspective of an inertial frame that is co-moving with the rocket (i.e., an inertial frame in which the rocket's instantaneous velocity is zero). The F=dp/dt versus F=ma imbroglio vanishes with this choice.

 

[arildno]

A very simple version, alon DH's reasoning, is the following:

Let us move along with the rocket through a tiny time Dt, in which a little mass Dm has been ejected with a velocity, relative to the rocket, v_e.

That little packet of mass has expreniced a momentum change, what we call an impulse, of Dmv_e

Impulse equals Dt*F, so the little packet of mass experienced the force:

F=v_e*Dm/Dt.

Now, by Newton's 3 law, you'll get the result for the thrust force for the rocket!
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[Emspak]

Thanks to you both. I felt like this was a much simpler problem than it looked...

 

[Nickie]

Hello, it seems like you are on the right track! The conservation of momentum is a key concept in understanding the thrust force of a rocket ejecting mass. However, there is one important factor that you have not yet considered: the change in velocity of the rocket due to the ejection of mass.

Let's start with the equation you have derived, -m v_e = k = Ft. This equation is correct, but it only considers the initial momentum of the rocket (mv) and the momentum of the ejected mass (m v_e). What we need to consider is the change in velocity of the rocket as a result of ejecting this mass.

According to Newton's second law, F = ma, where m is the mass of the rocket and a is its acceleration. In this case, the acceleration of the rocket is equal to the change in its velocity over time, or a = \frac{dv}{dt}.

Now, let's combine this with our previous equation, -m v_e = Ft. We can rearrange this to get F = -\frac{m}{t} v_e. Remember, t is the time it takes for the mass to be ejected, so \frac{m}{t} is actually the rate at which mass is being ejected (dm/dt).

Putting it all together, we get F = -v_e \frac{dm}{dt}, which is the desired equation for thrust force. This equation takes into account the change in velocity of the rocket as a result of ejecting mass, making it a more accurate representation of the thrust force.

I hope this helps clarify your understanding of the problem. Keep up the good work in applying conservation of momentum to solve problems like this!