Rocket F=dp/dt

1. Feb 10, 2004

RedX

I was looking through this forum, and I noticed one of those problems where a rocket ejects mass to get into orbit.

F=dp/dt

Now the arguments I saw went something like this:

p=mv

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

But that equation is for a point particle I believe. All you can say for a rocket is:

F=m*a

where a is the acceleration of the center of mass.

For one thing, in the equation

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

what is v? It's the relative velocity between the ejected mass and the rocket right? How does F=dp/dt "know" information about how the mass is going to be ejected? You could chuck the fuel with a small velocity or high velocity.

I think when you write

F=d(mv)/dt

you do that because relativity says m is a function of v (which is a function of t). But we are still assuming a point particle, and I don't think this expression can honestly be used for a rocket problem. However, I see everyone doing that, so I'm not too sure of myself on this one.

2. Feb 10, 2004

PrudensOptimus

Re: F=dp/dt

Well if you know a little calculus u would know that &Sigma;F = ma = dp/dt = dmv/dt = m(dv/dt) = ma. ==> if mass is constant.

3. Feb 11, 2004

krab

Re: F=dp/dt

No. v is the velocity with respect to an inertial frame.

4. Feb 11, 2004

RedX

Re: Re: F=dp/dt

Now I thought that the laws of physics are the same for different inertial frames. If you have

F=v(dm/dt)+ma

then the force depends on v, it depends on your frame of reference.

If you put the relative velocity (between the ejected fuel and the rocket) in the calculations, it works for rocket problems. However, I don't think it's legitimate physics to say that it works because of this equation:

F=v(dm/dt)+ma

and Newton meant for v to be the relative velocity.

You can show that the equation works when v is the relative velocity just by considering a mass dm changing its momentum from the rocket velocity to the velocity of the ejected mass, and divide this by dt to get the relative velocity times the mass flow rate.

5. Feb 11, 2004

ZapperZ

Staff Emeritus
Re: F=dp/dt

When a force acts on an object in SIMPLE, BASIC, classical mechanics, the force is applied onto the object's center of mass. When you draw a free-body diagram of the system, you don't draw the object, you represent the object simply as a point, because unless we are considering rotational motion here, the "size" of the object is irrelevant at this level.

In the above comment, you seem to be mixing the object in question. The system that is under consideration is the rocket. It has a mass m(t). It's mass is changing over time. It's velocity isn't, and this is measured based on some inertial reference frame. The stuff being spewed out of the rocket, as soon as it leaves the rocket, is no longer part of the "system" under consideration. The force F is the force that acts only on the rocket system, and not on the spewed gases, etc. The system here is just the rocket itself, and only the rocket.

You also can't say that

"F=m*a

where a is the acceleration of the center of mass."

because you need to specify the center of mass of what? The rocket? The rocket + ejected mass? Since the mass of the rocket is changing and being redistributed, is the location of the center of mass of the rocket also changing with time? Does this add an added "a" to the overall acceleration in the CM frame?

But what I don't quite understand is your statement:

"How does F=dp/dt "know" information about how the mass is going to be ejected? You could chuck the fuel with a small velocity or high velocity."

This is the issue of the chicken or the egg. When you push an object that is in contact with the floor, and you apply a force just so that it moves with constant velocity, did you know a priori just how much you need to push to get it to move this way? The question has been set up so that the conditions are: (i) it has no net acceleration and (ii) it is also losing mass. One can imaging of throttling the engine just enough to get to achive this, and so getting just the right level of rate of mass loss given the size of the mass "chunks" that the engine is spewing.

BTW, question like this isn't unique. Another popular scenario that is often encountered in intro physics is a conveyor belt moving with constant velocity, while a long length of chains with a constant mass per unit length is dropped onto it at a certain rate. Again, no net acceleration, but the conveyor belt (which is now THE system under consideration) is gaining mass. The treatment of this problem is identical to the rocket.

Zz.

6. Feb 11, 2004

krab

Re: Re: Re: F=dp/dt

Right. But you weren't using any inertial frame.

7. Feb 15, 2004

RedX

Okay I understand now. Thanks everyone.