Rocket equation without infinite small variables

[Germy]

Hey,
I'm tutor for theoretical physics for first year students and I found a question that I couldn't answer so far. It's about the rocket equation. I tried to derive the acceleration without using infinite small variables, but somehow there is one term left that shouldn't be there. In the following, the subscript r always means the rocket and the subscript g always means the gas or fuel that is expelled.
Let's start with the conservation of momentum $$p_r = p_s$$I use the point of view of an external person and ignore gravity. If I now insert the mass $M(t) = M_0 + m_g(t)$ of the rocket where $M_0$ is the mass of the rocket without fuel and $m_g(t)$ is the mass of the fuel I get $$-M(t) \cdot \dot{z}(t) = m_g(t) \cdot (v_g - \dot{z}(t))$$ where $\dot{z}$ is the velocity of the rocket and $v_g$ is the constant velocity of the expelled gas. Now if I take the derivative of the time I get $$-\dot{m}_g(t) \cdot \dot{z}(t) - M(t)\ddot{z}(t) = \dot{m}_g(t) \cdot (v_g-\dot{z}(t)) - m_g(t)\ddot{z}(t)$$ which after cancelling becomes $$ - M(t)\ddot{z}(t) = \dot{m}_g(t) \cdot v_g- m_g(t)\ddot{z}(t)$$Now the problem is that the actual solution of the rocket equation is $$\ddot{z}(t) = \frac{\dot{m}(t)}{M(t)}\cdot v_g$$ that means without the last term. Does anybody see my mistake? Is it maybe something with my point of view or the conservation of momentum? I know how to derive this equation using infinite small variables like dt and dm but I think it should be possible to do it without them and I just don't see where the solution above is wrong.

 

[Hiero]

Your momentum equation does not describe the situation. You do know that the ejection velocity of gas is taken to be constant relative to the rocket. right? So then, when the rocket is at rest, a bit of gas comes out with momentum $v_gdm$, but then when some time goes by and the rocket now has velocity $\dot z$, then a bit of gas is ejected with momentum $(v_g+\dot z)dm$ (or with a minus sign if you are working with magnitudes). So you see, the gas does not have a single velocity but a continuum of velocities and the gas-momentum must be found by integrating over the variable speed.

(I could be wrong since you didn't explain your momentum equation but,) I also get the impression that you're thinking of $m_g(t)$ as the total mass of ejected gas? But the way you've defined it, $m_g(t)$ is the mass of gas on board. The mass of ejected gas (since time t = 0) is $m_g(0)-m_g(t)$

Also you never said that you chose the frame with zero net momentum, but assuming that is how you chose it, then your momentum equation should technically be either $p_r = -p_g$ or $|p_r| = |p_g|$ (but I guess you are just defining them as magnitudes rather than 1 dimensional vectors).

Hope this helps. You see why it's preferable to work in the instantaneous-rest frame!

 

[kuruman]

I remember solving the following problem. Two men each of mass $m$ are on a platform of mass $M$ on wheels. They jump off the platform in the same direction with speed $u$ relative to the platform. Find the final speed of the platform if (a) they both jump off simultaneously and (b) if they jump off one after the other. The answers are different.

 

[Germy]

You do know that the ejection velocity of gas is taken to be constant relative to the rocket. right? So then, when the rocket is at rest, a bit of gas comes out with momentum $v_gdm$, but then when some time goes by and the rocket now has velocity $\dot z$, then a bit of gas is ejected with momentum $(v_g+\dot z)dm$ (or with a minus sign if you are working with magnitudes).

I think I already did exactly this in my second equation on the right-hand side, that's why it says $v_g-\dot{z}(t)$.

I also get the impression that you're thinking of mg(t)m_g(t) as the total mass of ejected gas? But the way you've defined it, mg(t)m_g(t) is the mass of gas on board. The mass of ejected gas (since time t = 0) is mg(0)−mg(t)

You're right there. Unfortunately it doesn't make anything better because when taking the derivative of the time this constant disappears. On the other hand, in the term with $\ddot{z}(t)$ it just adds to the total mass of the rocket which needs to be there anyway.

Also you never said that you chose the frame with zero net momentum, but assuming that is how you chose it, then your momentum equation should technically be either pr=−pgp_r = -p_g or |pr|=|pg||p_r| = |p_g| (but I guess you are just defining them as magnitudes rather than 1 dimensional vectors).

You're right there too, I forgot the minus sign there in the first equation, but I have it from the second equation on (that's the minus sign on the very left).
After all, you proposed some good points where I made some mistakes but unfortunately the big problem is still there. Maybe it actually is because of this problem that kuruman mentionend? But then I think my way is completely continuous and if you take dm this is infinite small, so basically continuous too. Does anybody have some other thoughts on this problem?

 

[Hiero]

I just happened to log in right when you replied, I promise I don’t spend all my time on this forum!

I think I already did exactly this in my second equation on the right-hand side, that's why it says $v_g-\dot{z}(t)$.

...

unfortunately the big problem is still there.

I think I see how you were thinking now. But I think you missed my main point! Re-read this part of my reply:

So you see, the gas does not have a single velocity but a continuum of velocities and the gas-momentum must be found by integrating over the variable speed.

I will try to explain it a bit differently;

The way you wrote the equation (if we use the correct mass of gas ejected) it’s basically saying that ALL of the ejected mass has speed $v_g-\dot{z}(t)$... but the gas that was ejected an hour ago does not care how fast the rocket is moving right now!

Your momentum equation is basically saying that the gas which was ejected an hour ago (or whenever) is magically being accelerated to the same speed as the gas being currently ejected.

In other words, you’re treating the whole mass of ejected gas as if it has the same velocity (that’s the only way p=mv works) but like I said, different mass elements have different speeds (so it’s more like p = ∫vdm with varying values of v).

I hope that makes sense now. Sometimes I explain things poorly /:

 

[Germy]

Hiero I think you got it there! I'm pretty sure that's where I would get this additional term out. I will try to solve this and come back to tell you my result.

 

[kuruman]

Hiero I think you got it there! I'm pretty sure that's where I would get this additional term out. I will try to solve this and come back to tell you my result.

You may wish to solve the problem I posted in #3 to see @Hiero's point at work. The rocket can be viewed as containing a whole lot of men (very small men of mass $dm$) jumping off the rocket one after the other with speed $u$ relative to the rocket.