Equations of motion for a rocket

I'm trying to figure out how Tsiolkovsky rocket equation is derived. I've tried following along in the Wikipedia page and from my book, but I'm not getting it.

First, are they assuming the $$v_{\text{exhaust}}$$ is the same for all of the exhausted fuel? Because that doesn't seem right. If the rocket is firing, each new bit of fuel should be moving faster than each bit already fired out the back.

And if they're not assuming that, how do this page get $$P(\Delta t) = m(V+\Delta V) + \Delta m V_{\text{exhaust}}$$?

Could somebody walk me through this derivation or point me toward a source that does a better job?

EDIT:

I've made some progress by just trying to work it on my own without looking at the Wikipedia page but I'm still not getting the correct final result. Can you tell me where I made my mistake?

Consider conservation of momentum to get the equation $$\frac {d}{dt} (m_r \vec v_r) + \frac d{dt} (m_e \vec v_{e_{\text{inertial}}}) = m_r\frac {d\vec v_r}{dt} + \vec v_r\frac {dm_r}{dt} + \frac {dm_e}{dt}\vec v_{e_{\text{inertial}}} + m_e\frac {d\vec v_{e_{\text{inertial}}}}{dt}=0$$

We also know that $$\frac {dm_r}{dt} = -\frac {dm_e}{dt}$$ and $$\vec v_{e_{\text{inertial}}} = \vec v_e + \vec v_r$$, where $$\vec v_e$$ is the velocity of the exhaust relative to the rocket -- and for this is assumed to be constant. Plugging these in we get:

$$m_r\frac {d\vec v_r}{dt} + \vec v_r\frac {dm_r}{dt} - \frac {dm_r}{dt}(\vec v_e + \vec v_r) + m_e\frac {d(\vec v_e + \vec v_r)}{dt}$$ $$=m_r\frac {d\vec v_r}{dt} - \frac {dm_r}{dt}\vec v_e + m_e\frac {d\vec v_r}{dt} = 0$$

Therefore $$M\frac {d\vec v_r}{dt} = \frac {dm_r}{dt}\vec v_e$$, where $$M$$ is the total mass.

This looks really close to what I should be getting, which is $$m_r\frac {d\vec v_r}{dt} = \frac {dm_r}{dt}\vec v_e$$ -- so apparently the last term in the above should have cancelled out somehow. Why?

Hint:

Draw a diagram of the rocket at time $t$ (with its speed and mass at time $t$), and then draw a diagram of the rocket and the exhaust produced at time $t+\Delta t$ (with the speeds and masses of both entities). Then apply conservation of momentum.

$v_{exhaust}$ is relative to the rocket.

Edit:

I'd advise considering a time $\Delta t$, rather than trying to do this in a differential manner. You can then let $\Delta t$ go to 0 in order to get a differential equation.

• Could you take a look at what I've done now? Mar 7 '15 at 17:58
• Right. Thanks. I figured it out. My equations hold for a single time (like $\Delta t \to 0$) and thus the amount of $m_e$ at that exact instant is just going to be $0$. Mar 7 '15 at 22:40
• Would Cliff or @BobDylan mind expanding slightly more on why, using OP's method, $m_e = 0$? I'm getting the exact same result as OP did, but I substituted $m_e = M - m_r$, and I got $$m_r \frac{dv_r}{dt} = -M \frac{dv_r}{dt} + m_r \frac{dv_r}{dt} - v_e\frac{dm_r}{dt}$$And I don't see why I can't just cancel out $m_r\frac{dv_r}{dt}$ on both sides... I understand that considering time $\Delta t$ I would get the correct result, I just don't understand why my method is wrong. May 8 '15 at 9:59
• @ace It's because the way I decided to set up the problem is flawed. Consider this: I said $\vec v_{e_{\text{inertial}}} = \vec v_e + \vec v_r$, but remember $\vec v_e$ might be constant, but $\vec v_r$ definitely is not. Which means that this equation couldn't possibly hold for every bit of exhaust, because the exhaust particles are not all moving at one speed. So this equation only really holds at a single instant in time, where then $m_e$ is the mass ejected from the rocket at this instant. But instantaneously no mass is ejected. Thus $m_e=0$. This is not the best way to do the problem. May 9 '15 at 21:54
• @BobDylan I agree that this isn't the best way, I was only curious about why it didn't work. Anyway, it makes a lot more sense now, thank you very much. May 9 '15 at 22:25

Necro post but since the author of the question solved his doubt himself I decided to include a detailed demonstration on why the result is wrong, following the same line of thought. I am one of those people who doesn't like the traditional derivation because of some considerations

To start solving the rocket problem, we calculate the velocity of the exhausted fuel as it leaves the rocket relative to an inertial reference (the ground, for instance). The fuel ejects with velocity $$v_{ex}$$ relative to the rocket, its velocity relative to the ground is $$v_{exg}=v-v_{ex}$$ .The rocket won't expel all the mass of fuel at once, instead it will, at each time, expel particles of fuel, each carrying some mass $$m_{ex}$$. The expelled fuel, thus, will be a system of particles.

Additionally, since each particle is being expelled at different times, each particle will have different velocities because $$v_{exg}(t)=v(t)-v_{ex}$$. So, the total momentum of the expelled fuel will be them sum of the momentums of each individual particle, that is:

$$p_{fuel}=\sum^n_{i=1}{m_{ex}v_{exgi}} \textit{}$$

However, the fuel is a continuous body and at each infinitesimally small time $$dt$$, the expelled mass will only be a infinitesimally small fraction of the fuel's mass $$dm_{ex}$$ and the total exhausted mass will be the sum of each infinitesimally small element, that is, an integral, so the total momentum of the fuel at any given time t is: $$p_{fuel}\left(t\right)=\int^{m_{ex}(t)}_{m_{ex}(0)}{dm_{ex}v_{exg}(t)}$$

Now we need to remember that each mass of fuel expelled means that a fraction of the total rocket's mass itself reduces while the mass of the expelled fluid increases, conserving the total mass. The total mass of the rocket plus fuel system is simply the initial mass, before exhaustion $$m_0=m\left(t\right)+\int^{m_{ex}(t)}_{m_{ex}(0)}{dm_{ex}}$$ . As the total mass is conserved the time derivative of the total mas is zero, so: $$\frac{dm_0}{dt}=\frac{dm\left(t\right)}{dt}+\frac{d}{dt}\left(\int^{m_{ex}\left(t\right)}_{m_{ex}\left(0\right)}{dm_{ex}}\right)$$ $$0=\frac{dm\left(t\right)}{dt}+\frac{dm_{ex}}{dt}$$ $$\frac{dm\left(t\right)}{dt}=-\frac{dm_{ex}}{dt}$$

The final equation needed to solve the system is simply the momentum of the rocket itself. $$p_{rocket}=m\left(t\right)v(t)$$ $$p_{total}=p_{rocket}+p_{fuel}=m\left(t\right)v\left(t\right)+\int^{m_{ex}(t)}_{m_{ex}(0)}{dm_{ex}v_{exg}(t)}$$ $$\frac{dp_{total}}{dt}=0=\frac{dp_{rocket}}{dt}+\frac{dp_{fuel}}{dt}$$

Dropping the t labels and calculating each time derivative we have: $$\frac{dp_{rocket}}{dt}=\frac{dm}{dt}v+m\frac{dv}{dt}$$ $$\frac{dp_{fuel}}{dt}=\frac{d}{dt}\left(\int^{m_{ex}\left(t\right)}_{m_{ex}\left(0\right)}{dm_{ex}v_{exg}}\right)$$

The differential element of an integral is simply the differential element itself, so: $$\frac{dp_{fuel}}{dt}=\frac{dm_{ex}}{dt}v_{exg}=\frac{dm_{ex}}{dt}v-\frac{dm_{ex}}{dt}v_{ex}$$ $$\frac{dp_{fuel}}{dt}=-\frac{dm}{dt}v+\frac{dm}{dt}v_{ex}$$ $$\frac{dm}{dt}v+m\frac{dv}{dt}-\frac{dm}{dt}v+\frac{dm}{dt}v_{ex}=0$$ $$m\frac{dv}{dt}+\frac{dm}{dt}v_{ex}=0$$ $$mdv=-dmv_{ex}$$

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