# 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? – Bob Dylan 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 $0$. – Bob Dylan 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. – user12205 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. – Bob Dylan 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. – user12205 May 9 '15 at 22:25