Deriving the Tsiolkovsky Rocket Equation using the rate at which the fuel burns

Question

I'm trying to understand the derivation of the Tsiolkovsky Rocket Equation. $$\Delta v= \int_{t_0}^{t_1} \dfrac{|T|}{m_0 - t\Delta m}dt$$ The Wikipedia page confused me here :

where $T$ is thrust, $m_{0}$ is the initial (wet) mass and Δm is the initial mass minus the final (dry) mass

Which means that $$m_0 - t\Delta m = m_0 -t(m_0-m_{dry})$$

If the rocket launched at $t_0=0$ seconds and reaches the dry mass at $t_1 = 60$ seconds, then at $t_1$ we should expect an acceleration of $$a(t_1) =\dfrac{|T|}{m_{dry}}$$ but we get $$a(t_1) =\dfrac{|T|}{m_{0} - t_1 \Delta m} = \dfrac{|T|}{m_{0} - 60s \Delta m} $$

I am unsure if this is an error or not. I then derived the equation using a slightly different approach I made up which seemed more intuitive to me, instead of using $\Delta m$, I used the rate $r$ (kg/s) at which the fuel exits the system.

$$\Delta v= \int_{t_0}^{t_1} \dfrac{|T|}{m_f - tr}dt$$

Now if I we take the integral we obtain

$$\Delta v =-\dfrac{|T|}{r}\ln\left|\dfrac{m_0}{m_{dry}}\right|$$

My derivation seems wrong because I have a negative sign in front which doesn't make sense. Is there something that I missed?

Answer 1

Yes, this is at present a low-quality derivation on Wikipedia. In particular the assembly $m_0 - t \Delta m$ is not dimensionally consistent if $t$ has any given units. One would instead write, say, $$\Delta v = \int_{t_0}^{t_1}\mathrm dt ~\frac{T}{m_0 - (m_0 - m_1)(t - t_0)/(t_1 - t_0)}.\tag{1}$$ And then this expression is properly $T/m_0$ at $t = t_0$ and $T/m_1$ at time $t=t_1.$

You are totally fine here to write $r = (m_0 - m_1)/(t_1 - t_0)$ to simplify; it also looks like you invented a new mass $m_f = m_0 + r~t_0$ which allows us to write this entire expression indeed as $$ \Delta v = \int_{t_0}^{t_1}\mathrm dt ~\frac{T}{m_f - r~t},\tag{2}$$ which is fine. That denominator is indeed still $m_0 + r~t_0 - r ~t_0 = m_0$ at $t=t_0$ and with some more work, $m_0 + r~t_0 - r~t_1 = m_0 - r(t_1 - t_0)$ which simplifies to $m_1$ at $t = t_1.$ Great, although I question the choice of the name $m_f$ since it might indicate to someone “final mass” instead of “fictitious mass at $t=0$,” which is what it is.

In fact I think it would be much easier if we simply defined $\tau = t_1 - t_0$ and took $t_0 = 0$ as an arbitrary zero, so that you simply have $\int_0^\tau \mathrm dt T/(m_0 - r~t)$ and you do not need to think about half of these details.

Anyway the proper evaluation of (2) involves a $u$-substitution where we define, say, $\mu = m_f - r~t$ and then $\mathrm d\mu = -r~\mathrm dt,$ which is probably where you are getting the minus sign from. The minus sign here is 100% correct; the endpoints are that $\mu(t_0) = m_0, \mu(t_1) = m_1$ as discussed above, so we will find$$ \Delta v = -\frac{1}{r}\int_{m_0}^{m_1}\mathrm d\mu ~\frac{T}{\mu} = -\frac Tr \ln\left(\frac{m_1}{m_0}\right),\tag{3}$$and the minus sign serves a crucial purpose here: this logarithm is a logarithm of a number between $0$ and $1$ and therefore it is a negative value; the minus sign makes it positive.

So as far as I can tell, your error was in finding these boundaries or some other elementary step after that; you effectively had that $m_f - r~t$ was $m_1 = m_\text{dry}$ at time $t = t_0$ and $m_0$ at time $t = t_1$ when it is exactly the opposite; the subscript $0$ marks the beginning instant and the subscript $1$ marks the ending instant.