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

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?

• I would like to add one thing to CR drost's answer that it should be m_final=m_initial-dm/dt(time elapsed) in the wiki article Nov 21 '19 at 21:51

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.