The usual proof of Rocket propulsion goes something like this (University of Central Florida, Rocket propulsion)

$$mv = (m-dm_g)(v+dv) + dm_g(v-v_{rel})$$

where:

• m - initial weight of rocket
• v - initial speed of rocket
• dv - increase in speed of rocket
• $$dm_g$$ - mass of ejected gas
• $$v_{rel}$$ - relative speed of gas

Then we get to this equation (by neglecting $$dm \cdot dv$$)

$$m \cdot dv = dm_g \cdot v_{rel}$$

Then by realising that $$dm_g$$ = -dm

$$m \cdot dv = -dm \cdot v_{rel}$$

This negative sign is off course necessary for the later integration. However I don't see where it follows from. What does it mean to have a mass that is negative and what is 'dm'? I understand it in the derivation sense:

$$\frac{dm}{dt} = \lim_{t_1\to t_0}\frac{m(t_1) - m(t_0)}{t_1-t_0} = \lim_{t_1\to t_0}\frac{(m_0-dm_g) - m_0}{t_1-t_0} < 0$$

However this feels 'after the fact' knowledge and some manipulation of notation. Where does it follow from formally when we only defined '$$dm_g$$' as the mass of the ejected gas?

Correction (2023.3.11, 13:19):

Would this be a correct derivation?

$$\frac{dp_R}{dt} + \frac{dp_G}{dt} + \frac{dp_{G_{prev}}}{dt} = 0$$

where:

• $$\frac{dp_{G_{prev}}}{dt} = 0$$ change of momentum of previously released gas, without adding the mass of the current change
• $$m_R + m_G + m_{G_{prev}} = m_{R0}$$

$$\frac{dp_R}{dt} = -\frac{dp_G}{dt}$$

$$\frac{dm_R}{dt} \cdot v_R + m_R \cdot \frac{dv_R}{dt} = -(\frac{dm_G}{dt} \cdot v_G + m_G \cdot \frac{dv_G}{dt})$$

$$\frac{m_R(t+ \Delta t) - m_R}{\Delta t} \cdot v_R + m_R \cdot \frac{v_R(t+\Delta t) - v_R}{\Delta t} = -(\frac{m_G(t+\Delta t) - m_G}{\Delta t} \cdot v_G + m_G \cdot \frac{v_G(t+\Delta t) - v_G}{\Delta t})$$

where:

• $$m_G$$ = 0
• $$\frac{dm_R}{dt} = -\frac{dm_G}{dt}$$
• $$v_G = v_R - v_{ex}$$

Which results in:

$$m_R \cdot \frac{dv_R}{dt} = \frac{dm_R}{dt} \cdot (v_G - v_R) = -\frac{dm_R}{dt} \cdot v_{ex}$$

I might be saying the same thing but this way it seems more obvious that '$$dm_R$$' and '$$dm_G$$' are infinitesimal changes rather than infinitesimally small masses.

• What does it mean to have a mass that is negative and what is 'dm' What level calculus have you studied? Mar 16 at 2:14
• As I said in the post, I understand what 'dm' means in the context of an infinitesimal change and I have no problem with a negative change. I had the problem because I thought from the text that $dm_g$ is defined as the mass of the currently expanded gas. (Original text in the mentioned article: 'During an infinitesimal time interval dt, the engines eject a (positive) infinitesimal mass of gas $dm_g$'). But now I understand it was just a misunderstanding from my side. Apr 28 at 16:49

$$m$$ is the mass of the rocket and $$dm$$ is the change in the mass of the rocket after burning a small amount of fuel. The negative sign means that the rocket is losing mass. In other words, the initial mass is $$m$$ and the final mass is $$m+dm$$ after burning a small amount of fuel, where $$m+dm because $$dm$$ is negative.

• No, the final mass is still $m+dm$, regardless of the sign of $\dot{m}$.
– J.G.
Mar 10 at 19:03
• Oops you are right. I will fix that
– Dale
Mar 10 at 19:09
• Thank you for the answer! Sorry, it's probably a basic question, but what would stop me from just replacing '$dm_g$' with 'dm' in the initial equation? It still feels like an intuitively correct assumption that I could see myself do naively (to say that the rocket loses the same general mass as the mass of the gas). From what can I know that I used an incorrect sign as an assumption? Is it necessitated by the direction of the integration later on? Mar 10 at 21:30
• @NemDenemam there is nothing mathematical that determines the sign. It is physical. That is from the conservation of mass. If the mass of the ejected gas increases then some other mass must decrease.
– Dale
Mar 10 at 22:10
• @Dale but this equation also satisfies mass conservation: mv=(m−dm)(v+dv)+dm(v−$v_{rel}$) and it produces a result with the incorrect sign ($m \cdot dv=dm \cdot v_{rel}$). So what warns me that I made a mistake, other than the fact that I know that the result will have to be decreasing mass and positive thrust? In other words, shouldn't this negative sign come from my initial assumptions, encapsulated in the first equation, rather than from an after-the-fact analysis? Mar 11 at 11:22

Actually $$dm_g = -dm$$ means that the mass that the rocket loses in the burned fuel, is the mass of the ejected fuel.The $$dm$$ is negative as the change is negative(The rocket is losing mass), whereas $$dm_g$$ has the same magnitude but it represents positive change(mass is add or change in the mass is +ve). Therefore,we used a negative sign while equating them.

(Sorry if there are mistakes in grammar,this is my first answer).

The total mass is the mass of the rocket plus mass of the fuel is

$$m_t=m_r+m_f=\text{constant}$$

hench

$$dm_r+dm_f=0\quad\Rightarrow\\ dm_f=-dm_r$$

• f fuel
• r rocket

The rocket equation

\begin{align*} \boldsymbol{p}_t&=(m_r+dm_f)\,\boldsymbol{v}\\ \boldsymbol{p}_{t+dt}&=m_r(\boldsymbol{v}+d\boldsymbol{v})+dm_f\,(\boldsymbol{v}+d\boldsymbol{v}+\boldsymbol{v}_{\text{rel}})\\ \boldsymbol{F}&=\frac{\boldsymbol{p}_{t+dt}-\boldsymbol{p}_t}{dt}= \frac{m_r(\boldsymbol{v}+d\boldsymbol{v})+dm_f\,(\boldsymbol{v}+d\boldsymbol{v}+\boldsymbol{v}_{\text{rel}}) -(m_r+dm_f)\,\boldsymbol{v}}{dt}\\&=m_r\,\frac{d\boldsymbol{v}}{dt}+\underbrace{ {\frac{dm_f\,d\boldsymbol{v}}{dt}}}_{=0} +\frac{dm_f}{dt}\,\boldsymbol{v}_{\text{rel}}\\ &\text{with:}\\ dm_f&=-dm_r\\\\ &\Rightarrow\\ \boldsymbol{F}&=m_r\frac{d\boldsymbol{v}}{dt}-\frac{dm_r}{dt}\,\boldsymbol{v}_{\text{rel}} \end{align*}

• Thank you for the answer, it's a nice way to show it. Although I think there is a mistake in it. In the first line of rocket equation if $m_r$ doesn't include $m_f$ at 't' then it is constant and $\frac{dm_r}{dt} = 0$, isn't it? The $m_r$ can only change in next iteration if it contains $dm_f$ at 't', right? So if we define $m_{Tot_t} = m_r + dm_f$ and use $m_{Tot_t} = m_r$ substitution from line 2 in your comment then the $dm_{Tot_t} = -dm_f$ becomes clear. Apr 28 at 17:24
• Probably you are right, but at the rocket equation you use $~dm_f=-dm_r~$ ?
– Eli
Apr 28 at 18:44
• Sorry I don't understand the question. Apr 29 at 23:57
• See my answer „the rocket equation „ $~dm_f=dm_r~$
– Eli
Apr 30 at 6:12
• Sorry I still don't understand what the question is. I will try to reiterate my question differently, maybe it helps a common understanding of what I mean. The $dm_f = -dm_r$ is only true if the $m_r$ is not constant. However at time 't' it looks like you already assumed the whole mass to be both the weight of the rocket and the mass that is going to be exploded out separately. If that is how you meant then I think $dm_f =-dm_r$ should be $dm_{rocket+fuel} = m_r - (m_r + dm_f) = -dm_f$, shouldn't it? May 1 at 17:32