Unclear negative sign in rocket propulsion equation

Question

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.

Answer 1

$m$ 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 $m+dm<m$ because $dm$ is negative.

Answer 2

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).

Answer 3

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*}