Rocket Propulsion Equation: Meaning of Net Force

Upon deriving the rocket equation, my textbook (Giancoli) states that the meaning of the net force on the final equation is the net force on M. Note that $M$ and $dM$ are used as the rocket (without fuel) and fuel respectively.

$$M\frac{d\mathbf{v}}{dt} = \sum{\mathbf{F}_{ext}} + \mathbf{v}\frac{dM}{dt}$$

In the previous section, they described the meaning of the center of mass and stated it as the point where all of the net forces can be considered. The CM - for two points that is- was never at an end point but rather in between.

Here, is the CM somewhere in between the centers of M and dM? If so, how do the net external forces act on M solely?

• "M and dM are used as the rocket (without fuel) and fuel respectively"? dM is just infinitesimal mass. $M$ must be total mass of rocket at any given instant, including fuel, and $\frac{dM}{dt}$ is the rate at which the rocket's mass is changing due to ejection of exhaust gases.
– Deep
Jul 13, 2016 at 5:42
• This question seems to arise solely because of a misunderstanding of the meaning of M and dM. There is no physics to explain here. Jul 13, 2016 at 21:04
• I mean, I suppose I knew what M and dM meant but my real question was the relation of the external net force to CM and in this case M. But okay :) Jul 13, 2016 at 23:38
• If I understand, your problem is, what happens to $dM$? Well it is the exhaust gas, we don't care, we are interested primarily in the rocket. And as others have written, the $dM$ is small/infinitesimal. May 27, 2019 at 13:57

The equation is built up from considering considering the mass $M$ moving with velocity $v$ splitting into $M+\delta M$ moving at ${\bf v}~+~\delta\bf v$ and a smaller piece with mass $\delta M$ moving with velocity ${\bf v}-\bf V$, for $\bf V$ the velocity of hot gases or plume from the rocket. The diagram below illustrates this
Conservation of momentum requires that $$M{\bf v}~=~(M~-~\delta M)({\bf v}~+~\delta{\bf v})~+~({\bf v}-~{\bf V})\delta M$$ We ignore the term $\delta M\delta\bf v$ and we get the equation $$M{\bf v}~=~M{\bf v}~+~M\delta{\bf v}~-~{\bf V}\delta M$$ We can simplify this by eliminating the $M\bf v$, take the limit as calculus to get the simple integration $$\int d{\bf v}~=~{\bf V}\int_{M_i}^{M_f}\frac{dM}{M}$$ that gives the rocket equation.
If we further include exterior forces this results in $$\sum{\mathbf{F}_{ext}}~=~M\frac{d\mathbf{v}}{dt}~-~\mathbf{V}\frac{dM}{dt}$$ If I multiply by $dt$, divide through by $M$ and integrate $$\int dt\sum{\mathbf{a}_{ext}}~=~\int d\mathbf{v}~-~\mathbf{V}\int\frac{dM}{M},$$ The left hand side gives the velocity of the center of mass, being that this is due to an external force. Therefore the velocity of the center of mass is $u_{cm}~=~\int dt\sum{\mathbf{a}_{ext}}$. If $\sum{\mathbf{a}_{ext}}~=~0$ the center of mass has constant or zero velocity.