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?

  • 2
    $\begingroup$ "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. $\endgroup$
    – Deep
    Jul 13, 2016 at 5:42
  • $\begingroup$ This question seems to arise solely because of a misunderstanding of the meaning of M and dM. There is no physics to explain here. $\endgroup$ Jul 13, 2016 at 21:04
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    $\begingroup$ 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 :) $\endgroup$
    – Ian L
    Jul 13, 2016 at 23:38
  • $\begingroup$ 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. $\endgroup$
    – patta
    May 27, 2019 at 13:57

1 Answer 1


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 enter image description here

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.

  • $\begingroup$ And if it is not equal to zero? :) $\endgroup$
    – Ian L
    Jul 13, 2016 at 4:41
  • $\begingroup$ Also can you clarify how the external net force is acted on M and not dm (or is it both???) $\endgroup$
    – Ian L
    Jul 13, 2016 at 4:44
  • $\begingroup$ Based on what I wrote this assumes the exterior force acts on the entire system. An example would be a gravity field from a planet the rocket is accelerating away from. $\endgroup$ Jul 13, 2016 at 21:15

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