During the launch of a rocket into space, the exhaust gases from the engine possess momentum. Outline, with the help of an equation, how this causes the rocket to be propelled upwards.

Can someone point me in the right direction? I understand how the exhaust gases are used to propel the rocket upwards, and based on the specific mention of "momentum" I would think that $p=mv$ is to be referenced, but I don't understand how to link this to the question.


2 Answers 2


It does follow from the conservation of momentum. Consider the diagram (from Wikipedia) of a rocket expelling gas of mass $\Delta m$:


At $t=0$, the initial momentum is $$ p(t=0)=\left(m+\Delta m\right)V\tag{1} $$ but at $t=\Delta t$, we've lost some mass and gained some velocity, $$ p(t=\Delta t)=m\left(V+\Delta V\right)+\Delta m \left(V-v_e\right) \tag{2} $$ where $V_e=V-v_e$ with the difference coming from reference frames (see the above linked Wikipedia page). So now we have the momentum before & after the exhaust velocity, which shows an increase in the velocity of the rocket. The implication of this equation can be found in the post Why are rockets so big?

For other aspects of this problem (momentum conservation in the rocket equation), see


The most important equation to remember to use is:

$F\Delta t = m\Delta v$

Where "F" is the magnitude of the thrust and "t" is for how long (in this case) constant force is being applied. If the thrust changes based on time, you have to use:

$\int F dt = m \int dv$

EDIT: Kyle pointed out a flaw in my answer, so I will try to remedy it. Mass depends on the amount of time that has elapsed ($m(t)$), since this is with respect to time. The new change in velocity is:

$ \int \frac {F(t) dt}{m(t)} = \int dv$

  • 1
    $\begingroup$ Well this only holds if you're not losing any mass while accelerating. Since you are (because that's how rockets work), you can't use $F=ma$ here. $\endgroup$
    – Kyle Kanos
    Aug 4, 2015 at 4:05

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