Is momentum conserved in the rocket equation? If so, why is there force? I'm studying the rocket equation and from what I get from it the momentum between the fuel and the rocket is conserved. But if that happens, how come the calculated force is not zero?
The rocket equation is:
$$
F = -\frac{dM}{dt}v_{ex}
$$
But if the change in momentum of the rocket-fuel system is zero. Shouldn't the force be zero as well?
$$
F = \frac{\Delta p}{\Delta t} = \frac{0}{\Delta t} = 0
$$
 A: By the law of conservation of momentum, the total momentum change in the system is indeed zero. However, that's not to say that the force on the rocket individually is zero.
Consider the analogous case of a bullet fired from a gun.

There is a force exerted on the bullet causing it to leave the gun chamber, while the gun experiences an equal and opposite force - the recoil. This is simply a consequence of Newton's third law or, equivalently, the conservation of momentum. It is a common misconception that the forces cancel each other out; if this were true, we would live in a static universe! The forces act on different objects, so both objects accelerate in opposite directions, whether this is a bullet and a gun, or a rocket and its exhaust.
Mathematically,
$$\vec{F_{system}} = \vec{F_{gun}} + \vec{F_{bullet}}\\
$$
By Newton III,
$$ \vec{F_{gun}} = - \vec{F_{bullet}}$$
so
$$ \vec{F_{system}} = \vec{F_{gun}} - \vec{F_{gun}} = 0$$
However, considering the acceleration of the objects individually,
$$ \vec{a_{bullet}} = \frac{\vec{F_{bullet}}}{m_{bullet}} \neq 0$$
Hence, the error in your argument above enters when you assume that $F$ is the force on the rocket; in fact, it is the force on the system as a whole, which must be zero for conservation of momentum, as you correctly note.
For a visual explanation of this fact, try the following video: https://www.youtube.com/watch?v=TVAxASr0iUY&ab_channel=Don%27tMemorise
