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This question already has an answer here:

What actually causes a rocket to move? Is it the pressure in the rocket engine or the amount and velocity of mass that is being ejected out.

The reason I am asking is, I found these two explanations for a rocket motion. Are they same or different. If same, is there any correlation between them?

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marked as duplicate by Carl Witthoft, JamalS, ACuriousMind, Qmechanic Jan 12 '15 at 18:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This question appears to be off-topic because it is shows a complete lack of effort and research. $\endgroup$ – Carl Witthoft Jan 11 '15 at 13:01
  • $\begingroup$ Outside of the atmosphere a rocket moves as an "equal and opposite reaction" to the momentum of the ejected exhaust. $\endgroup$ – Hot Licks Jan 11 '15 at 15:27
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The two explanations are different, but they do complement each other.

Conservation of linear momentum holds in the absence of external forces. Thus, it allows us to ignore any messy details of the internal forces when trying to determine the result. (Internal to the vessel-reaction mass system) "The reaction mass and the vessel were both at rest; now the reaction mass is going in one direction at a certain velocity, so the vessel must be going in the other direction at this velocity." We don't need to worry about how the reaction mass got it's velocity. A space walking astronaut may have fired a thruster, or thrown a tool, or unzipped his suit and.... But I digress.

On the other hand, we sometimes need to worry about where and how the internal forces are applied to the vessel. In a rocket, the force in question comes from the gas pressure against the forward wall of the combustion chamber. We need to consider this aspect if we are, say, designing the walls and attachment of the rocket engine.

This is the case in most conservation laws; they let us skip intermediate steps. if I were to tell you that I was going to drop a bowling ball on your head from the top of a ten-story building, you would quite rightly have some concerns! But how the bowling ball got up there would not be relevant; conservation of potential and kinetic energy lets you skip that part. But as the perpetrator, I might be concerned about the breaking strain of the elevator cables, or the strength of the stairs...

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A rocket moves because of the conservation of linear momentum. Imagine you have a rocket of total mass $M$ (which includes fuel) that expels exhausts at a speed $u$ relative to the rocket. As the rocket burns fuel, its mass $M$ reduces and becomes a function of time $M(t)$. Similarly, the rocket velocity $v_r(t)$ and the exhausts velocity $v_e(t)$ relative to a fixed observers are functions of time. Conservation of momentum requires then that $$\frac{\text d}{\text dt}\left[M(t)v_r(t) + \int_0^t v_e(t)\dot m(t)\text dt\right] = 0,$$ where $\dot m$ is the rate at which the fuel is being burnt by the rocket. The first term in square brackets is the rocket momentum, while the second is the total exhausts momentum. The latter involves an integral because bits of exhausts are released at different speeds with respect to the fixed observer, say on the ground. Of course, mass conservation enforces the relation $$\dot M + \dot m = 0.$$ Computing the above derivative one gets $$\dot M v_r + M\dot v_r + \dot m v_e = 0$$ and using mass conservation this leads to $$\dot M(v_r-v_e) + M\dot v_r = 0.$$ Now the velocity of exhausts is linked to the velocity of ejection $u$ by Galilean relativity $v_e = v_r - u$ and therefore the above equation becomes $$\dot Mu + M\dot v_r = 0,$$ which has the well known solution $$v_r = v_0 + u\log\left(\frac{M(0)}{M(t)}\right).$$ Observe that $M(t)$ is a strictly decreasing function in $t$ and therefore the logarithm is positive, leading to a boost in the rocket's speed according to the above law, which is known as Tsiolkovsky rocket equation.

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A rocket moves because of Newton's 3rd law.The rocket propels itself forward by spitting out fuel behind.

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Absent any external force, an object may accelerate by ejecting some of its mass in the direction opposite the desired motion. This is an easy application of conservation of momentum.

Before, we assume the object $O$ is at rest (no momentum) and has mass $M$.

After, by conservation of momentum, the sum of the momenta must be zero.

$$(M - \Delta m)\cdot v_O + \Delta m\cdot v_{\Delta m} = 0$$

Thus, by ejecting some its mass, the object is accelerated to some velocity relative to the initial frame of reference.

But realize that, in order to eject the mass, there must be an internal force (imagine, e.g., a compressed spring) that acts to both accelerate the object and the ejected mass, converting the stored potential energy to the kinetic energies of the object and ejected mass.

Now, a rocket is, according to Wikipedia:

... [a] vehicle that obtains thrust from a rocket engine. ... Rocket engines push rockets forward by expelling their exhaust in the opposite direction at high speed.

Delving deeper into the details of rocket engine operation, we find that pressure (force per unit area) is necessarily associated with the acceleration of the expelled mass and rocket.

Indeed, there is a relationship, for a given rocket engine, between the pressure and the momentum (amount and velocity) of the expelled exhaust. So yes, there is correlation between them.

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