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**Physics, Astronomy & Cosmology / Re: In deriving the Rocket Equation, why does this segment seem to be added twice?**

« **on:**12/06/2021 01:09:41 »

Hi @cmiles0509

You said:In deriving the rocket equation, we go through this step:Which is not true and isn't mathematically correct.

mv = (m-Δm)(v+Δv)

The derivation you are looking at uses the principle of conservation of momentum. This applies to the TOTAL (or sum of all) momentum in the entire system. It does not apply to the rocket on it's own.

Before burning a unit of fuel, the entire system was just the rocket with mass m and velocity v. So the total momentum was just mv.

After burning the unit of fuel, the system is more than just the rocket. It is the rocket PLUS the expelled gas (fluid). We tend to put both liquids and gases under the general name "fluids".

We have: Initial Momentum = Final Momentum

mv = momentum of rocket +momentum of expelled gas

This effectively jumps straight to the second equation you stated:

mv = (m-Δm)(v+Δv) +Δm(v-u)

on the right hand side, the bit inis the momentum of the expelled fluid. The first term (not in italics) is the momentum of the rocket (with the new mass of m-Δm and new velocity v+Δv ).italics

Does that make some sense?

Ahhhhhh, so that means that the equation (in its entirety) is referring to "All the momentum that is occurring between the fuel and the rocket"?

And, if I understand correctly, by "isn't true and is mathematically incorrect", you mean that the first part of the equation I put (which excluded exhaust velocity and all that) was incorrect (since it was incomplete)? Therefore the second section is, in fact, correct?

That's a little harder then, why does the equation account for the momentum of both fuel AND rocket, if the fuel's momentum doesn't really matter once it has been used to propel the spacecraft? I need to look into this further!!

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