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while deriving rocket equation we find change in momentum of rocket and this is the result: $$Δp=MΔV+Ve ΔM$$ from here we divide this by time to calculate force and since no external force(freespace) is acting we get: $A=Ve ΔM/M $ ($A $ is acceleration of rocket)

And now we integrate this with time to get change in velocity and we assume intial velocity is zero so we get real time velocity this is rocket equation. SO FAR EVERYTHING LOOKS CORRECT but in the equation : $Δp=MΔV+Ve ΔM$

since $ Δp=0$ and the equation becomes: $MΔV=-Ve ΔM$

dividing the whole by M we get:

$$ΔV=-Ve ΔM/M$$

this also tells change in velocity and since initial velocity is zero this will give me real time velocity but this is wrong I know because rocket equation is different what exactly is wrong with this thinking.

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Sep 29 at 17:44
  • $\begingroup$ @Community the question doesn't need any further details, is clear enough. I'll take this. But don't close it, please $\endgroup$
    – Pizzaguy07
    Sep 29 at 17:54
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    $\begingroup$ What is the subscript $e$ in $V_e$? $\endgroup$
    – Bob D
    Sep 29 at 19:32

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I think there is an abuse of notation. You should consider a differential increment. Then, the correct equation becomes:

$$ 0 = dp = m dv + v_e dm$$

$$ \int_{v_0}^v dv = -\int_{m_0}^m \frac{v_e}{m} dm$$

$$ v - v_0 = \Delta v = v_e \ln(\frac{m_0}{m_f})$$

Your equation $0 = \Delta p = m \Delta v + v_e \Delta m$ considers that the increments will remain in the same proportion as they become bigger. But the rocket equation considers differential increments, not huge variations.

See the complete procedure here

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  • $\begingroup$ What's the meaning of $v_e$? (i.e. what does the subscript $e$ stand for?) $\endgroup$
    – Quillo
    Sep 29 at 20:14
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    $\begingroup$ @Quillo Is the effective exhaust velocity. The velocity at which exhaust gasses leave the nozzle of the rocket's engine $\endgroup$
    – Pizzaguy07
    Sep 29 at 20:18
  • $\begingroup$ Starting off with $m\mathrm{d}v+v_e\mathrm{d}m$ seems to be missing quite a lot of description. Almost so much that this is more abusive of notation than using $\Delta$. $\endgroup$
    – Kyle Kanos
    Sep 30 at 19:04

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