Linear Momentum: The Rocket Equation 

Thank you so much. It is nice to have someone who likes the way I approached this classic problem.

Nicely done. There are devils in the details but I always preferred this derivation compared to ones that have terms like (m+dm)(v+dv) and then hand-wavingly say they are going to "ignore the dm*dv term". That darn minus sign though! You could have introduced it earlier since impulse is a vector, so one could (should?) write JR = - JG (which comes from Newton's 3rd law) or even JR + JG = 0 (from conservation of momentum of the rocket+exhaust system). Also note that _your_ exhaust speed is in the center-of-mass frame, and not "relative to the rocket" as others have used since Tsiolkovosky. But for m>>dm the two are indistinguishable. In practice since not all exhaust elements reach the same speed, rocket scientists use "specific impulse" anyway, which has the same units m/s and can be thought of as _effective_ exhaust speed.

Thank you! This video was fun to make.

​@@INTEGRALPHYSICS You're welcome! Glad you fun haha I thoroughly enjoy your teaching style. Wish you taught me back in college instead of my professors then 😜

Ultimately an ion drive (which 'throws' an atomic particle at a huge speed) still obeys the rocket equation. The benefit of an ion drive being that the engine produces a much more impulse per unit mass of 'fuel' expended.

Thanks!... Anything specific?

Everything about physics is interesting, all the best to you

Not really, you get the same thrust in any direction; gravitational force could affect the "V_esc" variable, but that question is addressed on the engineering side of the rocket's engines and fuel lines

@@MalumNecessarium He is not driving the thrust. He is deriving the change in velocity of the rocket. The rocket accelerates greater if it is "flying" horizonatlly than when it is working vertically against gravity.

I see your logic, but M is a variable which changes over time. Yes, at any point M will equal Mi minus dM*t , and this problem can be solved using that (possibly more intuitive) method. In the method I did, the math is built around looking at both impulses simultaneously... I've been meaning to record this problem using that (more intuitive) method for some time now... Perhaps I need to get on it.

kg m/s^2 * s = kg m/s

F=ma=m(v-u)/t, so Ft=m(v-u). Now of course v-u is DeltaV and boom Ft=mDeltaV

@@Miftahul_786 or you could do f=ma and a=dv/dt then just move dt onto force leaving fdt=mdv or ft=mdeltav

@@LT72884 correct :)

Acceleration is dv/dt, not just dv.

@@INTEGRALPHYSICS Righttt thanks lots, small oversight huge difference