Linear Momentum: The Rocket Equation

[@keywestskoko]

This is perhaps the best explanation for the derivation of the rocket equation I have seen. In many of the cases of alternative vids, the math gets in the way, and it's usually through some kind of mathematical sleight of hand that terms change signs. The author's explanations here are right on and intuitively understandable. Perhaps the only place he could have been clearer is towards the end, where the natural log of the mass fraction (ln Mf - ln Mi) becomes ln (Mi/Mf) but the velocity of the escaping gas loses its negative sign (this is because the negative of ln(Mf/Mi) becomes ln(Mi/Mf) - which is where a lot of vids fail to explain the mathematical gymnastics. Note you can substitute the idea of impulse for momentum and still arrive at the same place.

[@INTEGRALPHYSICS]

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

[@redbaron07]

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.

[@BERRUEZA]

Such an underrated channel!
Your explanations are top notch! So easy to understand, thank you!

[@INTEGRALPHYSICS]

Thank you! This video was fun to make.

[@BERRUEZA]

​@@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 😜

[@freeofbug4761]

So nice. Big thanks. As I understood with impulse (with you good example of cannon), the impulse is function of mass, ejected at a certain speed. The rocket equation shows the differential mass, as the difference is ejected at a speed. What could be the equation, if we suppose that we throw atomic particle at huge speed? (I suppose that the differential mass will be very small in this case).

[@INTEGRALPHYSICS]

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.

[@user-rj9fq1yb5n]

It would be slightly better to avoid any questions about gravity force, to draw the rocket going horizontally?

[@MalumNecessarium]

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

[@user-rj9fq1yb5n]

@@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.

[@Rami-bakri]

Simple and amazing explanation. want more👍

[@INTEGRALPHYSICS]

Thanks!... Anything specific?

[@Rami-bakri]

Everything about physics is interesting, all the best to you

[@plazmaguy13yago9]

Isnt the mass of the rocket supposed to be M-dM?

[@INTEGRALPHYSICS]

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.

[@derrickbecker9856]

But need to add in mg and drag to actually model a rocket launch

[@Mod3bola]

You're the best!

[@mark2727]

Sorry but I wouldn't be able to explain this to my 10 yr old grandson who's in 4th grade to save my life. I searched for "Model Rocket equation explained and got the Einstein's explanation. Can you explain it for a 10 yr old's to understand what goes up must come down! Please!?
Just a grandpa trying to help my grandkids with the science.

[@LT72884]

I thought impulse was F(time)

[@INTEGRALPHYSICS]

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

[@Miftahul_786]

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

[@LT72884]

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

[@Miftahul_786]

@@LT72884 correct :)

[@khaledsh8506]

Impulse of rocket is = (m dv) which is (m * a) which is just force not impulse. I do not follow here.

[@INTEGRALPHYSICS]

Acceleration is dv/dt, not just dv.

[@khaledsh8506]

@@INTEGRALPHYSICS Righttt thanks lots, small oversight huge difference