text: # Doubt regarding derivation of rocket equation
  
  I'm having quite some issues deriving the ideal rocket equation, and I suspect it is due to lack of identifying scalar terms and vector terms. Here is how I derive it :
  
  Let $$v$$ be the velocity of the rocket and $$u$$ be the velocity of the exhaust, relative to us, the observer.
  
  $$P_i = mv$$
  
  $$P_f = (m-dm)(v+dv)-udm \space\space\space$$ (Here I've put the negative sign before $$u$$ as it is in the opposite direction)
  
  $$\Delta P = P_f-P_i = mdv-vdm-udm\space$$(ignoring the term $$dmdv$$)
  
  $$F_{ext}\space dt = mdv-(u+v)dm$$
  
  This is where my confusion begins. What exactly is relative velocity? Is it $$(u+v)$$ or is it $$-(u+v)$$ ?
  
  If I consider the rocket is moving as $$v\hat{i}$$ and the gas is moving at $$-u\hat{i}$$, then their relative speed is $$u+v$$. From the rocket's perspective, the relative velocity of the gas is $$v_{rel} =(u+v)(-\hat{i})$$.
  
  Then we can say, $$u=(v+v_{rel}) = v\hat{i}+(v+u)(-\hat{i})=-u\hat{i}\space,$$ which is exactly what we wanted. So, we can write (along $$\hat{i}$$, taking that to be positive) :
  
  $$F_{ext}dt=mdv + v_{rel}dm$$
  
  We can also do this in another way, where we take the direction of the gas to be positive. In that case, our expressions would have an extra negative sign on both sides, but after that, the final expression would be exactly the same.
  
  My confusion is regarding relative velocity. If the speed of the rocket is $$v\hat{i}$$ and the exhaust is $$-u\hat{i}$$, shouldn't we say $$u=v+v_{rel}$$ ?
  
  Many sources write $$u=v-v_{rel}$$ instead. For example, the comments below this [question]:Rocket equation derivation mistake by my professor
  
  Is it because, they are considering $$v_{rel}$$ a scalar here, and using the extra negative sign to denote the opposite direction, but since I'm considering $$v_{rel}$$ a vector here, and so, I don't have to plug in an extra negative sign here ?
  
  Can I say :
  
  $$\vec{u} = \vec{v}+\vec{v_{rel}} = v\hat{i} + (v+u)(-\hat{i})$$
  
  Then this will automatically give me the negative sign in front of $$u$$.
  
  I suspect what the others do is consider the scalar version of the above equation. Can anyone tell me if I'm correct, is my expression for relative velocity correct, or should there be a negative sign as most sources, including Wikipedia claim there is.
  
  • you do not know that $u$ is in the opposite direction in the frame of observer, because that depends on the motion of the observer. Sep 6 '21 at 8:56
  • @Umaxo $u$ is the velocity of the gas coming out of the rocket. If the observer is on the ground, at rest, shouldn't the velocity of the gas be in the opposite direction to the velocity of the rocket ? Sep 6 '21 at 8:59
  • No. One sensible scenario is that exhaust gas is repelled from the spaceship with constant speed relative to the spaceship. Be it 800 m/s. Then, if the spaceship moves away from the Earth with velocity 2000m/s, the exhaust gass moves also away from the Earth, with velocity 1200m/s, i.e. they move in the same direction relative to Earth. Sep 6 '21 at 9:02
  • Also, I do not understand why is your $F_{ext}$ nonzero. $P_i$ seems to be momentum of the whole (spaceship+gass) system at certain time $t$ and $P_f$ to be momentum of the same system at time $t+dt$. Since this system is isolated, shouldn't conservation of momentum apply and $F_{ext}$ be zero? Sep 6 '21 at 9:07
  • @Umaxo yeah it is non-zero, I was writing the general form, so I didn't plug that in yet. I'm just thinking about the relative velocity expression, like how do I reconcile whatever you've said ( correctly so ) with my general expression. As in how do I derive a general expression in a vector way. Sep 6 '21 at 9:10