We derived the equation of motion of a rocket this way: all the velocities are taken with respect to inertial observer standing where the rocket starts. As we take upwards direction to be positive. The initial momentum of rocket is mv. That's fine. Now at a certain time $\delta t$ later the momentum becomes $(m - \delta m) (v + \delta v) - \delta m (v + \delta v_0) $ {where $\delta m$ = amount of decrease of mass of rocket or amount of gas emitted. $\delta v$ = increase of velocity of rocket due to mass loss and conservation of momentum. Now my question is why we take the velocity of the gas to be $v + v_0$ . As we measuring with respect to inertial observer standing at the starting point of rocket. And if we are measuring with respect to the rocket then the initial momentum should be zero as well as the velocity of the rocket relative to itself is zero. So in my opinion the relative velocity of the gas with respect to the inertial observer should be $v_0 -v$ ;$v_0$ = velocity of the gas.
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$\begingroup$ I don't quite get your questions? The way i see it, observer standing at the launch pad -> V0 is nonzero, observer at the rockets frame of reference -> V0 = 0. Wouldn't the relative velocity of the gas be dependant on your coordinate system? v-+V0? $\endgroup$– DakkVaderCommented Aug 14, 2018 at 5:24
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$\begingroup$ @DakkVader As if I think the gas has inertia and the rocket is going up with velocity v then the velocity of the gas should be equal to $-v +v_0$ this is my thinking. I just wanna know my mistake. $\endgroup$– Nobody recognizeableCommented Aug 14, 2018 at 6:17
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$\begingroup$ You have not defined $v_0$. $\endgroup$– FarcherCommented Aug 14, 2018 at 7:30
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$\begingroup$ @Farcher thanks for the suggestion I've edited the question. By the way this is velocity of emission of gas. $\endgroup$– Nobody recognizeableCommented Aug 14, 2018 at 7:41
1 Answer
The proper derivation says that we are in an inertial reference frame, and at some time $t$, we see the rocket traveling at speed $v(t)$ with total mass $m(t)$ ejecting mass backwards at a constant exhaust speed $s$ relative to the rocket. A momentum balance therefore says, $$ m(t) v(t) = m(t+dt) v(t+dt) + [m(t)-m(t + dt)](v(t) - s). $$ The term on the left is the total momentum of the rocket at time $t$, the first term on the right is the new momentum of the rocket at time $t+dt$, and the second term on the right is a chunk of fuel which has been ejected at speed $s$ backwards relative to the rocket, which is traveling forward at speed $v$.
Expanding we have$f(t+dt)=f(t) + f'(t) dt + \dots$. Thus, this becomes$$ m~ v = (m + m'~ dt)(v+v'~dt) - m'~ dt(v - s). $$ Discarding the $dt^2$ term on the right yields $$ 0= m'~v+ v'~m - m'~v +m'~s. $$ And so finally, $$ v'~m = -m'~s. $$This integrates directly to $$ v(t) - v(0) = - s\ln\big(m(t)/m(0)\big). $$ You can get there somewhat more easily by just using a reference frame that happens to be moving forward at speed $v$ but that always seemed like cheating for me: it is important to understand that the two $m'~v $ terms cancel each other out.
