heyhey281 said:
It's really important to know your sign conventions. In particular, which of the following are positive and which are negative quantities: ##g##, ##w##, and ##\dot m(\tilde t)##?
Most textbooks take ##g## to be a positive number (the magnitude of the acceleration of gravity).
If ##m(t)## is the mass of the rocket and fuel inside the rocket, then ##\dot m## would be negative.
You call ##w## the speed that the fuel particles get thrown out. Speed is usually considered positive. So, the velocity of a fuel particle relative to the ground just after the time ##\tilde t## that it is ejected would be ##v(\tilde t) - w##.
Anyway, you can see how important it is to clearly define the signs of your symbols.
The integral in your equation looks OK to me except for possible sign issues and I think ##v(t)## inside the integral should be ##v(\tilde t)##.
The general principle is that the final momentum ##P_f## of the system should equal the initial momentum of the system ##P_i## plus the impulse ##J_{ext}## due to external forces: $$P_f = P_i + J_{ext}$$ Rearrange as $$P_i = P_f - J_{ext}$$The impulse in this case is due to the force of gravity, so ##J_{ext} = -M_{sys} g \xi##, where ##\xi## is the time interval and ##g## is taken to be positive. The mass of the system is the mass ##m(t)## of the rocket and fuel at the initial time ##t##. So ##J_{ext} = -m(t)g \xi##. Then, $$P_i = P_f + m(t) g \xi.$$ The initial momentum of the system is ##P_i = m(t)v(t)## and the final momentum of the system is ##P_f = m(t+\xi)v(t+\xi) +\int_t^{t+\xi}## , where ##\int_t^{t+\xi}## is your integral (with possible corrections) that represents the momentum of the ejected fuel.
So, $$m(t)v(t) = m(t+\xi)v(t+\xi) + \left(\int_t^{t+\xi} \right) + m(t) g \xi$$
Now, I'm not sure what the use of this equation would be.