 A rocket is launched vertically by combusting a solid propellant, which includes oxygen. The exhausting gases create thrust, pushing the rocket directly upwards. If the rocket, including all rocket parts and propellant, has an initial mass of m-dot, the exhaust gases leave at a velocity of v subscript e. The mass flow rate of the exhausting gases is m-dot, gravitational acceleration is a constant, written as a g, and you neglect drag, determine an expression for the velocity of the rocket relative to the ground as a function of time. So far, we have only worked with control volumes in an inertial frame of reference. We will now relax this restriction to consider an accelerating but non-rotating control volume. For that, we now have a new term appearing in our conservation of momentum equation. In this equation, the velocity vectors are measured relative to a non-inertial, that's the accelerating reference frame. The acceleration is acceleration of control volume relative to an inertial reference frame, otherwise the problem is worked pretty much as before. We know our control volume is going to have a body force in the downward direction, that's the weight of the actual rocket, and then we can account for this new term by writing the acceleration, or relative acceleration here, as the velocity, the change in velocity of the rocket with respect to time. So I'm neglecting surface forces because the only one that would be relevant here would be drag, the body force would be the weight of the rocket, which I'm going to write as the mass times gravity, and that's in the opposite direction of our axis, which I'm defining in the upward direction. And then I am going to subtract that density times volume that is left over after acceleration comes out of the integral is going to be written as mass, and then referring to the rocket. I will point out that acceleration comes out of the integral because we don't have a change in the acceleration with respect to our volume, our volume is moving. And as a result of being steady state, that volume is the same, it's just that it's moving now, the same volume is just flying upwards. I have one opportunity for mass to cross the boundary of my control volume that is at the bottom. So I will write the Z component of velocity here as being negative VE times density, and I will collapse the velocity vector by assuming uniform flow at the outlet, at which point I will have the magnitude of the velocity, the average velocity VE again, times the area. And because the exiting velocity and area vector are both in the same direction, I can leave that as a positive term. So this would be negative VE times density times VE times area. Then I have, see, in factor of the mass, and I can get rid of all three negative terms. Mass times gravity plus dV dt is equal to density times VE squared times area. And I recognize that the mass here is referring to mass of the rocket at some point in time. It's losing mass because it's combusting the stuff that it brought along with it. That's the propellant and the oxygen. So I could describe mass as a function of time, and I could say that mass would be, whatever the initial mass is, minus the rate at which mass is exiting times whatever time has passed so far. So I have quantity m naught minus m dot times t minus, I'll write that one more time. And that's equal to the density times the velocity at the exit squared times area. And to make the math a little bit easier, I'm going to write density times velocity at the exit times area as the mass flow rate of the exiting gases. So that gets rid of one of my exiting velocity terms, but not both. So I still have to multiply by VE. Now if I solve for dV dt here, I have dV dt is equal to m dot times VE divided by m naught minus m dot times t plus gravity. And now it's just a matter of integrating. So I can bring dt over to the right and write this as dV is equal to, let's say, m dot times VE divided by m naught minus m dot times time. And then I am going to integrate that with respect to time plus gravity times dt. That's again, integrated. And when I integrate the left hand side, I get our zero to velocity at whatever point I get V function of time. And this is integrated from zero to time, zero to time at any instance in time. So the right hand side is going to simplify a little bit easier. So I'll do that first, zero to t m dot times VE divided by big m not minus m dot times t dt. And I do that gravity times whatever time has elapsed because gravity comes out of the integral and I'm left with just the integral with respect to time, which is just going to be t evaluated from zero to t, which is just t. Now I am going to have to use integration by parts here, so I'm going to call u m naught minus m dot times t. And then I'm going to write du as negative m dot times dt at which point I have velocity as a function of time is the integral from m not to m not minus m dot times t. And that would be of m dot VE divided by u du over negative m dot and that entire quantity is going to be before I subtract gravity times time. So this is going to simplify down to let's see that would be negative negative VE times natural log of u evaluated from m not to m not minus m dot t minus gravity times time. And then that would be minus VE times the natural log of m not minus m dot times t divided by m not and that entire quantity has minus dt at the tail end. So I can say that my velocity as a function of time is this relationship here. And if I wanted to, I could simplify that a little bit further, which I will do on another sheet of paper here. When I simplify that further by writing that as negative VE times the natural log of one minus m dot times time divided by m not minus gravity times time. So this equation is the rocket equation, which is pretty neat that rocket equation gets used all the time is a simple way of determining a rocket's velocity. If you look up the Wikipedia article, you can see this equation written just like this, and it isn't something special that applies to only rockets is just a simplification for a rocket that has a moving frame of reference. And if m dot t is a large in a portion of m not, you can even simplify further by writing the velocity at the end of the function as exceeding the exit velocity, which is kind of cool. I will point out that this problem is a little bit outside the scope of what I actually expect you to be able to do on an exam. I don't want you to get too distracted by this concept of a moving frame of reference. I just wanted to give you an idea of the sort of things that you can do with that in your conservation of momentum tool set.