# What does "m Δv" stand for in the rocket equation?

I'm currently trying to understand how the Rocket Equation works.

I've gotten this far:

mv = (m+Δm)(v+Δv) − Δm(v−ve) Left side is the momentum of the rocket+fuel BEFORE, and right side is the momentum of the entire system AFTER the fuel has been expelled out the back.

After this, the brackets are removed causing the equation to become this:

mv = mv + Δmv + Δmv + ΔmΔv - Δmv + Δmve

ΔmΔv is ignored due to it being too small, and when we clean up the equation, it becomes:

mΔv = -Δmve

Now, what is mΔv in the equation? I can see that -Δmve is "momentum of the expelled fuel", since -Δm is the mass of the expelled fuel (still a positive number) and ve is the velocity of the fuel being expelled.

If we just calculate it, mΔv is [original mass]*[change in velocity for rocket]. That.. doesn't really result to anything important, does it?

So, the question summed up is: Is mΔv an important part that I should understand, or should I simply skip over that as-is and start integrating?

I'm sure this must have a simple answer, but I can't wrap my head round it! Right after this is integration. I'd like to go forward but can't, because I can't understand this one bit! Can someone help? :)

• Do the know the importance of delta-v?
– Fred
Jun 19 at 12:37
• @CliffMiles: every orbital maneuver requires a certain change in velocity which is largely independent of spacecraft mass, thrust, acceleration, or burn time (things like ion thrusters being an exception due to their extremely low thrust and long burn times). The main quantity you need to budget for in order to be sure your spacecraft can perform the needed maneuvers is thus delta-v. Jun 19 at 15:28
• If ve is the velocity of the exhaust, then v must be the velocity of the rocket. mΔv is the small change in momentum of the rocket associated with expelling that small amount of mass Δm. Momentum is extremely important, that's just not the term you're trying to solve for. Divide by m to get Δv=-veΔm/m to get the change in velocity. If you were using calculus you would take the limit as Δ -> d (that's why the double Δ term disappears), write dv=-vedm/m, and integrate. It's harder to do without calculus because m changes continuously.
– Greg
Jun 19 at 20:31
• @AJN Agh, this is tantalizingly close to my being able to understand. OK, so if we laid out the equation as you did, (m+Δm)Δv, we'd get the momentum that the expelled fuel added to the rocket. As I did above, ΔmΔv was ignored due to its insignificance. But how does the +Δm get added onto (m)Δv? Greg seems to be suggesting that I need not bother with the equation in its current form (mΔv = -Δmve) and just move on, divide by m, to get Δv. Should I just do that? Jun 20 at 6:39
• From your original question, you seems to have already proceed to the next step which is integration and that you have understood that part. You may have to re-write the question to be more clear as to what you are asking for.
– AJN
Jun 20 at 7:48

$$m\Delta v$$, as you state it, would be the impulse gained by your rocket.

To see more clearly what's going on, we can rearrange your:

$$m\Delta v = -\Delta mv_e$$ to:

$$\Delta v = \frac{-\Delta mv_e}{m}$$

That is, we can take the impulse of the exhaust, $$-\Delta mv_e$$, and divide it by the mass of the rocket to see what change in velocity this impulse would result in.

Is mΔv an important part that I should understand, or should I simply skip over that as-is and start integrating?

You now have a change in one variable, expressed as the change in another variable. That seems like an excellent place to start integrating.

• Oh, so does that mean that in my explanation (since I'm doing a project on this), I won't have to explain that stage, just simply divide by M, integrate, and be done with it? ALSO, what exactly would we be achieving by dividing the equation by M before integrating? Jun 20 at 14:28
• The analysis starts with the conservation of momentum. That being worked out, presumably what you really want is velocity versus time, so you need to isolate the v term. At that point it's just math. Formally you're "separating the variables". You pretty much always want to do that. It doesn't make any sense to integrate dm on the right side if m is sitting there on the left, because m is not a constant, and the reason it's not a constant is because of the dm.
– Greg
Jun 20 at 17:29

Delta V represent a change in the craft's velocity.

Typically this will be used to represent how much velocity has to be added or removed from a craft in order to attain a specific change in its heading (for example the velocity change required to move a craft from a planetary orbit to one that will bisect a orbit around another planetary body).

The "m" normally represents the mass of the craft.