To be truthful, even I have a difficult time wrapping my mind around how (dm/dt)v actually plays into the equation, but the math doesn't lie, and the math that derives it is very simple, and makes it plain that it cannot be ignored. (Provided you know Calculus, that is.) But physically, it is pretty unintuitive. (Again, as in my last post, bolded values are vectors. Just for bookkeeping and clarity.) So, to start off, we have commonly taught version of Newton's 2nd law: ÃŽÂ£F = ma. This is read as "The sum of all external forces on an object (ÃŽÂ£F) is equal to the mass of the object multiplied by the rate at which it accelerates." As I said in a previous post, this is not 100% true in every possible situation, but it works in the case of a constant mass. (In fact, as you'll see later, it actually IS this in the case of constant mass. The general case will simplify down to this.) However, to get the whole story, you need the equation ÃŽÂ£F=dp/dt. This is read as "The sum of all external forces on an object is equal to the time derivative of its momentum." Now, I'm sure you know what momentum is, it is mass multiplied by velocity! Or mv. But, what the heck does "time derivative" mean? Well, in calculus, there are three big concepts: limits, derivatives, and integrals. Limits hare extremely useful tools, but they're not relevant here. All you need to know about them for now is that they are used in the construction of the next two tools we gain from calculus: derivatives and integrals. Integrals are also extremely useful, but also very confusing, and not needed here, so we will throw them out for now too. That leaves us with derivatives, which is the relevant topic. A derivative, in short, is the rate of change of something. It can be the same everywhere (as is the case in a linear function), it can change smoothly according to its own function (as with any polynomial that isn't a linear function), or it can even jump around pretty wildly in some weird cases. But, for any function which has a derivative, the derivative will tell you how fast that function is changing at any point you want! (Well, not any point you want, there are some rules. But they're not relevant to us here.) That is what a derivative is, it is a function that tells you how fast another function is changing at any point you need to know. In the case of a time derivative, we are concerned with how our function changes as time changes. You can take a functions derivative with respect to any variable you want, but time is the one we are looking for here. Whenever you see something like dm/dt, that is what it means. You want to know how the top variable (m, in our case) changes as the bottom variable (t) changes. Don't worry about the d's, those are just part of the notation. So, for the sake of brevity, we are going to skip a lot of steps here, stuff like calculating derivatives by hand, and go right to the shortcuts! As it turns out, derivatives of certain types of functions always follow the same rules, and instead of doing everything out the long way, you can use these rules to calculate your derivative much quicker. The rule we are interested in is called The Product Rule. The product rule says that whenever you have a function (which we will call h(t)) that is equal to two other functions multiplied by each other, (e.g. h(t) = f(t)*g(t)) the derivative of h(t) is equal to the derivative of the first function, multiplied by the second function, plus the first function multiplied by the derivative of the second function. (dh/dt = df/dt * g(t) + f(t) * dg/dt) Now that we have our rule, let's see how it applies to what we were talking about: we know that ÃŽÂ£F = dp/dt. And we know that p = mv. And we know how to take the derivative of two things multiplied together, so let's do that! Our result ends up being dp/dt = (dm/dt) * v + m * (dv/dt). Now, it just so happens that dv/dt (the rate of change of velocity as time changes) has its own name: acceleration. It also has its own symbol: a. So we'll plug a in for dv/dt so things get less cluttered. Now we have ÃŽÂ£F = dp/dt = (dm/dt) * v + ma. This is the general form of Newton's 2nd law. But, most of the time when you're using it, the mass of the thing your using it on (such as a billiard ball, to pull an example out of a hat) doesn't change at all. Ever. Period. In this case, dm/dt (the rate of change of mass as time changes) is 0. Everywhere. Always. So, if we plug that in, we get ÃŽÂ£F = dp/dt = 0 * v + ma = ma. In this case, the bit with dm/dt goes away, and the law simplifies down to what you were taught in school. But, in our case, we have rockets that do little else but throw their mass away really fast so they can get to space. (Or the ground 100 feet from the launch pad.) In this case, our mass is always changing, so dm/dt is never zero, and so that part of the equation never goes away, and so we can't pretend it doesn't exist, we have to take it into account. That's the math, and the math doesn't lie. TL;DR: I don't understand it either, but the math says it has to be there, and the math doesn't lie. Also, unfortunately I haven't had much time to play with KSP and experiment with this, as much as it's been gnawing at me. So I have no data to report.