We've got that. We're going to use calculus to calculate the gravitational force for non-spherical distributions of mass. So let's recap everything that we've learned about Newton's law of gravity. Whenever we had point masses like m₁ and m₂, and we had the distance involved, the gravitational force was just given by Newton's law of gravity. Now, even when we had planets, like in the second example over here, we kinda just pretended that these planets or these objects could be treated as points, and all the mass was just concentrated at the center. So that's big M, and this guy over here is sort of just like little m. And as long as we had the center of mass distance, which is that little r, it basically worked the exact same way. So instead of m₁s and m₂s, we just had big M's and little m's. So we just used Newton's law of gravity, and we just replaced it with the appropriate variables. The difference is that when we have a non-spherical mass distribution, we can't just imagine that all the masses are concentrated at the center here. It doesn't work that way. So, we need a new technique to solve these problems. And one of the things that we can do is we can sort of pretend that this rod of mass M, big M that I have here, can be thought of or broken up into tiny little pieces of rectangles or mass right here. And the smaller they become, they become a differential mass. So I'm going to call that dm, and we can treat that dm as a point mass. So that means that this point mass dm generates a tiny amount of force in this direction, and that's going to be df, and it's due to that center of mass distance of little r right here. df=g•dm•m/r2 But remember that this is only one tiny piece of the force from one tiny piece of the mass. So, to solve for this total force, we have to basically go along the rod, and we have to add up all of the tiny amounts of forces that are generated from all these tiny amounts of masses. So, think back to calculus. What do we call it when we add up a bunch of tiny infinitesimal things? We call that an integral. So the way that we're going to solve these problems is we have to integrate along whatever mass that we're given. So that means that the total amount of force is going to be given as the integral of df, which is just going to be the integral of g•dm•m/r2. But one of the things that we can do with this equation is we can actually pull out the big g and the little m as constants to the outside of the integral. Right? Because those things don't change. So the more useful form that we're always going to use here at Clutch when we solve these problems is we're going to use the g•m•∫−∞dm/r2. So we're always going to start from this equation when we're solving these problems.

Alright, guys. That's basically it. So we actually have a list of steps that we're going to need to solve any one of these mass distribution problems. But rather than telling you, I actually want to go ahead and show you. So we're going to work out this example together. This is actually a very classic common type of problem that you'll see if you're doing this topic right here. So we've got a hollow ring. It's got a mass m. It's got some distances over that radius. We have a distance which a little mass over here is sitting and we need to find out what the gravitational force is. So, the first thing that we do in all of these problems is we write out the equation for f, which is right over here. So we have f is equal to and then we have g•m•∫−∞dm/r2. So that's the first step. Okay? So the second step is we have to pick Let's see. It says pick 2 DMs and we have to write an expression for r. So I've got this DM that's going to be over here and then I'm going to pick another DM like this over here and then we know that these DM's act like point masses and they produce forces on this mass over here. They're going to point in this direction like that. That's going to be 1 df and then this guy is going to be in this direction and that's going to be another df. Okay? So, now we actually have to figure out what the r's are, the center of mass distance is. So, that is this right here, these pieces right there, and then I've got one over there. So, this is going to be r. Okay. So, we have to use, write an expression for f using the problem's geometry. That means is that we're just going to use the length variables that are given to us, r and d. Now, this little r, if you think about it, is actually just the hypotenuse of this triangle that we've made here. So we can actually use the Pythagorean theorem to come up with that expression for little r. It's actually going to be the square roots of r² plus d². Okay? Now, we're just going to save this for later. We're not actually going to start plugging it in yet because what happens is if we start plugging this in, we're going to have to write this a bunch of times. It's going to get really annoying. Okay? So we have this second step right here, so we have the expression for r. Now, let's take a look at this third spep here. Step 3 says we have to break, the integral into its x and y components. So in other words, what happens is that f which is the integral of df actually gets split into 2 things. The f x component is going to be the integral of all the DFXs, and the f y component is going to be the integral of all the DFY's. Now, where do these components actually come from? Well, remember that these D F's actually are 2-dimensional vectors. Right? They point in opposite directions or different directions. So, we have to use vector decomposition to break them up. So what I'm going to do is I'm going to, sort of, draw an angle right here relative to the x axis. That's my angle theta. And what happens is now I can break this df into its components. So I've got dfx and then I've got dfy over here. And by the way, the same exact thing happens for this df vector. So I have another one of these components that's going to be here and then another one of these components that's going to be over here, so dfy. So what we can see here is that my dfx components are always going to be sort of pointing to the left, and they're always going to be together, and they're going to be adding together. Whereas these dfy's here are always going to be equal and opposite. So, because these things are equal and opposite, what happens is that their components of dfy will always end up canceling out. So, for any mass here, for any little point, I can think of the mirror opposite point on the opposite side of the ring as directly symmetrical and those y components will always end up canceling out. So that means that this integral just goes away, and I don't have to deal with it anymore. Now remember that if I have this angle right here, I can write the dfx. So now that I've actually sort of split this and canceled out the components, I have to expand this into sines and cosines. So remember that this dfx components can be written with this angle as the integral. Actually, I'm going to have g•m•∫−∞dm/r2, and this is actually going to be the cosine of that angle right here. So, now what I have to do is I have to expand this into sine and cosine, which I've done. Now I have to rewrite this cosine in terms of the side lengths that are given because what happens is I'm integrating dm and I've got this r variable here, so I don't want this cosine sitting in here. So what I have to do is I have to relate it using the triangle. Now, remember that this cosine angle here is always the adjacent over the hypotenuse. So, given this triangle right here, in which I have the d as the adjacent side, this and I've got the r as my hypotenuse, this is actually just going to be d / r. So we're actually just going to replace that cosine of θ in there for d / r. So that means that for 4, my equation becomes f equation becomes f x is equal to or we could just write f, is equal to the integral, which actually is going to be g•m•∫−∞dm•d/r3. Okay? So that is step 4. So we're done with that. So now what we have to do is for step 5, we have to plug in the expression for R from step 2, and then pull all the constants out of the integral. Okay. So we've got f=g•m•∫−∞dm. Now we have the integral, and then we have dm. And then what happens is we're going to plug in our expression for r. Now remember that expression for r is this guy over here. So, actually, what this r³ becomes is it becomes r2+d2. And then because this is square roots, which really means that this is, to the one-half power, this actually becomes to the 3/2 power over here. And then we have this d. Now, what we have to do is we have to pull all of these letters that are constants outside of the integral. But if you take a look at this, remember that d is an integral is just a constant. Right? This is just a constant horizontal distance up here. So that is, that just gets pulled out. And then this r² and then d², those are both constants as well because they are capital letters. Remember that this radius of the ring never changes, and the d, the distance, also never changes. So, in fact, everything out of here is just a constant. All of these things here are constants and they get pulled out. So these are constants. Alright. Now, we're done with that, so we pulled all the constants out of the integral, and then we're going to rewrite this. So, this actually ends up being, f is equal to, and then we've got g•m•d/rd3. Okay? And then we've got to just write that last little remaining integral piece. So you have the integral of DM. Okay. So now what happens is if we look at step 6 a, if we're only left with DM, the integral of dm, then the integral of dm here is just m. Right? So if we're just integrating our differential dm, that's just basically the whole entire mass. So then what happens is that we're done. Right? So we just replace that with big M and we're done here. So that means that the force is equal to G•M•M•d/r+d23/½, and that is actually our force. So we don't have to do the integral because we already did it in this step. Okay, guys? So let me know if you guys have any questions, and I'll see you guys in the next one.