Hey, guys. So, as you know, the Earth is rotating. Therefore, it has a moment of inertia. And if we make some assumptions about the shape of the Earth, we can actually calculate the moment of inertia of the Earth. Let's check it out. It says here the Earth has a mass and radius given by these big numbers. And then I also tell you that the radial distance between the Earth and the Sun is this. What I mean by radial distance is that, if the Sun is here, the Earth spins around the Sun at this distance here; little r is \(1.5 \times 10^{11}\). Squeeze it in there. Cool? That's what I mean by that. And then I gave you the mass of the Earth and the radius of the Earth as well. I want to know the moment of inertia of the Earth as it spins around itself and as it spins around the Sun. As you know, the Earth has 2 oceans, and we can calculate a moment of inertia about, or relative to those 2 motions or for those 2 motions. Remember, moment of inertia depends on the axis of rotation. That's why these numbers will be different. So if you want to know the moment of inertia of the Earth around itself, you would have to treat the Earth as an object with significant size. You can't treat it as a tiny object. So what we do here is we're going to treat the Earth as a solid sphere. Okay? As a solid sphere. So the Earth is a big ball that spins around itself. Now technically, it's at an angle like that, but it doesn't really matter. You can just do this. Okay? So it's spinning around itself. And your book would show you that solid spheres have a moment of inertia given by this equation right here. So when I tell you solid sphere, I'm indirectly telling you, hey, use this equation for I. Okay? So for part a, we're going to do, part a is over here, we're going to say \(I = \frac{2}{5}mr^2\). And all we've got to do is plug in the numbers here. So m is the mass of the Earth, which is \(5.97 \times 10^{24}\), and r is the radius of the Earth, which is this, and not the radial distance. It's the Earth going around itself, so it's the radius of the actual object of the sphere. \(6.37 \times 10^6\)^2. Okay. If you look at this number, I got a 24, and then I got a 6 squared. So you should imagine that this is going to be a gigantic number, and it is. I multiplied everything. I get \(9.7 \times 10^{37}\) kilograms meter square. The Earth has a lot of inertia. And what that means is that it would be incredibly hard to make the Earth stop spinning. Okay? Now if you were to Google this number, you would see that it's actually a little bit off. The actual moment of inertia is a little bit off. And that's because the Earth is not a perfectly, a perfect sphere. It's got different layers. It's not even a sphere. So but this number is a pretty good approximation. For part b, we want to find out what is the moment of inertia of the Earth as it spins around the Sun. Now in this case, relative to the sun, the Earth is tiny. So we're going to treat it as a point mass, which is crazy. The Earth is a huge thing, and you're going to just treat it as a little point mass of negligible radius. And that's because, relative to the sun, the Earth is negligible in size. Okay? So I'm going to put the Earth here as a tiny m, earth, and it's going around the sun. And the distance here, the radial distance, which is little r, big r is the radius of an object, and little r is the distance to the center, is \(1.5 \times 10^{11}\) meters. In this case, we're going to use, instead of \(\frac{2}{5}mr\), we're going to use \(mr^2\) because the Earth is being treated as a point mass here. M is the mass of the object itself. Right? So it's the object that's spinning. It's not the sun. So it's going to be \(5.97 \times 10^{24}\), the same thing. But r is going to be the distance to the center, which is 1.5. So \(1.5 \times 10^{11}\)^2. I got a 24. I got an 11 squared. This is going to be, again, a gigantic number, \(1.34 \times 10^{47}\). This number is, like, a billion times bigger than the other number. Right? So as hard as it would be to stop the Earth from stopping, to get the Earth to stop spinning, it would be way harder. Right? It would be \(10^{10}\) times harder to make the Earth stop going around the Sun. And that's it. So that's it for this one. It's a very typical, classic problem. Hopefully, it makes sense. You should try this out on your own and let's keep going.