Now we have ways to get the torque to be zero either the shape can make it go to zero or for any shape particular orientations if we locked up one of our principal axis and, you know, be, make it equal with the position vector. So what we've just identified is if you look at this equation, if you have a circular orbit at 7,000 kilometers or a circular orbit at 42,000 kilometers Leo versus Geo, you're going to get different gravity gradient torques. Kind of makes sense, right? The further out you go the weaker gravity gets, that we could effect the gravity gradients becomes as well. So as you change your orbits, it actually affects your attitude, because you get different torques acting on it. So is the inverse true? That's what we want to study here. So if you take your spacecraft and it's pointing at the sun and you decide to rotate it with reaction wheels and point now at the Earth, you've taken all this finite mass and moved this distribution around. Since orbits affect attitude, do attitudes affect orbits? If they have different orientations, do I change my orbit type? No? The answer is yes. You do. The more important question is do you care? How big is that term? So we're going to step through this math a little bit quicker than what I did with the other stuff, but it's the same kind of principles, but you'll see it would have to go to second order expansions that are set up first because the first order expansions actually vanish and then you get this expression and I'm going to rewrite it. So now, you get the gravity force much more precisely assuming it's not just a point mass, but it's a finite shape. And now, orientation couples into it. Where is your mass? Do you have more mass lined up close to the Earth versus all the mass lined, you know, lined up evenly away from the earth? That means different gravitational accelerations. You can have different orbits out of it. So let's go through the same math again. It's the same blob, the same force, but instead of doing a body integral of R across the four step, which gave us torques about center of mass we just want the net force acting on this body. So we just treat the body in a roll of all the little D.N. forces that we're going to have. So we just dropped R cross parts. So several terms you'll see will be very similar. That's why I'm kind of going through this quicker. Now we have to do in the end, you plug in the stuff like before. It's very, very similar. Just you don't have that R cross term and then we have to do an expansion of one over R cubed, because that's RC plus little R, but I'm keeping the first order term, which we had before. These two's would cancel, in that case. But there's also a second order term here and here, those end up being critical because, again, first order vanishes. You plug this in, manipulate. Similar arguments as before, we can see these body integrals here, center of mass makes that vanish, center of mass will make this vanish. Then we have other terms. This is one we just expanded earlier with vector cross-product identities. That ended up being the inertia tensor plus something. Relative, parallel to RC which cancel because we cross with RC, but here we're not crossing with stuff so it's going to be a little bit different and that's going to be total mass. This just screams for more vector identities just because, and that's just going to be the R squared. It will give us a polar moment of inertia of the stuff. So we apply, again, the same identity we used earlier, plug this stuff in, right. It's the same process and you start to identify, ah, look this with the minus sign that becomes the inertia tensor. This too, we can relate to inertia tensors and these R squares bodying with R squares are actually polar moments of inertia and that's defined here. And this is something I'm just giving you. You can look this up into a classic statics mechanics book, but the polar moment is the same thing as one half of the trace of the stuff. You can actually prove this to yourself pretty quickly. If you look at the I matrix, the diagonals are all X square plus Y square. Y square plus Z square and X square plus Z square. You start adding them up. You basically get two times X square Y square C square and X square Y square, C square, is always little R squared. So that's where you've just proven it, right? That's the same thing. So I can go from the inertia tensor, get the trace and do that half. I'm also going to start factoring out more RC's here. If I divide both by RC magnitude, I get just these unit direction vectors. It will be a nice convenient form, but I have to factor out an extra RC squared over there. So it's the same math steps. It's just a lot of extra details and let's see what we get. We use inertia tensor definition and in the end this is your more precise gravity force for an orbit. All right. Where if you look at this, and this is, in orbits class you tend to deal with gravity accelerations. R dot is equal to minus as you ever are cube times R. That was the classic orbit equation here and getting it to in terms of force everything is multiplied times M but it's basically the same. And so you got minus MU over RC cubed times RC, right, for that one here. That's the point mass approximation. That's what 50\50 was all about, that orbits class. Then, first order terms dropped and all we have left afterwards are all the second order terms, which every one of them we're able to write somehow is a function of the inertia tensor, which is kind of cool and I factor things out. So, have this unit direction vectors and had extra RC squares that I had to factor out. So now, this is the formulation. So you can see, as you change your attitude, all of a sudden you will have different results and you have a small impact on your orbit. So if you fly, flying this way and then you go this way, you will have a slightly different orbit. But for Earth, man-made spacecraft on Earth. This is a really, really small effect. If you start flying around small asteroids and boulders, this may not be a small effect. Right? And this is stuff you have to actually account for when you do these extra expansions. The classic point mass model is not going to work necessarily. So in this class we're not doing asteroids, so we're going to look at this now quickly and do some dimensional analysis. Inertia's are all mass of the spacecraft times distance squared for every element. Kind of the large, good opera estimate just make it a sphere. Then all the radii, all the mass, is as far out as you can do it. We don't build spacecraft that way, but that would be kind of an upper bound. So it's mass times the radius of the spacecraft squared divided by mass times the radius of the orbit squared, the masses cancel. So only need to look at the ratios of spacecraft squared to radius of orbit squared. Orbits, let's make Leo easy 7,000 kilometers squared at 7 million kilometers that's 10, 49 times 10 to the 12. So it's about 10 to the 13-ish order of meters. Man-made objects. Around one-ish average radius. If look at the mass distribution and everything. So this term is one, plus something, one over 10 to the 13. That's pretty small. You're going to have to account for, you know, Lawrence forces, drag forces, origin pressure forces, out gassing forces probably of the material. There's all these other disturbances that are way bigger than 10 to the minus 13. You know, J2 that you were talking about does 10 to the minus three J four and five and those six is, okay, maybe, there are three, two, three orders of magnitude smaller again. Right?. But still about that level, we're nowhere near 10 to the minus 13. So for Earth Application that's why typically you don't see this, but it is there and with these tools you can develop it. But again, now we're getting into pretty cool missions too around other small bodies, trying to get around boulders and flying stuff and your crafts might be, you know, this might be much more of a significant effect. Asteroids still tend to be basically lumps of stuff. We don't build spacecraft as cannonballs or lumps of stuff. This spacecraft has lots of empty space, right? So the inertia ratios are still going to be big, it might still be quite small, but it might not be negligible either. So it's kind of a cool result people, good prelim questions, you know. Does your attitude affect your orbit? And the answer is yes. Always, Spencer?. Still a little confused as to how like small asteroid, like uneven bodies like that, of the gravity gradient work as more of a factor? Because isn't it. Use a metrical force. Force. It's proportionately a bigger influence, because with this big square bracketed term as basically normalized to point mass approximation of that object is just one. And then how much is this going to be? So you just have to look at, well what's my inertia? And if we're still about size one, meter-ish. You know size spacecraft and this isn't 7 million meters, but now all of a sudden 50 meters, because you're flying around the five meter boulder, you what I mean? It's still going to be small, but not 10 to the minus 13 small, right? And then, there's bigger binary systems and other kind of stuff people are studying. It definitely makes an effect. There's some binary asteroid systems, where you have, you know, Shear's done a lot of work on this stuff. He's got a whole book on this. But then, it affects it more. That makes sense. So good. So we've covered now how to derive gravity gradients, we need that one approximation that the spacecraft is small compared to the orbit. We got the nice vectorial way to write it and in some different coordinate frame versions. This is the force version. This is more FY, but it's just so you can see the answer.