Next thing you want to do is now that we have a model of gravity gradient torque, we want to talk about equilibrium. And in this application, I have the word relative equilibrium state. We talked about equilibria simply meaning, if we have x dot equal to f as a dynamical system, equilibrium means at what states x is x dot going to be zero? All right, that's it. And then, those are fixed points in your states phase or flows phase, whatever you might want to call it. A relative equilibrium means we're not looking for stuff where the spacecraft rotation relative to the inertial is constant. But actually, it's relative to a rotating frame. And what you typically have in orbits is, we care about the motion of a spacecraft as seen by a rotating orbit frame. We have a frame that line, one axis lines up with orbit radial, the old frame is an old radial, alright? One is the long track, one is a cross track stuff. That's a rotating orbit frame. And we often care, if I'm flying to space station like that, always pointing at the Earth, I'm actually interested in the motion of the space station, relative to this orbit frame. How does gravity, torques and everything and gyroscopics effect that? So we're going to look at relative equilibrias here. The equations of motion is the same. It's a rigid body, we derived it with b relative to n, still, right? And then this is the gravity gradient torque which we just derive. We can now plug that in in different forms and do stuff with it. And omega BN is the angular motion of B relative to the orbit frame plus the orbit frame motion. In this analysis, we're assuming a circular orbit, which means omega o relative to n is simply a constant, it's your mean orbit rate about your cross axis or orbit normal axis momentum axis. Whatever you call it, right. So that's just a constant orbit rate that we have. We're rotating at so many degrees per second. That's a nominal rotating frame motion, but what we also have with equilibria is, we're looking for conditions where omega B relative to O go to zero. That means, if I have my rotating frame that I'm flying, how can I put my craft in here, such that gyroscopics and torques, and everything don't cause any change in omega BO. So we want omega BO time derivatives to be zero as well. And omega BO the motion should be zero. This must not have a battery. Okay. So that's the overarching goal, right? So we have the orbit frame o1, 2, and 3. And now, we have b1, 2, and 3 drawn in. And right now, it's in a general orientations for what attitudes. So we have two parts now, for what attitudes will my torque go to zero? That's one thing that can cause non-zero accelerations here. And the other question is, for what attitude will my gyroscopic, relative to the orbit frame, that term has to vanish as well. And so, we're going to look at both of them individually. For the gravity gradient torque to be 0, we said we could have certain shapes, we could make it a cube, we could make it a sphere. Then, as always the case, as kind of the trivial case. So let's put that one aside. Now we're going to look at the more general inertia tensor case, and say hey, with kind of a stage craft, how do I orient it? And what we found was, as long as you line up one of your three principle axes with the orbit radius, that whole torque vector goes to zero. All right. So if I do that without loss in generality here, I'm going to line up B3 with O3. It's just easier to line up. It could be really be any of them. And if that's lined up, if you look at the picture, what happened, well, actually, that comes in the next step. What happens then is, in the O frame, my inertia tensor of the space craft, and as seen by general orbit frame, depending how the attitude, that's going to be a general three by three matrix. The orbit frame typically isn't a principal frame of your spacecraft. But because of this condition, we know that O3 must be a principle frame. That means the off diagonal parts with the third axis have to all be 0. But I still don't know what the rest of the spacecraft does. This means the three axis has to line up with my gravity vector but there is an infinity of rotations, right, that you could rotate about that vector. And if you have your orbit frame, if it's tilted like this, 45 degrees, you would have a general two by two inertia tensor. It's only if it's lined up, like you were talking about, David, perfectly with the orbit frame, then, it would be perfectly diagonal. So, right now, this doesn't require that. So we get this block diagonal form. And this is now how they illustrated that, O3 is equal to B3, but there's B1 and B2. Just because my third axis is lined up with gravity my one and two axes, I don't know yet what they have to do, right? So, this condition simply makes my gravity gradient torque go to 0. So now, we also want to look at this term, because that can cause angular accelerations. So if we make a B/0 by definition of being a relative equilibria, we want to stay there, we don't want to tumble and move relative to this orbit frame. The orbit motion itself is this constant, O times O2. All these accelerations have to be 0 to stay at such a condition. Now, we can plug these conditions into here. Omega B/N is Omega B/O plus Omega O/N. Classic stuff that you've done. And so, we can combine this at the equilibrium condition, plug it into that gyroscopic term, and if you do this, in one of your homeworks, you actually go through this math yourself, it goes really quickly. It's really this utility matrix times the diagonal inertia, times the 3-by-1, not very hard. You do this because this one is only 0 n 0 in the O frame, a lot of things vanish. And in the end, you end up with this vector. That's your gyroscopic acceleration that you get from a body that had zero initial conditions. So, in all cases here, the spacecraft is not twisting and gyrating relative to the orbit frame. We're enforcing that at the equilibrium we expect this to be zero. But depending on the attitude you're kicking in, the gyroscopics might give you additional accelerations, which means it's not an equilibrium, right? You would now have a non-zero dots, omega dots, in the system, so how can we make this to zero? One option is >> [COUGH] >> could you make N go zero? Matt, what do you think? What does N mean again? >> The average rate. >> Right. If N goes to zero, where's your spacecraft relative to the earth? >> Going straight in. >> No. As you get closer to the earth, what happens to that? Yeah, it goes faster and faster. So at the center, you've probably reached an infinite speed. Slight challenges with that. Except for Hollywood, Hollywood [INAUDIBLE] movies [INAUDIBLE] travel. Yeah that's not [INAUDIBLE] Hollywood makes it work, but for the rest of us not so lucky. If your n goes to 0, the further out your orbits are, the slower your orbital rates go. And to make it go to 0 you have to be infinitely far out. So you're really not in orbit anymore. So making this 0 is really not a practical thing. The other thing we can do is, we can say this term can be 0. Now, what does that mean? I want to go back to this block diagonal form that we had. Keep in mind, this is the inertia tensor, we typically always write it in the body frame. The key thing to realize in this and the following analysis is, this is always written relative to the orbit frame. That means, I3 is the inertia about the O3 part. I1 is the spacecraft inertia about the O1 axis, and I22's about the O2. And, if you have certain angles, you may have some cross coupling inertia, right? Now, we're making these terms go to 0. What does that mean now, David? About, to be at a gravity gradient relatively equilibria? What orientation must we have? >> That the body frames align with the other frame. >> Yep. That means, basically mathematically, the result we get. This stuff only happens if the inertia tends to description, as seen by that rotating orbit frame, and it's a constantly rotating orbit frame here. It has to be a diagonal form, that's the only way this is going to work. So your earlier statement is right. For an equilibria, we do need all three principal axes to actually line up. There's still finite permutations, you could have B1 line up with +03, -03, +02, -02, +01, -01, right. And then different permutations to keep it as a right handed frame. But there's more than just one answer. That's the same as saying, look at the craft, it could be this way, it could be this way. If I'm flying this way, I could flip it this way. I just can't have it fly at a skewed angle, right? The axes have to line up with a long track, both for vertical and cross track. That's what this condition means, and that's what we found here. So that's the result, this is not the requirement for gravity grading torque to be zero. This is the requirement for these relative equilibria. Yes sir? >> So congruent that [INAUDIBLE] cylindrical spacecraft [INAUDIBLE]? >> Yes. You could. You could. Very good. All right. So now, the different combinations of this, if you have a cylindrical spacecraft, you line up one of these axes, the other two axes, there's an ambiguity on what is the principle is saying. So you could make it cylindrical, if your mission allows it, and you want to exploit gravity gradients, that's a great thing. Often, structures people have issues with solar panels and sizes, and they don't want to make it cylindrical and darn structures people. But, from the dynamics point of view, yes, that would actually make it easier. But it also actually loses any, you will look later, we'll see three axis controls that happens. If you make it cylindrical, a downside is, if I have a three dimensional shape, I can look at it. And if anything is perturbed around any of the axes, it will be stable. That means these angular departures don't just grow. If I given a little bit of twist there's something gyroscopically bringing that again. If you make it a cylinder, we will lose that. because then all of a sudden, because of the symmetry there never would be a restoring force about that access.
