0:05

Next thing you want to do is now that we have a model of gravity gradient torque,

Â we want to talk about equilibrium.

Â And in this application, I have the word relative equilibrium state.

Â We talked about equilibria simply meaning, if we have x dot equal to f as a dynamical

Â system, equilibrium means at what states x is x dot going to be zero?

Â All right, that's it.

Â And then, those are fixed points in your states phase or flows phase,

Â whatever you might want to call it.

Â A relative equilibrium means we're not looking for

Â stuff where the spacecraft rotation relative to the inertial is constant.

Â But actually, it's relative to a rotating frame.

Â And what you typically have in orbits is,

Â we care about the motion of a spacecraft as seen by a rotating orbit frame.

Â We have a frame that line, one axis lines up with orbit radial,

Â the old frame is an old radial, alright?

Â One is the long track, one is a cross track stuff.

Â That's a rotating orbit frame.

Â And we often care,

Â if I'm flying to space station like that, always pointing at the Earth, I'm actually

Â interested in the motion of the space station, relative to this orbit frame.

Â How does gravity, torques and everything and gyroscopics effect that?

Â So we're going to look at relative equilibrias here.

Â The equations of motion is the same.

Â It's a rigid body, we derived it with b relative to n, still, right?

Â And then this is the gravity gradient torque which we just derive.

Â We can now plug that in in different forms and do stuff with it.

Â And omega BN is the angular motion of

Â B relative to the orbit frame plus the orbit frame motion.

Â In this analysis, we're assuming a circular orbit,

Â which means omega o relative to n is simply a constant,

Â it's your mean orbit rate about your cross axis or orbit normal axis momentum axis.

Â Whatever you call it, right.

Â So that's just a constant orbit rate that we have.

Â We're rotating at so many degrees per second.

Â That's a nominal rotating frame motion, but what we also have with equilibria is,

Â we're looking for conditions where omega B relative to O go to zero.

Â That means, if I have my rotating frame that I'm flying,

Â how can I put my craft in here, such that gyroscopics and torques, and

Â everything don't cause any change in omega BO.

Â So we want omega BO time derivatives to be zero as well.

Â And omega BO the motion should be zero.

Â This must not have a battery.

Â Okay.

Â So that's the overarching goal, right?

Â So we have the orbit frame o1, 2, and 3.

Â And now, we have b1, 2, and 3 drawn in.

Â And right now, it's in a general orientations for what attitudes.

Â 3:01

And so, we're going to look at both of them individually.

Â For the gravity gradient torque to be 0, we said we could have certain shapes,

Â we could make it a cube, we could make it a sphere.

Â Then, as always the case, as kind of the trivial case.

Â So let's put that one aside.

Â Now we're going to look at the more general inertia tensor case, and

Â say hey, with kind of a stage craft, how do I orient it?

Â And what we found was, as long as you line up one of your three principle axes

Â with the orbit radius, that whole torque vector goes to zero.

Â All right.

Â So if I do that without loss in generality here, I'm going to line up B3 with O3.

Â It's just easier to line up.

Â It could be really be any of them.

Â And if that's lined up, if you look at the picture, what happened, well,

Â actually, that comes in the next step.

Â What happens then is, in the O frame, my inertia tensor of the space craft, and

Â as seen by general orbit frame, depending how the attitude,

Â that's going to be a general three by three matrix.

Â The orbit frame typically isn't a principal frame of your spacecraft.

Â 4:03

But because of this condition, we know that O3 must be a principle frame.

Â That means the off diagonal parts with the third axis have to all be 0.

Â But I still don't know what the rest of the spacecraft does.

Â This means the three axis has to line up with my gravity vector but

Â there is an infinity of rotations, right, that you could rotate about that vector.

Â And if you have your orbit frame, if it's tilted like this, 45 degrees,

Â you would have a general two by two inertia tensor.

Â It's only if it's lined up, like you were talking about, David,

Â perfectly with the orbit frame, then, it would be perfectly diagonal.

Â So, right now, this doesn't require that.

Â 5:20

The orbit motion itself is this constant, O times O2.

Â All these accelerations have to be 0 to stay at such a condition.

Â Now, we can plug these conditions into here.

Â Omega B/N is Omega B/O plus Omega O/N.

Â Classic stuff that you've done.

Â And so, we can combine this at the equilibrium condition,

Â plug it into that gyroscopic term, and if you do this, in one of your homeworks,

Â you actually go through this math yourself, it goes really quickly.

Â It's really this utility matrix times the diagonal inertia,

Â times the 3-by-1, not very hard.

Â You do this because this one is only 0 n 0 in the O frame, a lot of things vanish.

Â And in the end, you end up with this vector.

Â That's your gyroscopic acceleration that you get from a body

Â 6:12

that had zero initial conditions.

Â So, in all cases here, the spacecraft is not twisting and

Â gyrating relative to the orbit frame.

Â We're enforcing that at the equilibrium we expect this to be zero.

Â But depending on the attitude you're kicking in, the gyroscopics might

Â give you additional accelerations, which means it's not an equilibrium, right?

Â You would now have a non-zero dots, omega dots, in the system, so

Â how can we make this to zero?

Â 6:47

What does N mean again?

Â >> The average rate.

Â >> Right.

Â If N goes to zero, where's your spacecraft relative to the earth?

Â >> Going straight in.

Â >> No.

Â As you get closer to the earth, what happens to that?

Â Yeah, it goes faster and faster.

Â So at the center, you've probably reached an infinite speed.

Â Slight challenges with that.

Â Except for Hollywood, Hollywood [INAUDIBLE] movies [INAUDIBLE] travel.

Â Yeah that's not [INAUDIBLE] Hollywood makes it work, but

Â for the rest of us not so lucky.

Â 7:40

Keep in mind, this is the inertia tensor,

Â we typically always write it in the body frame.

Â The key thing to realize in this and

Â the following analysis is, this is always written relative to the orbit frame.

Â That means, I3 is the inertia about the O3 part.

Â I1 is the spacecraft inertia about the O1 axis, and I22's about the O2.

Â And, if you have certain angles, you may have some cross coupling inertia, right?

Â Now, we're making these terms go to 0.

Â What does that mean now, David?

Â About, to be at a gravity gradient relatively equilibria?

Â What orientation must we have?

Â 8:18

>> That the body frames align with the other frame.

Â >> Yep.

Â That means, basically mathematically, the result we get.

Â This stuff only happens if the inertia tends to description, as seen by

Â that rotating orbit frame, and it's a constantly rotating orbit frame here.

Â It has to be a diagonal form, that's the only way this is going to work.

Â So your earlier statement is right.

Â For an equilibria, we do need all three principal axes to actually line up.

Â There's still finite permutations, you could have B1 line up with +03,

Â -03, +02, -02, +01, -01, right.

Â And then different permutations to keep it as a right handed frame.

Â But there's more than just one answer.

Â That's the same as saying, look at the craft, it could be this way,

Â it could be this way.

Â If I'm flying this way, I could flip it this way.

Â I just can't have it fly at a skewed angle, right?

Â The axes have to line up with a long track, both for vertical and cross track.

Â 9:19

So that's the result, this is not the requirement for

Â gravity grading torque to be zero.

Â This is the requirement for these relative equilibria.

Â Yes sir?

Â >> So congruent that [INAUDIBLE]

Â cylindrical spacecraft [INAUDIBLE]?

Â >> Yes. You could.

Â You could. Very good.

Â All right.

Â So now, the different combinations of this,

Â if you have a cylindrical spacecraft, you line up one of these axes,

Â the other two axes, there's an ambiguity on what is the principle is saying.

Â So you could make it cylindrical, if your mission allows it, and

Â you want to exploit gravity gradients, that's a great thing.

Â Often, structures people have issues with solar panels and sizes, and

Â they don't want to make it cylindrical and darn structures people.

Â But, from the dynamics point of view, yes, that would actually make it easier.

Â But it also actually loses any, you will look later,

Â we'll see three axis controls that happens.

Â If you make it cylindrical, a downside is,

Â if I have a three dimensional shape, I can look at it.

Â And if anything is perturbed around any of the axes, it will be stable.

Â That means these angular departures don't just grow.

Â If I given a little bit of twist there's something gyroscopically

Â bringing that again.

Â If you make it a cylinder, we will lose that.

Â