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Gravity Gradients.

Â This is kind of the last big thing we're doing in dynamics.

Â We've looked at single rigid body, we've looked that now, you know, torque free,

Â with a little torque, some homework, you know, a little torque.

Â They move the dual spinner just in the rigid body.

Â That's a way to passively stabilize, there's no active feedback control,

Â you have to make sure you designed it with the right spin rate, so the system is stable.

Â The other way we can passively stabilize the system is with gravity gradients, right?

Â This is all about, if you look at this pen,

Â this lower part of the pen is closer to the Earth than the upper part of the pen

Â and the gravitational acceleration depends on one over R squared.

Â So things closer to the center of the Earth weigh more.

Â Even though it's the same mass here and here, they have a stronger force

Â and that gives you the gradient that acts on it, right?

Â There's a gradient across the body and how strong the gravity forces are.

Â So, now we want to look at these gradients on general objects

Â and figure out what are the forces, what are the torques,

Â what are the equilibrium's, and what are the stabilities.

Â Again, the same classic things that we did with the dual-spinner,

Â getting these torques though is a little bit more interesting.

Â So, gravity gradients tend to be tall and slender.

Â This is the typical gravity gradient kind of this configuration.

Â Why?

Â Because if this thing is the normal condition,

Â it turns out here the gravity gradients still give you zero torque.

Â They all act, all the gravity forces act through the center of mass,

Â so there's no torque about the center of mass.

Â But if I twist it, and this one gets pulled stronger than this one,

Â you get a restoring force, which actually helps stabilize it.

Â This will end up being another gravity gradient where everything has essentially

Â the same distance from the center of the Earth.

Â So you'll get the same force, and they all just balance out, there's no torque.

Â But if I twist this one slightly,

Â this one now weighs more than this one and you get a destabilizing effect, right?

Â So, as with the dual-spinner, sometimes equilibrium's can be stable

Â and sometimes unstable and we like to find this way this for anti-symmetric.

Â What we want to derive is for general inertias at the space shuttle.

Â It has a very particular orientation when it was flying in a gravity gradient orientation.

Â Why?

Â Where does that come from?

Â So, it's the tidal forces.

Â So, we're going to get to mass and flying blobs again.

Â So, we're going to say, this is a general object.

Â It's going to be a rigid body in this case, we're not flying Jello,

Â we're just flying rigid spacecraft.

Â But we have our typical R,

Â that's the inertial position vector of your little infinitesimal mass element, every bolt,

Â every screw, every panel, right, we're accounting for.

Â And it's position, inertially, can be written as the position of the center of mass

Â of the spacecraft,plus the bolts position relative

Â to the center mass of the spacecraft, right?

Â Just what we did earlier.

Â Now, the gravity acting on this differential element,

Â if you look at Newton's Universal Law of Gravity you've got that G,

Â the mass of the Earth, the mass of the elements, right?

Â G times N one, N two, but N one is Earth in and N two is DM divided by distance squared.

Â That would be this distance times a direction and it has to be in the minus R direction.

Â I probably should have drawn this arrow, this down here, that would have made more sense.

Â Anyway, next time.

Â But that's the mathematics, that's just Newton's Gravity Law acting on a DM particle.

Â Now, we have these classic definitions.

Â So what we're going to do now is we're going to first look at torques,

Â gravity gradient torques, about the center of mass.

Â So, I have this point here, I need to get the moment arm crossed with this force.

Â That's how we did torque, right,

Â position where we applied a force crossed with the force itself,

Â about the moment, about the point C.

Â So, this would be easy.

Â That's little R crossed with this DF that we have.

Â And DF is repeated here.

Â So, we start to plug it in and substitute stuff.

Â The first step, we put this in here and R is RC plus little R.

Â Quickly you'll see while there's an R cross RC, and a little R cross little R.

Â What does that become?

Â Zero, right?

Â And we love zero by now.

Â Zero is really nice.

Â Zero is our friend.

Â So, immediately this term is going to drop out and it won't contribute to the torque.

Â So, we like that.

Â We end up with RC, we can take outside the integral,

Â as we argued earlier and, you know, in blobs and space.

Â For the system every point in that system has the same center of mass location.

Â So it's not impacted by this body integral, But that leaves you with this.

Â Now, R does depend, right?

Â R is RC plus little R, so it's got little r imbedded.

Â You can't move that R outside.

Â This is where you end up with.

Â Now, this is where all up to now.

Â This is rigorous, and no approximation.

Â What we're going to do now is this is R cubed.

Â That's where all the complications come in

Â and we're going to do a first order approximation.

Â R cubed is bloody close to RC cubed, right?

Â 7,000 kilometers from the center of the Earth,

Â and then we're looking at a half a meter distance.

Â That's really not going to change at 7,000 kilometers too much, right?

Â As can be very, very close, which motivates,

Â let's take a first order approximation of this and you get a really good analytical answer

Â to these gravity gradient torques.

Â So, we have to do a linearization,

Â essentially and I'm showing here how you can take this, plug in this stuff.

Â I'm doing a binomial expansion, factor things out,

Â and in the end you're dropping off,

Â you have to do this linearization of this term and you end up with this.

Â This is your zeroth order term, one over RC cube,

Â which kind of makes sense, minus this times this.

Â This is your first order term, the first order here is in terms of R, little R, right?

Â You may have done Taylor Series Expansion in terms of X's and Y's and Z's.

Â I expect you not to do a Taylor Series Expansion terms of vectorial quantities.

Â So, I'm going to do this once with you here.

Â But this is definitely a skill on an exam I would expect you to be able to do.

Â Okay.

Â Let me get my notes up.

Â Quickly.

Â Yeah, that will work.

Â So, what we're trying to do is, we have this one term.

Â The thing I'm trying to expand about

Â is I'm treating little R as the small quantity, right?

Â RC is the big quantity, that's 7,000 kilometers to your left.

Â But then, I only have little wiggles of plus\minus a fraction

Â of a meter across the spacecraft,right?

Â So little R is a small thing,

Â and the function I'm trying to approximate is one over R cubed,

Â which is the same thing as R to the minus third.

Â R is just a scalar.

Â R is always defined that way.

Â So, now, if you look at a Taylor Series Expansion, if you have F of X,

Â and we're doing an expansion about XR, just X is a scalar in this case,

Â you would say Taylor says, 'Okay, that is the F at the reference point,

Â plus one over one, vectorial times, the partial of F with respect to this state'.

Â And then this partial is evaluated at the reference times Delta X.

Â And Delta X would simply be X minus XR, right?

Â And then you can go second order and higher.

Â We're not going to go second order here, right?

Â That's the classic Taylor Series Expansion.

Â Let's apply this now here,

Â where this function depends on a vector not a scalar.

Â Now, the function itself is a scalar function that makes life a little bit easier.

Â So, we have to figure out these terms.

Â So, applying this here we're going to say F

Â of R is approximated as F at the reference which is F,

Â at the reference is the center of mass, that means R is equal to zero.

Â Plus, one over one factorial, which is just one over one which is one,

Â times the partial of F with respect to R evaluated at the reference,

Â that's R is equal to zero.

Â That's the center of mass location.

Â Everything is about the center of mass now times

Â instead of this is the vector dot product with the small quantity,

Â which in this case is just R.

Â R is the small quantity.

Â So, we're doing the expansion about R is equal to zero, essentially.

Â This part, if R is equal to zero then R magnitude is equal to RC magnitude,

Â and that's just going to be RC to the minus third, right?

Â That one's trivial.

Â Let's look at this.

Â This part we have to figure out first is partial derivative of it.

Â So, if F is R to the minus, third the partial of F with respect to little R,

Â chain rule, you know, that's going to be minus three times R to the minus fourth times

Â the partial of the radius with respect to this little R vector.

Â So, good?

Â We get there.

Â Now, we need to find this quantity.

Â So, to do that relationship I'm just going to go back to this stuff.

Â R squared is really this vector dotted with itself,

Â which is the same thing as RC plus little R dotted with RC plus little R, right?

Â The magnitude squared of a vector is just a dot product of the vector itself,

Â and the vector is written in terms of a song.

Â So, we've got that, good.

Â Now, I have the scalars here, this is a scalar at the end,

Â but it's a scalar in terms of these vectorial quantities and some vector math.

Â Now, taking the derivative of this these, these partials,

Â hopefully you remember how to take partial derivatives, on the left hand side.

Â I'm just going to get two times R, times the partial of R with respect to little R.

Â On the right hand side, if I carry this out,

Â I need to take the partial derivative of, let's see.

Â Here you're going to have RC dotted with RC,

Â plus two times RC dotted with an R, plus little R dotted with itself.

Â This part doesn't depend on little R, so that partial is going to drop out.

Â Here two RC times R, the partial of a vector with a with respect to itself,

Â which is going to give you identity operator.

Â So, like the partial of X with respect to X just gives you one, right?

Â So, in this case you can end up with two times RC

Â from this partial and then here this dotted with itself.

Â Like we talked about earlier, you know, when we had R transpose.

Â R take the time derivative,

Â you get to two times R transposed R dot or taking partial derivatives.

Â This just gives you two times the R vector, in that case.

Â Yup.

Â So, if you're fuzzy on this, practice with this, approve these identities to do so,

Â it just takes one or two steps.

Â Now, I'm looking for this partial, so I can... I'm just going to rewrite this.

Â These twos all cancel.

Â So the partial of R with respect to little R is just one over R times RC plus little R.

Â All right?

Â That's what we have.

Â Now, we can plug this back into here and we'll see what we're going to get.

Â We have minus three R to the four, minus fourth is another one,

Â so R to the minus fifth, times RC plus little R.

Â You go, okay, getting close.

Â Now, we have to evaluate this sensitivity here at the reference where R is equal to zero.

Â So, if I compute that partial of this with respect to little R at R is equal to zero,

Â then this R just becomes RC, and this little R just becomes zero.

Â And what you will end up with is minus three over RC to the fifth,

Â times RC vector, essentially.

Â And now, we're just about there.

Â So, the first order expansion of this function is simply RC to the minus third,

Â plus this term dotted with the little R vector.

Â And if I go look at my final result here,

Â that's exactly what I had, one over RC to the third,

Â minus three RC over RC to the fifth because three times two is five,

Â three plus two is five, dotted with that first order expansion.

Â Right?

Â So, we've gone through this reasonably quickly.

Â This is something it's not going to sink in unless you do it yourself,

Â and I think in one of the hallmarks, I'm asking you to do this.

Â Derive this, make sure you can do these different binomial expansions

Â and then linearize this or just start from front, which I showed you with that,

Â you know, make sure you know how to take partials of, you know, typical gradient.

Â This is your sophomore level calculus or freshman level calculus,

Â whenever you took Calc II that kind of stuff like, right?

Â Partials of scales with respect to vectors, just gradients that we have to get.

Â That's all we have to compute because now we can plug this.

Â Now we're making this assumption that this R,

Â this spacecraft is small compared to the orbit.

Â This is not the Star Wars, Death Star orbiting Endor or something.

Â There this approximation wouldn't hold.

Â For current man-made object it's a pretty good approximation.

Â So, if you plug that in, this is where we had that R over this other stuff,

Â this is what that stuff combines.

Â I've taken some of the constants outside.

Â You carry out this multiplication, you will see the body integral of RDM,

Â which we know from the center of mass definition goes to zero.

Â We still have zero one term less to worry about and that leaves you with this part.

Â Here, this one doesn't quite look convenient,

Â but in essence you can see it's mass times distance squared.

Â At this point, hopefully,

Â your spidey sense is tingling and this looks like something

Â that could be related to inertia.

Â And so, we have to re-manipulate these equations, again, to get to that form.

Â So, if you do that we have lots of different vector identities,

Â one of them is double cross product and if I solve for minus this term,

Â you just bring this over, this over to the other side, and then you got it,

Â you can relate that, this is what's inside that integral term, precisely.

Â So, A is equal to little R, C is equal to RC, B is equal to little R.

Â You apply the vector identity, you end up with this stuff, right?

Â So, doing that though, now you can see immediately we've got little R, till D,

Â minus little R till the little R till D, that will go back into our body integral.

Â Hopefully, you already at this part recognize that's going to be the inertia term.

Â We have something else we need to identify.

Â So we plug that in.

Â You get to here and, so the only assumption,

Â so far, still is that little R is small compared to RC,

Â spacecraft is small compared to the radius of the orbit.

Â That first term, you can factor out RC to the left, which is done already,

Â you can factor in RC to the right.

Â Because it doesn't matter on the body integral, that means you move the DM inside,

Â and then you end up with the classic definition.

Â The other one is RC here, you can take that outside,

Â and then you just have the body integral of little R dotted with itself,

Â which is just R squared, which we can do.

Â So this term is what we can recognize as the inertia tensor, right?

Â That appears all these different places.

Â This one here is actually a polar moment of inertia.

Â But here we don't care about it, because the RC's you factored out,

Â give you an RC crossed RC, which of course, again, goes to zero, which we like, right?

Â So that second term vanishes completely with this mathematics and now you end up with,

Â I use in this inertia tensor definition.

Â We end up with this expression of, this is basically the linear, you know,

Â this is the first order.

Â I shouldn't say linearized, we'll do that differently later.

Â This is the first order approximation of the gravity gradient torque, right?

Â When we expanded one over R cubed, we only kept first order term.

Â If you want higher order terms, you'd have to include those in expansions,

Â and then it would be this plus some other stuff.

Â That would happen.

Â Now, when we write this, have I specified a particular coordinate frame

Â that you have to express everything in?

Â Horace?.

Â No.

Â No, it's just vectors, vectors, even the tensor stuff, right?

Â We know when we evaluate this we have to pick probably somebody fixed frame.

Â That's how we do it mathematically.

Â But generally we can write the tensor just as a coordinate frame independent way.

Â If you have it in the B frame, and you need in the C frame,

Â we know we have to pre- and post- multiply

Â with the DCM to do a coordinate transformation, right?

Â But we treat the inertia tensor just like a body, a position vector or velocity vector.

Â So, we can write them in a very agnostic ways, which is nice.

Â It just means when you evaluate this make sure you have your orbit in the same frame

Â as your inertial tensor.

Â What coordinate system do we typically use for orbits?

Â Mandar?

Â Inertial?

Â You could do inertial, right?

Â You might even use an orbit frame, we'll see later on, a rotating orbit frame, right?

Â You don't typically use a body frame.

Â That's probably the least you scream for an orbit

Â because then it couples everything in crazy ways, right?

Â We want something different.

Â But the inertia tensor, we tend to express that one in the body frame.

Â So, this is just an equation where, right up front,

Â I can tell that you probably are going to be given this information, in real life,

Â using different frames, and that's fine.

Â So, implied in here is maybe a DCM rotation that you have to do.

Â You know, the tensor of the vectors are mapped from one frame to another, that's all.

Â So, this is the most compact way I can write it.

Â But again, implied might be some important transformation when you actually evaluate it.

Â So, looking at this, let's just jump ahead quickly.

Â Can anybody tell me for what type of inertia tensor will LG go to zero?

Â T-Bo?

Â It's diagonal, enough, probably not, just a identity scale.

Â Yes, diagonals running off.

Â If you have distinct three different inertias, this is not going to go to zero,

Â because you actually scaled this vector in a way

Â that it's no longer co-linear with itself, right?

Â But if it's an identity times a scalar,

Â so what kind of shapes give us those kinds of tensors?

Â Sphere.

Â Sphere and?

Â Cubed.

Â Cube, right?

Â Keep that because we'll see that result again in other mathematics as well.

Â So, here just in a vectorial form you can see with the cross-product.

Â If it's a cross-product times itself, whatever shape makes that happen,

Â that would make the gravity gradient torques,

Â actually, to be zeroregardless of attitude.

Â