1: Gravity Gradient Torque Development

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Del curso dictado por University of Colorado Boulder

Kinetics: Studying Spacecraft Motion

19 calificaciones

University of Colorado Boulder

Kinetics: Studying Spacecraft Motion

19 calificaciones

Curso 2 de 4 en SpecializationSpacecraft Dynamics and Control

As they tumble through space, objects like spacecraft move in dynamical ways. Understanding and predicting the equations that represent that motion is critical to the safety and efficacy of spacecraft mission development. Kinetics: Modeling the Motions of Spacecraft trains your skills in topics like rigid body angular momentum and kinetic energy expression shown in a coordinate frame agnostic manner, single and dual rigid body systems tumbling without the forces of external torque, how differential gravity across a rigid body is approximated to the first order to study disturbances in both the attitude and orbital motion, and how these systems change when general momentum exchange devices are introduced. After this course, you will be able to... *Derive from basic angular momentum formulation the rotational equations of motion and predict and determine torque-free motion equilibria and associated stabilities * Develop equations of motion for a rigid body with multiple spinning components and derive and apply the gravity gradient torque * Apply the static stability conditions of a dual-spinner configuration and predict changes as momentum exchange devices are introduced * Derive equations of motion for systems in which various momentum exchange devices are present Please note: this is an advanced course, best suited for working engineers or students with college-level knowledge in mathematics and physics.

De la lección

Gravity Gradients

The differential gravity across a rigid body is approximated to the first order to study how it disturbs both the attitude and orbital motion. The gravity gradient relative equilibria conditions are derived, whose stability is analyzed through linearization.

Module 3 Introduction1:17

1: Gravity Gradient Torque Development19:16

1.1: Gravity Gradient Torque in Body Frame7:45

1.2: Gravity Gradient Net Spacecraft Force9:45

2: Gravity Gradient Relative Equilibria Orientations10:45

3: Gravity Gradient Linear Stability about Equilibria22:40

Extra Example: Gravity Gradient Polar Pear Mission5:51

Conoce a los instructores

Hanspeter Schaub
Glenn L. Murphy Chair of Engineering, Professor
Department of Aerospace Engineering Sciences

Gravity Gradients.

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

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

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

They move the dual spinner just in the rigid body.

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

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

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

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

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

and the gravitational acceleration depends on one over R squared.

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

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

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

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

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

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

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

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

getting these torques though is a little bit more interesting.

So, gravity gradients tend to be tall and slender.

This is the typical gravity gradient kind of this configuration.

Why?

Because if this thing is the normal condition,

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

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

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

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

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

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

the same distance from the center of the Earth.

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

But if I twist this one slightly,

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

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

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

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

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

Why?

Where does that come from?

So, it's the tidal forces.

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

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

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

we're just flying rigid spacecraft.

But we have our typical R,

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

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

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

of the spacecraft,plus the bolts position relative

to the center mass of the spacecraft, right?

Just what we did earlier.

Now, the gravity acting on this differential element,

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

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

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

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

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

Anyway, next time.

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

Now, we have these classic definitions.

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

gravity gradient torques, about the center of mass.

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

That's how we did torque, right,

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

about the moment, about the point C.

So, this would be easy.

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

And DF is repeated here.

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

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

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

What does that become?

Zero, right?

And we love zero by now.

Zero is really nice.

Zero is our friend.

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

So, we like that.

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

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

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

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

Now, R does depend, right?

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

You can't move that R outside.

This is where you end up with.

Now, this is where all up to now.

This is rigorous, and no approximation.

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

That's where all the complications come in

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

R cubed is bloody close to RC cubed, right?

7,000 kilometers from the center of the Earth,

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

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

As can be very, very close, which motivates,

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

to these gravity gradient torques.

So, we have to do a linearization,

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

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

and in the end you're dropping off,

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

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

which kind of makes sense, minus this times this.

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

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

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

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

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

Okay.

Let me get my notes up.

Quickly.

Yeah, that will work.

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

The thing I'm trying to expand about

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

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

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

of a meter across the spacecraft,right?

So little R is a small thing,

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

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

R is just a scalar.

R is always defined that way.

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

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

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

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

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

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

And then you can go second order and higher.

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

That's the classic Taylor Series Expansion.

Let's apply this now here,

where this function depends on a vector not a scalar.

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

So, we have to figure out these terms.

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

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

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

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

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

that's R is equal to zero.

That's the center of mass location.

Everything is about the center of mass now times

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

which in this case is just R.

R is the small quantity.

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

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

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

That one's trivial.

Let's look at this.

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

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

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

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

So, good?

We get there.

Now, we need to find this quantity.

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

R squared is really this vector dotted with itself,

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

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

and the vector is written in terms of a song.

So, we've got that, good.

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

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

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

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

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

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

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

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

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

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

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

which is going to give you identity operator.

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

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

from this partial and then here this dotted with itself.

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

R take the time derivative,

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

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

Yup.

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

it just takes one or two steps.

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

These twos all cancel.

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

All right?

That's what we have.

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

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

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

You go, okay, getting close.

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

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

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

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

times RC vector, essentially.

And now, we're just about there.

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

plus this term dotted with the little R vector.

And if I go look at my final result here,

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

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

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

Right?

So, we've gone through this reasonably quickly.

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

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

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

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

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

This is your sophomore level calculus or freshman level calculus,

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

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

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

Now we're making this assumption that this R,

this spacecraft is small compared to the orbit.

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

There this approximation wouldn't hold.

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

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

this is what that stuff combines.

I've taken some of the constants outside.

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

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

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

Here, this one doesn't quite look convenient,

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

At this point, hopefully,

your spidey sense is tingling and this looks like something

that could be related to inertia.

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

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

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

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

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

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

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

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

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

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

We have something else we need to identify.

So we plug that in.

You get to here and, so the only assumption,

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

spacecraft is small compared to the radius of the orbit.

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

you can factor in RC to the right.

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

and then you end up with the classic definition.

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

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

which is just R squared, which we can do.

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

That appears all these different places.

This one here is actually a polar moment of inertia.

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

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

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

I use in this inertia tensor definition.

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

this is the first order.

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

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

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

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

and then it would be this plus some other stuff.

That would happen.

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

that you have to express everything in?

Horace?.

No.

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

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

That's how we do it mathematically.

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

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

we know we have to pre- and post- multiply

with the DCM to do a coordinate transformation, right?

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

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

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

as your inertial tensor.

What coordinate system do we typically use for orbits?

Mandar?

Inertial?

You could do inertial, right?

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

You don't typically use a body frame.

That's probably the least you scream for an orbit

because then it couples everything in crazy ways, right?

We want something different.

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

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

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

using different frames, and that's fine.

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

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

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

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

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

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

T-Bo?

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

Yes, diagonals running off.

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

because you actually scaled this vector in a way

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

But if it's an identity times a scalar,

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

Sphere.

Sphere and?

Cubed.

Cube, right?

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

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

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

that would make the gravity gradient torques,

actually, to be zeroregardless of attitude.

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