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.
1.1: Gravity Gradient Torque in Body Frame
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來自 University of Colorado Boulder 的課程
Kinetics: Studying Spacecraft Motion
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從本節課中
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.
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Hanspeter SchaubGlenn L. Murphy Chair of Engineering, Professor
Department of Aerospace Engineering Sciences
So good.
This is the vector form, the coordinate independent form.
Let's now actually apply some particular frames and
there's different answers that we get out of this.
It's always the same answer, but it's different expressions,
different formulations of that same answer.
So if I'm using an orbit frame, and here I've used an aircraft-like frame,
where if an airplane is flying you have your nose pointing in the one direction,
your wings may be to the left or right.
I have yaw axis up, aircraft probably would have had it down.
But with orbits we tend to have our third axis or
one of the axis up in the radial direction.
So, 01 is my long track, 02 is my cross track orbit normal, and
03 is my orbit radius direction vector.
They're all unit direction vectors.
That means Rc by definition is always going to be 7,000 kilometers in that
direction, all right?
And that direction is defined here by 03.
So, in the O frame, your position vector on a circular orbit in particular,
is just going to be constant.
It's just going to be 0, 0, Rc, all right, which is kind of nice.
So we can write that.
But we've seen the spacecraft part, the inertia tensor,
we tend to write in the body frame.
So while the orbit positions could be written in very simple manners,
we tend to need it in the body frame.
And now, so if this is your position vector pointing up to your satellite, and
you've got a coordinate frame on your satellite, depending on your orientation,
you can have an infinity of different variations, and most generally,
you're going to have to account for three different vector components, for
your Rc inertial position vector, always.
So, Rc 1, 2, 3, nothing but
B frame components of my position vector, different ways to look at it.
So we have that.
Now we plug this in, I added these this morning.
Everything's in the body frame just to make it explicit.
So we had Rc, cross I Rc.
I've picked a principal coordinate frame here just to make,
I can always do that for a rigid body, all right.
It exists, it makes your analysis a little bit easier.
Now you can do this math, this multiplication is what talking about.
You get a little bit of scaling, then the cross product, and
this is going to be your answer.
So this box, the answer's the same as the earlier one after it's
expressed everything now in body frame components.
But we can again look at more conditions now and
try to figure out when will the gravity gradient torque go to zero?
Why do I care about it going to zero?
For equilibrious that we're about to study,
that's thing if you have these conditions and this is the one disturbance you're
modeling, if that disturbance acts on it, you won't be at an equilibrious.
This is like gravity gradients we talked about last time.
This would be a gravity gradient equilibria.
If you disturb it, then,
this one weigh more than this point makes it restore itself.
So what orientations or what shapes give us that?
And as you guys mentioned earlier, one way to make these things go to zero
is simply these differences all have to vanish.
And that means, as you guys already identified, is that all the principal
inertias must be equal which is either a sphere or a cube, those kinds of shapes.
So, there are particular shapes that can make your attitude go to 0.
That's not or so your gravity gradient torque go to 0.
And that's important, there was a satellite that we built here called Dandy.
Anybody know that one?
Heard about it a little bit, right?
It was a sphere for reason like that so we could do this stuff, and
didn't have any other gravity gradient torques disturbing because I really wanted
to measure atmospheric force effects, and
I didn't need all the disturbances acting on it very much.
So that was really leading to this decision, all right, that gave us zero
gravity gradient torques, also how the atmospheric drag effects so
you didn't have weird panels sticking out, there were other reasons too.
So the other way we can make it go to zero is we can look at these components.
Are there particular orientations?
Remember, Rc one and two and three,
the position vector always points to straight outward but
then depending on my attitude, I will have different Rc 1, 2, 3 components.
Are there different orientations that will force this LG to go to zero?
And the answer I'm showing you is Rc is equal to Rc times bi.
So, is it David?
What does that mean for your orientation, if Rc is equal to Rc bi hat?
>> Does it mean you're aligning your body axis with the O frame that you created?
>> Be more specific, what part of the O frame?
>> The radius vector.
>> Right, because if you say you're adding a body axis relative to the O frame-
>> If they're principal inertia axis?
>> Then that implies all my principal inertia axis
have to line up with the O frame, and that's not true.
So your second argument is correct, and
it's what we really care about is the Rc part here that we have, right?
That's in the O3 direction here, it's basically the orbit to radium to axis.
And so that what it says is your position vector on that direction has to
be a principal axis of your body.
And that could be B1, B2, or B3.
But it's just locking one of your R axis in.
I'm going to use these to have three distinct axis, right?
So we have, this is a principal axis, this is a principal axis, and
this is a principal axis.
It just says that if I line up one of those things as a principal axis.
Let's do the simple example, right?
Here I have my skinny axis in the vertical direction.
That satisfies that condition.
If I rotate about that axis,
this is still in equilibrium, because of the symmetry of the stuff, right?
So you can see, that orientation doesn't matter.
It just matters that one of the principal axes is lined up.
This is another one that we identified,
where right now the gravity gradients are even and they cancelled each other.
But I can rotate about that about the gravity direction.
And I still have a zero gravity gradient torque, right?
That's what this condition means.
So it only locks one of my body axis to be aligned with the orbit direction vector.
That's it.
But it still leaves an infinity of orientations because you can rotate about
that local vertical axis.
Okay, good, yes?
>> [INAUDIBLE]
>> Thinking more like J2 and that stuff?
>> [INAUDIBLE] >> They're pretty small.
J2 is about a thousand times smaller than the regular stuff.
So if you care about those accuracies, you could do that and you'd have to account
for that then in these developments, that would be actual terms.
I don't typically see those because this is already a pretty small torque.
Many simulations don't even include this, but if you do include it, great,
and J2 is much, much smaller.
If you're going around asteroids, that's a whole different problem.
If you're looking at an asteroid binary system,
I know some of you this year might be looking at those.
That's a full on thing.
Then you do these expansions like I've done but
on two bodies not just treating the Earth as a big point mass, as a big thing.
But for Earth, it is very spherical as far as that goes, right?
But yep, no, great question.
So, this would be modified.
The process is the same, but you might have extra conditions to line it up with
the local area, and what does that mean?
