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Now we have ways to get the torque to be zero either the shape can make it go to zero

Â or for any shape particular orientations if we locked up one of our principal axis and,

Â you know, be, make it equal with the position vector.

Â So what we've just identified is if you look at this equation,

Â if you have a circular orbit at 7,000 kilometers

Â or a circular orbit at 42,000 kilometers Leo versus Geo, you're

Â going to get different gravity gradient torques. Kind of makes sense, right?

Â The further out you go the weaker gravity gets,

Â that we could effect the gravity gradients becomes as well.

Â So as you change your orbits,

Â it actually affects your attitude, because you get different torques acting on it.

Â So is the inverse true?

Â That's what we want to study here.

Â So if you take your spacecraft and it's pointing at the sun

Â and you decide to rotate it with reaction wheels and point now at the Earth,

Â you've taken all this finite mass and moved this distribution around.

Â Since orbits affect attitude, do attitudes affect orbits?

Â If they have different orientations, do I change my orbit type? No?

Â The answer is yes.

Â You do.

Â The more important question is do you care?

Â How big is that term?

Â So we're going to step through this math a little bit quicker

Â than what I did with the other stuff, but it's the same kind of principles,

Â but you'll see it would have to go to second order expansions that are set up first

Â because the first order expansions actually vanish

Â and then you get this expression and I'm going to rewrite it.

Â So now, you get the gravity force much more precisely assuming

Â it's not just a point mass, but it's a finite shape.

Â And now, orientation couples into it.

Â Where is your mass?

Â Do you have more mass lined up close to the Earth versus all the mass lined, you know,

Â lined up evenly away from the earth?

Â That means different gravitational accelerations.

Â You can have different orbits out of it.

Â So let's go through the same math again.

Â It's the same blob, the same force,

Â but instead of doing a body integral of R across the four step,

Â which gave us torques about center of mass we just want the net force acting on this body.

Â So we just treat the body in a roll of all the little D.N. forces

Â that we're going to have.

Â So we just dropped R cross parts.

Â So several terms you'll see will be very similar.

Â That's why I'm kind of going through this quicker.

Â Now we have to do in the end, you plug in the stuff like before.

Â It's very, very similar.

Â Just you don't have that R cross term and then we have to do an expansion

Â of one over R cubed, because that's RC plus little R,

Â but I'm keeping the first order term, which we had before.

Â These two's would cancel, in that case.

Â But there's also a second order term here and here,

Â those end up being critical because, again, first order vanishes.

Â You plug this in, manipulate.

Â Similar arguments as before, we can see these body integrals here,

Â center of mass makes that vanish, center of mass will make this vanish.

Â Then we have other terms.

Â This is one we just expanded earlier with vector cross-product identities.

Â That ended up being the inertia tensor plus something.

Â Relative, parallel to RC which cancel because we cross with RC,

Â but here we're not crossing with stuff so it's going to be a little bit different

Â and that's going to be total mass.

Â This just screams for more vector identities just because,

Â and that's just going to be the R squared.

Â It will give us a polar moment of inertia of the stuff.

Â So we apply, again, the same identity we used earlier, plug this stuff in, right.

Â It's the same process and you start to identify, ah, look this with the minus sign

Â that becomes the inertia tensor.

Â This too, we can relate to inertia tensors

Â and these R squares bodying with R squares are actually polar moments of inertia

Â and that's defined here.

Â And this is something I'm just giving you.

Â You can look this up into a classic statics mechanics book,

Â but the polar moment is the same thing as one half of the trace of the stuff.

Â You can actually prove this to yourself pretty quickly.

Â If you look at the I matrix, the diagonals are all X square plus Y square.

Â Y square plus Z square and X square plus Z square.

Â You start adding them up.

Â You basically get two times X square Y square C square

Â and X square Y square, C square, is always little R squared.

Â So that's where you've just proven it, right?

Â That's the same thing.

Â So I can go from the inertia tensor, get the trace and do that half.

Â I'm also going to start factoring out more RC's here.

Â If I divide both by RC magnitude, I get just these unit direction vectors.

Â It will be a nice convenient form,

Â but I have to factor out an extra RC squared over there.

Â So it's the same math steps.

Â It's just a lot of extra details and let's see what we get.

Â We use inertia tensor definition

Â and in the end this is your more precise gravity force for an orbit.

Â All right.

Â Where if you look at this, and this is,

Â in orbits class you tend to deal with gravity accelerations.

Â R dot is equal to minus as you ever are cube times R.

Â That was the classic orbit equation here

Â and getting it to in terms of force everything is multiplied times M

Â but it's basically the same.

Â And so you got minus MU over RC cubed times RC, right, for that one here.

Â That's the point mass approximation.

Â That's what 50\50 was all about, that orbits class.

Â Then, first order terms dropped

Â and all we have left afterwards are all the second order terms,

Â which every one of them we're able to write somehow is a function of the inertia tensor,

Â which is kind of cool and I factor things out.

Â So, have this unit direction vectors and had extra RC squares that I had to factor out.

Â So now, this is the formulation.

Â So you can see, as you change your attitude,

Â all of a sudden you will have different results and you have a small impact on your orbit.

Â So if you fly, flying this way and then you go this way,

Â you will have a slightly different orbit.

Â But for Earth, man-made spacecraft on Earth.

Â This is a really, really small effect.

Â If you start flying around small asteroids and boulders, this may not be a small effect.

Â Right?

Â And this is stuff you have to actually account for when you do these extra expansions.

Â The classic point mass model is not going to work necessarily.

Â So in this class we're not doing asteroids,

Â so we're going to look at this now quickly and do some dimensional analysis.

Â Inertia's are all mass of the spacecraft times distance squared for every element.

Â Kind of the large, good opera estimate just make it a sphere.

Â Then all the radii, all the mass, is as far out as you can do it.

Â We don't build spacecraft that way, but that would be kind of an upper bound.

Â So it's mass times the radius of the spacecraft squared divided

Â by mass times the radius of the orbit squared, the masses cancel.

Â So only need to look at the ratios of spacecraft squared to radius of orbit squared.

Â Orbits, let's make Leo easy 7,000 kilometers squared at 7 million kilometers that's 10,

Â 49 times 10 to the 12.

Â So it's about 10 to the 13-ish order of meters.

Â Man-made objects.

Â Around one-ish average radius.

Â If look at the mass distribution and everything.

Â So this term is one, plus something, one over 10 to the 13.

Â That's pretty small.

Â You're going to have to account for, you know, Lawrence forces, drag forces,

Â origin pressure forces, out gassing forces probably of the material.

Â There's all these other disturbances that are way bigger than 10 to the minus 13.

Â You know, J2 that you were talking about does 10 to the minus three J four

Â and five and those six is, okay, maybe,

Â there are three, two, three orders of magnitude smaller again.

Â Right?.

Â But still about that level, we're nowhere near 10 to the minus 13.

Â So for Earth Application that's why typically you don't see this,

Â but it is there and with these tools you can develop it.

Â But again, now we're getting into pretty cool missions too around other small bodies,

Â trying to get around boulders and flying stuff and your crafts might be, you know,

Â this might be much more of a significant effect.

Â Asteroids still tend to be basically lumps of stuff.

Â We don't build spacecraft as cannonballs or lumps of stuff.

Â This spacecraft has lots of empty space, right?

Â So the inertia ratios are still going to be big, it might still be quite small,

Â but it might not be negligible either.

Â So it's kind of a cool result people, good prelim questions, you know.

Â Does your attitude affect your orbit?

Â And the answer is yes.

Â Always, Spencer?.

Â Still a little confused as to how like small asteroid, like uneven bodies like that,

Â of the gravity gradient work as more of a factor?

Â Because isn't it.

Â Use a metrical force.

Â Force.

Â It's proportionately a bigger influence, because with this big square bracketed term

Â as basically normalized to point mass approximation of that object is just one.

Â And then how much is this going to be?

Â So you just have to look at, well what's my inertia?

Â And if we're still about size one, meter-ish.

Â You know size spacecraft and this isn't 7 million meters,

Â but now all of a sudden 50 meters,

Â because you're flying around the five meter boulder, you what I mean?

Â It's still going to be small, but not 10 to the minus 13 small, right?

Â And then, there's bigger binary systems and other kind of stuff people are studying.

Â It definitely makes an effect.

Â There's some binary asteroid systems, where you have,

Â you know, Shear's done a lot of work on this stuff.

Â He's got a whole book on this.

Â But then, it affects it more.

Â That makes sense.

Â So good.

Â So we've covered now how to derive gravity gradients, we need that one approximation

Â that the spacecraft is small compared to the orbit.

Â We got the nice vectorial way to write it and in some different coordinate frame versions.

Â This is the force version.

Â This is more FY, but it's just so you can see the answer.

Â