 What we're going to look at now is what's called central force motion. It's the dynamic that goes into orbital and mechanics, why things go in the orbits that they do. And it's because there's always a force directed toward a central point, which we looked at when we looked at circular motion and the like. We really didn't take that much beyond circular motion, maybe with a little bit of acceleration. We didn't look into all the possibilities of what that business could do. So let me lay out just a little bit of a background reminder. Gravitational force between two masses is an attractive force, and it's a situation where it's equal in opposite forces. And given Newton's law of gravitation, you probably remember that it's also a factor of the distance squared between the two objects, and then that G, the universal gravitational constant. I think it's called 673. Very small, the gravitational force is not a very big one. It's the weakest of the fundamental forces. And to make everything work, it's got kind of odd units. It's experimentally determined. I don't know that anybody's ever been able to derive that from empirical work. And it's possible, I guess. So if we take that to be, then let's let M2 be maybe the mass of the Earth. And then if M1 is U, for example, then the weight is M1 Mp, and then D squared becomes the radius of the Earth, the distance from the center. Now this is all assuming, remember, that the Earth is a uniform homogeneous sphere, which it isn't quite. So a lot of this is approximate, but it does lead to then, if you put in the values, cancel the M1, put in the values that are just right out of the book, you do get something about like the strength of the gravitational field that we use for all of our classes. And then the mass of the Earth, and what we use in these numbers, 5976, and then the radius of the Earth is 6, oh, I have the diameter of the Earth, 12,742 kilometers. So we'll divide that by 2 to get the radius, and then put it in meters. And if you use those numbers, you get the gravitational acceleration that we're so used to for what we do. It comes out actually a little bit higher. That comes out very close to what we use, so the numbers are all good. At locations above the Earth's surface, which is what we need to look at when we're talking about orbital mechanics, then the acceleration itself is reduced a little bit because you're getting farther from the Earth, and it turns out that it's as the square, the ratio where r then is the distance from the Earth's center, which then would be of course the radius of the Earth plus the height of the distance above the surface, I guess, if you will. So we'll call that h. So we can start looking at orbits at different heights and the different speeds that go into them. All right. That's physics, too, stuff, I believe, right? That's chapter nine of physics, too. Did you do anything with orbital mechanics? Well, you probably didn't do what we're going to go through, so we're going to beef it up a little bit. All right. So that's just a bit of a reminder of the background. Now, the setup that we're going to use, let's see, the Earth, the Earth is blue, of course. Blue marble, but it is spring, so we need to draw in there's Florida, all the people coming back from Florida now that the weather's nice. That's pretty generalistic, that looks like a picture from the space station or something. So we're going to start things from a point where our satellite, whatever it might be, will have velocity, v zero at a particular point where the velocity is perpendicular to the line connecting the two, and so we'll call that our fundamental distance, r zero. And then the path of the satellite, which may or may not be circular, may or may not be elliptical, we'll look at all those possibilities as we go here, then the rest of the definitions we need is the angle will be taken from this initial spot where the two are perpendicular, because after that they're not quite perpendicular anymore. So if the velocity and the line joining the object in the center are always perpendicular, then you can only get circular motion, and we have to allow for the possibility of other kinds of orbit, as we'll see once we get all the way through it, as you probably remember from some of what you set up in physics too. Alright, so that's the basis for our setup for what we're going to do, and oh no, we need a little bit more, just a little bit more, because we do have then, let me, for a coordinate system, so we'll use polar coordinates where we have a radial component straight out and a tangential component in the direction of motion, oh that's not tangential, for polar coordinates we use e theta there, we're not going to have those long, we're going to break it into two different directions in a second anyway. Okay, so that's our setup using polar coordinates, so we'll need to pull a little bit of that out of our memory as well. So we sum the forces in the r direction, the radial direction, well that is the gravitational acceleration, so that's a minus because the force is directed towards the center in the opposite direction of our radial direction, and so we have the original piece, that's just essentially the weight at places other than the earth's surface. If we're at the earth's surface, this would be r e over r e, that would cancel, we just have m g, and the minus just means it's directed opposite to our r direction. But if you remember our radial component of the acceleration, we established this the day we did polar coordinates, so it should look a little bit familiar, and we did that, a lot of this came about if you remember, we also had to have the time rate of change of the unit vectors themselves, and a lot of that led to the pieces that we have here if you remember, because unlike x y coordinates, we have changing unit vectors in polar coordinates when we're doing this kind of motion. And then in the theta direction, we have no forces in the theta direction, so that side is just zero, and then the other side is the acceleration component in the theta direction that we also established on that day a couple months ago when we did the polar coordinates. Plus 2 r dot, theta dot. We've got to be real careful, we can't lose any dots. Okay, we're on. In the book, just for reference sake, this is equation 214, and that's back when we did it, we did that in probably February or so. Alright, in the second part though, we can simplify a little bit in that, well actually for both of them, notice the masses cancel all through the mass of this object, does not matter, does not have anything to do with any bearing on what the orbit will be. And so the second part that can simplify to 1 over r times the time rate of change of r squared theta dot. And that r squared theta dot should also look a little bit familiar, so that all equals zero. Alright, so that in itself, I don't think there's much new here other than we're allowing for the fact that gravity, the gravitational field strength diminishes as we get above the Earth's surface a significant distance. Alright, since this business equals zero, then this piece right here must be a constant. So r squared theta dot must itself be a constant, but we also know that the velocity in the theta direction is r omega, and omega is theta dot. So we then have, we can put that in here so that we know that r, we take one r, one theta dot out, we know that r v theta equals a constant itself. And we can find out precisely what that constant is by going to the initial position where we have v zero and r zero, and they will also equal that constant because that's what the product of those two are at that initial position. At the very initial place where our distance r zero away from the center of the Earth, our theta velocity is v zero, our initial velocity, and so we know that then to be a constant. So we're going to need that several times with the remaining, so that's one of the little things we can't lose as we go through this. Alright, so the next little step, we're going to take, we're going to take this, let's see, we're going to take all that. So theta dot equals that constant r zero v zero over r, sorry over r squared, yeah we have the square there. So that's just saying that this side equals the constant and that constant also equals this side. So we just solve for theta dot because that we can put in over here on the first equation. So that's our next step. We've got minus g over r e r quantity squared then equals r double dot minus r theta dot squared which is that piece up there. Alright, that's combining this little piece here with that piece right there. Oh, r squared on the bottom, yeah. Yeah, we can't lose any of these things over on a side track for powers. Okay, so we can reduce this other side a little bit, just clean things up because we've got a lot of r's around here, a bunch of squares, r zero squared v zero squared over r cubed then. And now we're going to set aside four a little bit and we'll come back to it. That's a bit, we've got a little more work to do until we come back to that. Alright, it's going to look like this is at a left field but it comes back in the end. So we're going to introduce this new quantity, we'll call it u, is just one over r because we have a lot of one over r's over there and so it's going to help us bring something back together, put them together here and there. So now we can work on the, it's too bad I have to erase those equations. I always have those boards that slip up so we can leave them here. But we have that theta component, so I mean the r component that we're working on. Oh, this one, this one's here so we're working on it more now we want to work on that r double dot part. Let's see, that's d theta dot's just d r d theta. So d, not d r d theta, d r d t. But r is one over u, introducing our new term there and so that's minus u squared, sorry one over u squared, just doing the derivative of this with respect to time. So just taking that derivative so that's what the chain rule is. But we're not so much interested in how u varies with time, we're much more interested in how u varies with theta. So we want to go to the u d theta, d theta d t. Because d theta d t we have and now we're going to see how u varies with theta which is much more useful to us. And then that is minus r squared theta dot which is the n part there, d u d theta. Remember though that r squared theta dot which we have here is equal to that constant and that constant is also equal to r zero v zero. So then we can say that r dot equals minus r zero v zero, our initial conditions at the place where we started. That's the r squared theta dot v u d theta. And that is going to be equation number two that we need to set aside, we'll bring it back in a minute. So there's, there's, now we've got r dot which is the first step here, we can do now r theta dot, sorry r double dot which is that piece right there that we just did. So this is d d t of minus r zero v zero. Now remember those are constants, those are not variables, those are constant values that's where we started our problem. So this just then means we need to take the chain rule on this part, the minus r zero v zero comes out and we have d d t of d u d theta which is d theta d t theta of d u d theta. Okay that's the little bit of the chain rule in there. But we can clean that up a little bit too. This front part just constant theta dot for that part and then d squared u d theta squared. So that's r double dot. Oh wait we can do even a little bit more with it because we've got, we've got, we've got this piece here that we can use here because we have r zero v zero as part of it and then the r square and then the theta dot part that what we want, let's see, yeah. So then this becomes, remember this is r double dot. This becomes minus r u v r zero v zero squared. We're just putting in theta dot for that. That piece over there is theta dot. We're just putting that in so that squares that term. We've got r squared on the bottom which is u squared and then the d squared u d theta squared is untouched. So all I did was took this piece here for theta dot and put it in there for theta dot. And now we can put that piece, remember this is r double dot. We can put that piece in right there. Maybe I'll just continue that right below it. We can put that r double dot in there minus r zero v zero quantity squared u squared d squared u d theta squared minus the part we have on the back there minus r zero v zero quantity squared over r cubed. That came from taking this and putting in the r double dot we just came up with. We just came up with r double dot here. That's what that string is. So now I've just put it in there. There we go. I explained it. It's going to simplify greatly even though checking the algebra can take hours. Well at least it did me. So there's a couple places where I'm going to wave my hands. I'm saying simplifying gives and then you go home because you have hours and you check that. So let's see where I am. So we've got this whole business here. We've got minus g. I'm going to be rewriting that r e over r quantity squared equals minus r zero v zero quantity squared u squared d squared u d theta squared minus that constant squared. Remember r zero v zero is a constant over r cubed. Alright and here's one of the places where I'm going to say it simplifies to and then you go home, make a pot of coffee and spend three hours checking this like I did yesterday. Unless you're faster at this stuff than I am. It simplifies to g over quantity r e r zero v zero squared. So that's just a matter of dividing through by the minus r zero v zero squared which remembers a constant which makes all of this stuff on this side a constant equals d squared u d theta squared plus u squared. No more than a sentence missing. No squared missing. We're okay. And you guys are all in Diffie Q, right? So you've got this solved already. This is just you? Yeah. Because you've got r cubed and you're left by which is u cubed. Oh yeah, just you. It is right there in front of you. Yeah. It was u squared but we took one of the one over r squareds and used that. Alright. So this has solution which I'm sure you've all done already in your heads. a sine theta plus b sine b cosine theta plus that quantity on that side which is a constant remember. Alright. And so to find out what those constants are a and b we need to look at initial conditions. So that theta equals zero degrees we have u equals one over r zero. So that whole side has equal one over r zero. Well that part that gives us the a equals zero. If I got the right piece it is zero. We know that v are the velocity in the radial direction. At this point remember it was perpendicular to the radial direction itself. So that is itself zero. And so we need to erase the second equation. I've got one equation there. The second equation I needed remember was r dot d r dt equals minus r zero v. I'll just write down the second one that I had to erase. Okay. So that means then see that's what we got right here. So that all equals zero which means du d theta equals zero. Alright. That's using that second equation over there. We now know or at that point r dot is zero so du d theta is zero at the same as one spot. So now we can finish that because that gives us v the second constant in the solution to the differential equation is one over r zero minus g quantity re over r zero to zero squared. So that gives us the full solution to the differential equation. We're getting down to the meat of it. I'm going to get down to the solution that comes. Okay. So let's get all those things in. And this I couldn't derive directly but I could take it put in what I've got here and come back to this. So maybe you can derive it directly when you put this in. I couldn't but maybe you can. Where we take the constant terms and combine them all into this one piece called the eccentricity which is a term probably remember. So it's r zero v zero squared not quantity squared over r e minus one. And this is the eccentricity. Did you talk about that in physics too? A little bit. Okay. It gives what it gives us is the orbital shape. It's heading up on what its value is. If the eccentricity is zero then it's a circular orbit. And if you put it in there as one you get essentially all the way back you get a uniform circular motion. If it's between zero and one then we get an elliptical orbit. It limps. If it is one we get a parabolic orbit. And then if it's greater than one we get a hyperbolic orbit. And if you remember your conic sections that's the difference between those. No r naught is not squared. See I told you it's simplified greatly. It's just going to take you a couple hours to prove that it does. But like I said the easiest way out or the way I finally did it is I just took this put it back in there and then I can back out the solution to the differential equation. I couldn't do it directly from the differential equation. Alright the most interesting orbits are the elliptical ones. The parabolic ones are very useful. They use that when they slingshot satellites past other bodies like Mars or something to give them some pickups of speed. So there's the earth with a nice elliptical orbit. I'll make it a little bit bigger. With the earth at one focus of the ellipse which if you remember was one of Kepler's laws. Kepler's first law I believe. Alright so we have a couple things we need to define here. This distance remember was our original r zero. That was our initial point. This point here of closest approach is called the perigee. It's the point of closest approach. It's also the point of highest velocity. This is the apple g. The point farthest away. And then the axes, the major axes. That distance we'll call 2a. So a is just a semi major axis. And then across here, across the minor axis is b. So b is the semi, b itself is the semi major axis. Remember this assumes uniform masses or at least we're far enough away from the earth that can be taken as uniform masses. No other forces so we don't have perturbation effects from other planets. And of course it's far enough. There's no drag. And assuming too that this mass is fixed in space. If we want to do it in an external frame of reference we have to add in other effects that we have to play with. Alright so the last little bit is the orbital period itself. And this you can get from playing further with the equations. And I think you've had enough algebra for now. So there it is. 2 pi a to the 3 halves. Remember a is the semi major axis over r e square root of g. And if you move that around a little bit you'll see that's Kepler's third law. Alright so we'll do a quick problem just so we get an idea of the kinds of things we're talking about. So imagine it's pretty easy to make sure you get those initial conditions from which all this stuff came by launching such that right when it reaches a particular point it then has that velocity. And the orbit itself is a ballistic orbit meaning it's non-powered. So this would be the place where you want to have burnout where the rockets themselves are out of fuel and usually jettison. You've seen those pictures as they do that with the space shuttle. So we'll give a couple values to this and then figure a couple things and then we're done. So we'll take a speed at that point, the speed of burnout of just under 37,000 kilometers per hour and an altitude of 500 kilometers. And we'll take it to find a couple things. Find the maximum altitude which of course is at this point over here at apogee and also find the orbital period. It's when we can expect it to come back into view and closest approach later. Okay first thing we can do is even check that it is an elliptical orbit. So calculate the eccentricity, make sure that it's in those bounds for an elliptical orbit, which it will be since that's what I was talking about here. Kepler's second law, if you remember, is equal areas in equal time. So we could also calculate the rate at which areas are swept out and prove that but the metal gymnastics get even more severe. So check the eccentricity. Remember it's R0V0 squared which you are given. G is 9.81. We came up with something close to 9.82 with the numbers before but that was, remember, not using actual measurements like the 9.81 is minus 1. So check that eccentricity. Make sure it is an elliptical orbit and then use the solution that we came up with a little bit ago and find the maximum altitude using that eccentricity. So check those and then that will give you A because that's related to this maximum altitude and then you can find the orbital period using Kepler's third law there. The orbits with the main body at one focus is Kepler's first law. The second law is equal areas in equal times and the third law is the square of the period is directly proportional to the cube of the semi-major axis which I believe Kepler came up with first by pure observation of orbital times and positions. Do you know the radius of view of them? I do which I'll sell you for a small price. These are numbers in the book. 637, here's 6370 kilometers. What else do you got all the other pieces? After that just make sure your units work. R, are you asking what R is? R is just the distance from the Earth's center at any one particular point. And then the theta is the angle made from perigee because only a perigee and apogee are the velocities perpendicular to the radius. It's moving much slower at apogee than it is perigee. Essentially after apogee it falls towards the Earth and falls around it. So for R not, you just have a factor of 7.5 plus the radius of the Earth. No, R not is the radius of the Earth plus the height of the orbit which is the altitude of the 500 kilometers. So R equals the radius of the Earth plus the height of the orbit. And this is a bit different than the way the book did it. I thought the way the book did it wasn't very easily followed. So what's zero? Is it E back? For the eccentricity? Yeah 0.699999 Does anybody else have 0.399? I got 0.813 That's what I got. Are you watching your units? What's up? Because you've got kilometers per hour and kilometers. So I guess you could leave it in kilometers, I guess. Now, what G is in meters, so it's easier to put everything in meters in seconds and then I get 0, 8, 3, 8, 1, 3, the traffic's just confirmed, too. Anybody else? It's the 500 plus the height, the 500 plus the radius of the years, and then when you find out maximum amount to remember, you've got to take the radius of the earth out of that. Can it positive this time? The units even work on this piece over here. The higher the eccentricity, the more elliptical the orbit. And at maximum amount, you're going to have to put in theta equals 180 degrees for the rest of the work. You can use either one, which depends on how your calculator is set. Then you want to fix that. You can see how the freehand ellipsis look. That's pretty good. How much is this? I'll just use a 90 for theta, whatever, zero. No, theta for the maximum altitude would be all the way over here, 180 degrees. 180. Yep. In the book, David, I've got this down. It's in the book somewhere. I don't remember. I didn't take note of it, I guess. It's probably somewhere near the end of chapter three, maybe. What was chapter three? Kinetics of particles. Let's see. Maybe it's in kinetics. It wouldn't be rigid body though. It'd be for particle kinetics. To check on the index and the back for orbits. We'll wait for it. I don't know. That's right. That's work energy. Well, maybe here. Special applications. Yeah, central force motion right there. We got some of the pieces then. Not only this problem, but the class of that. What do you get? Okay, you got that. Now that you've got the eccentricity, you can put it in here. R zero is the altitude plus the radius of the earth is R zero. So you can then put in the eccentricity, solve for R at 180 degrees. That'll give you the full distance from the center of the earth. So you take out the radius of the earth, and then you'll have the original R zero plus the radius you just calculated. The two of those together is 2a. You can solve for R max using theta equals 180 degrees. And then R max plus R zero is 2a. You're still not getting eccentricity? Okay, let's space and try that. Okay, so R zero is the radius of the earth plus the altitude. I see my problem. What was it? Not square. Oh, yeah. You square the velocity, but not the tool. Just the velocity. And as long as your units are all the same, it's just huge. Yeah. What kind of max height? About 60,300 and 32 kilometers. 60,000 what? 332,333. I had 6,200 so it may just be round off. You had 60,200. Yeah, as the height. You had the maximum radius as 66. 702. Yeah. And then you have to take the radius of the earth out of that and get the maximum altitude. Oh, that's probably good. Yeah, I think so. It's probably some round off. So I got an R max of 66,600. Times 10 to the sixth meters. Somebody else just said they got that? Okay. So the altitude then to take the radius of the earth off of that is 60,200 kilometers. So that's maximum R, maximum H, 60,200 kilometers. And then 2A is R0 plus R max. It's 2A. And then that A is what goes in here. And R e raised to here must be in meters for that to work. Yeah, that's right. Is that 19? That's taking the radius of the earth off of R max. So R max minus R e. You had this in meters. That's course 9.81. Let's then check what you have for A. A equals R max plus R0 over 2. Because 2A is all the way across. R0 I think is very small compared to that. So it's actually not much significant. 36.7. Really? I had 33.3 for A. That's 10 to the sixth meters. That's what I had. That's R max plus R0. Yeah, I don't think that's, mine 33 is right because that would be just R max divided by 2. What do you have for that then? 36.7. But it shouldn't take the time down to whatever you had. 14 hours? 19.5. 19.5 hours? 19.7? No, it really makes a big effect. Because with the 33 in here, instead of 36, I had 42 hours. That is a huge, huge effect, I guess, because of three halves. Anyway, so you had an orbital period of, you said, what, 19 hours? 19.5 hours. How is it at the end of four hours? Let's look all this together. You only had to go an hour. So that's it. Any questions about that? Anything else on the term for next week? Remember, next week will just be the last three chapters. So the last two. So there will be some bit on the relative, oh, I know, Joey's at the Clarkson campus, isn't it? That's what Joey is. Okay, so some relative velocity. Remember, we used a method in the center. It's just one way to do it. You don't have to do it that way. Two a hand. I have two, if you have two a hand again. Is that 42? You do it for four hours a week. Because that was after all the double checking, all the algebra in there. Road down? Oh, no. Road down, eh? We're done anyway.