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>> Hello, class. Professor Anderson here. Welcome to another edition of Learning Glass Lectures on physics. We're going to pick up our discussion with Kepler's laws, and let's take a look specifically at Kepler's second law. So just to remind you what Kepler's second law was, he said that planets sweep out equal areas in equal times. What does that mean? Well, here's the sun, okay. Planets go around the sun in an elliptical orbit. And if you set your watch and you measure some amount of time, say a month, the planet will sweep out a certain arc, and you can calculate what that area is. It turns out, if you set your watch for a month when the planet is farthest from the sun, that area is exactly the same. This area is exactly equal to this area. And that should tell us a few things right off the bat. Since this is very far from the sun, this has to mean that the planet is moving slower out here. A longer radius means that if that air is going to be equal, it's got to be long and thin, which means the planet is moving slower. And then here it is, of course, moving faster. So the Earth is, of course, nearly circular orbit, so its speed around the sun is basically the same. But if you have a planet that has a very eccentric elliptical orbit like this, it will definitely be moving faster there than it is out there. How do we get to this from Newton's Law of Gravitation and what we know about angular momentum and torque and things like that? Let's think about it. If the planet is right here, here's our planet, then there is a force on that planet. What force is acting on the planet? I'm asking you guys. What force is acting on the planet? >> Gravity. >> Gravity, right we know that, gravity. Gravity acts in that direction directly towards the sun. All right, but there is also a line here, which we can call R, which is the position vector of the planet. It doesn't really matter if you draw it out from the sun or towards the sun, okay, it's not going to matter, as you'll see in a second. So what we said about torque was the following. Torque, tau, is equal to R cross F, right. This was a cross product. R is the position. F is the force that you apply. And that cross product has a very particular definition. It's equal to the magnitude of R, the magnitude of F, the sine of the angle between them. And then there is some direction vector that we would tack onto the end of it, okay, and it would be either into the page or out of the page. But let's think about this for our problem. In our case, that torque is position vector R force due to gravity sine of the angle between Fg and R. What is phi? What should I put in there? What's your name? >> Stephanie. >> Stephanie, grab the mic, Stephanie. Stephanie clearly made the mistake of volunteering to talk. What should I put in here for phi? What's the angle between R and Fg? >> Zero. >> Zero. Why? >> Because they're going in the same direction. >> Because they're parallel, right. Anytime you have two vectors that are parallel, the angle between them is zero. Is the sine of zero, zero? Or is the sine of zero 1? Or is it something else? >> It's zero. >> Zero. Sine of zero degrees is zero. Again, you can go back to the unit circle and convince yourself that yeah, it's zero, right. So tau is equal to zero. There is no torque. There is no torque at all on the planet due to the sun, specifically the sun's gravity. Okay, but what we said before was, torque relates to another -- torque relates to another principle in physics, another value, which we call the angular momentum L. Right, this thing L is the angular momentum. And this is similar to what we did with the linear momentum case. Remember, we said that force was DP/DT, where P is linear momentum. In the rotational case, torque now is equal to DL/DT, where L is angular momentum. All right, if T is zero, if tau is zero, then what does it say about L? L has to be equal to a constant. If torque is zero, L is constant, the angular momentum of this planet is constant. It's some number, and it is just always that number. All right, let's use that now to calculate something about speeds. Okay, the angular momentum of the planet is constant, but we know what angular momentum is. Angular momentum is R cross P, right. And R is the position vector, P is the momentum of the planet. And we know exactly what momentum is. Momentum is mass times velocity. That MP can in fact come out in front. And so we get MP is just R cross V. All right, let's go back to this picture right here and think about this picture for a second. If I think about how far you move here, we can call that our distance, which is DR. And that's just VDT. And if I think about this area, and we call the DA, what can we say? DA is 1/2 of R cross DR, but DR is VDT. And now I can tie all that back to this thing right here, the angular momentum. I get 1/2 L over the mass of the planet, multiply that by DT. So if that is equal to 1/2 L over MPDT, I can divide by the DT and I get DADT is 1/2 L over MP, and this is a constant. And this is exactly what that statement means. Equal areas and equal times. Okay, when you see DA's and DT's, just think deltas, if you're not comfortable with derivatives, right. DA is really delta A. DT is delta T. Just in the limit of small deltas.

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>> Hello, class. Professor Anderson here. Welcome to another edition of Learning Glass Lectures on physics. We're going to pick up our discussion with Kepler's laws, and let's take a look specifically at Kepler's second law. So just to remind you what Kepler's second law was, he said that planets sweep out equal areas in equal times. What does that mean? Well, here's the sun, okay. Planets go around the sun in an elliptical orbit. And if you set your watch and you measure some amount of time, say a month, the planet will sweep out a certain arc, and you can calculate what that area is. It turns out, if you set your watch for a month when the planet is farthest from the sun, that area is exactly the same. This area is exactly equal to this area. And that should tell us a few things right off the bat. Since this is very far from the sun, this has to mean that the planet is moving slower out here. A longer radius means that if that air is going to be equal, it's got to be long and thin, which means the planet is moving slower. And then here it is, of course, moving faster. So the Earth is, of course, nearly circular orbit, so its speed around the sun is basically the same. But if you have a planet that has a very eccentric elliptical orbit like this, it will definitely be moving faster there than it is out there. How do we get to this from Newton's Law of Gravitation and what we know about angular momentum and torque and things like that? Let's think about it. If the planet is right here, here's our planet, then there is a force on that planet. What force is acting on the planet? I'm asking you guys. What force is acting on the planet? >> Gravity. >> Gravity, right we know that, gravity. Gravity acts in that direction directly towards the sun. All right, but there is also a line here, which we can call R, which is the position vector of the planet. It doesn't really matter if you draw it out from the sun or towards the sun, okay, it's not going to matter, as you'll see in a second. So what we said about torque was the following. Torque, tau, is equal to R cross F, right. This was a cross product. R is the position. F is the force that you apply. And that cross product has a very particular definition. It's equal to the magnitude of R, the magnitude of F, the sine of the angle between them. And then there is some direction vector that we would tack onto the end of it, okay, and it would be either into the page or out of the page. But let's think about this for our problem. In our case, that torque is position vector R force due to gravity sine of the angle between Fg and R. What is phi? What should I put in there? What's your name? >> Stephanie. >> Stephanie, grab the mic, Stephanie. Stephanie clearly made the mistake of volunteering to talk. What should I put in here for phi? What's the angle between R and Fg? >> Zero. >> Zero. Why? >> Because they're going in the same direction. >> Because they're parallel, right. Anytime you have two vectors that are parallel, the angle between them is zero. Is the sine of zero, zero? Or is the sine of zero 1? Or is it something else? >> It's zero. >> Zero. Sine of zero degrees is zero. Again, you can go back to the unit circle and convince yourself that yeah, it's zero, right. So tau is equal to zero. There is no torque. There is no torque at all on the planet due to the sun, specifically the sun's gravity. Okay, but what we said before was, torque relates to another -- torque relates to another principle in physics, another value, which we call the angular momentum L. Right, this thing L is the angular momentum. And this is similar to what we did with the linear momentum case. Remember, we said that force was DP/DT, where P is linear momentum. In the rotational case, torque now is equal to DL/DT, where L is angular momentum. All right, if T is zero, if tau is zero, then what does it say about L? L has to be equal to a constant. If torque is zero, L is constant, the angular momentum of this planet is constant. It's some number, and it is just always that number. All right, let's use that now to calculate something about speeds. Okay, the angular momentum of the planet is constant, but we know what angular momentum is. Angular momentum is R cross P, right. And R is the position vector, P is the momentum of the planet. And we know exactly what momentum is. Momentum is mass times velocity. That MP can in fact come out in front. And so we get MP is just R cross V. All right, let's go back to this picture right here and think about this picture for a second. If I think about how far you move here, we can call that our distance, which is DR. And that's just VDT. And if I think about this area, and we call the DA, what can we say? DA is 1/2 of R cross DR, but DR is VDT. And now I can tie all that back to this thing right here, the angular momentum. I get 1/2 L over the mass of the planet, multiply that by DT. So if that is equal to 1/2 L over MPDT, I can divide by the DT and I get DADT is 1/2 L over MP, and this is a constant. And this is exactly what that statement means. Equal areas and equal times. Okay, when you see DA's and DT's, just think deltas, if you're not comfortable with derivatives, right. DA is really delta A. DT is delta T. Just in the limit of small deltas.