Kepler's Second Law: A Planet's Orbit Sweeps out Equal Areas in Equal Times by @mes on 3Speak
3Speak - YouTube - Telegram - Summary - Notes - Sections playlist - Vector Functions playlist - MES Links In this video, I derive Kepler's 2nd law of planetary motion and show that the line joining...
Tags: #3speak #math #science #calculus #vectors #chessbrothers #scifimultiverse #threespeak #keplerslaws
So Jonas Kepler stated the following three laws of planetary motion on the basis of massive amounts of data on the positions of the planets at various times. So Kepler's laws, law one, a planet revolves around the sun in an elliptical orbit with the sun at one focus. Again, that's what we just derived. Law 2, the line joining the sun to a planet sweeps out equal areas in equal times, and the square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit. So Kepler formulated these laws because they fitted the astronomical data. He wasn't able to see why they were true or how they were related to each other, but Sir Isaac Newton. in his famous, yeah, in his famous book, Principia Mathematica of 1687, showed how to deduce Kepler's law, three laws of motion, I mean, Kepler's three laws of orbital mechanics, from two of Newton's own laws, which are the second law of motion, and the law of universal gravitation. So earlier in this video, we proved Kepler's first law using the calculus of vector functions. In this project, this applied project, we guide you through the proofs of Kepler's second and third laws and explore some of their consequences. And now let's take a look at the solutions to the applied project and start off by looking at the solution to Problem 1, which was proving Kepler's second law. All right, so I'm just going to read this again, and then basically go over parts A, B, C, and D, and so on. So, this one states, use the following steps to prove Kepler's second law, which was, again, I just wrote here for reference. So the line, joining the sun to a planet sweeps out equal areas in equal times. Basically, the area that's being swept by the planet. or a line joining the sun to the planet, it's going to be basically the rate change of that is going to be constant. So the notation is the same as in the proof of the first law, in particular, use polar coordinates so that r equals r cosine. Theta i and then plus r sine theta j and then basically we're asked first to show this part show this equation and then b we're asked to show this deduce that we have this over here and then if a equals to a of t is the area swept out by the radius vector r equals r of t in the time interval t not to t as in the figure below show that this rate change of the area is equal to this and then deduce that this rate changes constant. So basically here's that area that's being swept across, and deduce that it's constant, and then this says that the rate at which A is swept out is constant and proves Kepler's second law. So, let's take a look at the solution to that. So first I'm going to do is we're going to use these polar coordinates set up, so R equals R cosine theta i plus r sine theta j. So with this notation, so we have with R, Yeah, R equals to, and again, that's the magnitude of R cosine, there's R vector equals to R magnitude cosine Theta I plus, and then this is going to be R sine theta, and then this is going to be J. And again, to recap why we have this setup is if we have just a vector, so we have X, y, quorance is X, is Y, and we're going to change this to polar coordinates. Let's say we have a vector here, and it has coordinates x and y. Basically, that means then is, well, if we drop a perpendicular line like this, then this x point right here, well, I'm going to draw theta, that's our angle theta, let's say we had our r vector like that. And then, well, we know basically using basic trigonometry, cosine theta, is equal to adjacent of right partner, so that's going to be our x. over R, X over the magnitude of R, that's just going to be well, X over R. Again, that's just that, X over a magnitude like that. And then we move this over to the other side, and the same thing, it's going to be equals to R, cosine, theta. Right, and likewise, this will be the y, and then sine theta, yes, sine, theta is equal to opposite over hypotenous, y over R. move this over and then what we get is y equals to r sine theta so again that's this part that's the j or the y component this is the i or the x component of the end that's polar coordinates All right, and now let's take a look at the solution to part A. So this part, initially, was just a notation part over here when we're told, just using these basic polar coordinates notation. So now part A, we're asked to show that H vector equals R squared D over DT times K standard basis vector there. So what I'm going to do is this is what we want, and let's go back here. So recall from the proof of law 1 that R cross V equals H, therefore we have, well, I'm going to write this right here, so this is what we want. I'm just going to put it over here, so we want this right here. Okay, so we want this like that, and I'll just say we have, we have R cross, Like, R cross V equals H, like that. All right, and yeah, this is from our earlier proof of law. I'm just going to scroll all the way back up to it, and you could see that it should be right over here somewhere. Yeah, this is right here. So remember here, we showed that the H vector is that vertical 1 equals to R cross V, and there's derivation and so on just equals, and H is a constant vector. So yes, fascinating stuff, let's cool. all the way back now. All right, so this is what we have, and now what I'm going to do now is, well, we know that, well, this right here, V, is this equals to, well, R, this is the velocity, it's R prime. So we have H equals to, yeah, this is going to be equal to R cross R prime, like that. So then this is going to equal to, well, let's put this all inside. is actually going to get a giant, I'm going to put it over, we need a lot more space, I'm going to put, well, actually, I was going to add more space, I think I'll just put it in two lines, so I was going to put it in one line, but anyways, it's going to be R, and we know what R is, that's just this part right here, so that's what R is, then we take the derivative of it by a component, so R, cosine, theta, yeah, this is going to be I, vector, plus, and then this is going to be R, R, sine, Theta J vector, like that. And then we're going to do cross, and then we'll also put it underneath here. And now we'll write it over here, the cross, and we're going to need a lot more space. And basically, so we have now, this is derivative of the position vector, and remember, it's derivative D over DT. So this is DR over DT. So these are overall going to be taken derivative with respect to time. So in other words, and in R, the magnitude, or position vector, all functions of time, even the theta, too. So as you move across time, the theta changes. So in other words, we're going to use product rule inside here, inside this part right here, that's going to be R prime, cosine, theta, and then we have to take the derivative of, well, R is going to be constant, derivative of cosine is going to be negative. sine theta, so put the negative there. And now we have to apply chain rules. Derivative of the inside function is going to be, well, D theta over DT, or theta prime, like that. And again, it's all functions of, of, of, of, uh, of, uh, of, uh, T. So now it's going to be, this is going to be now i. Now the next setup is going to be plus. Yeah, that's going to be plus. Now, this component here, we're taking the derivative, so we're going to have, this is going to be R prime, sine, theta, and then this is going to be now, yeah, then we're going to have to go plus, and then R, the derivative of sine is cosine, theta. All right, and now we're going to have a, well, then using chain rule, D theta over DT. Again, that's this part here. We're going to try to get this equation. So that's what we're trying to get. Alright, now this is going to be bracket like that. This is going to be our J component. And this is going to be like that, bracket, like that. All right, and here I quickly just move this, shrunk and fit it all into one place. So now what we can do is, well, we just use a cross product. So now this equals two. So we're going to have to write a giant thingy here. If I'm going to put the I component. So put the i component there. Well, actually, instead of writing this first, I want to put it in the center, I'm going to put, so now this is the I put one of the other one. I'm going to put that here. I'm trying to make it centered. So r cosine theta minus r sine theta. So that's going to be this part right here, and then d theta. And then times it by d theta, d t. And that's going to be now our i, put this in the middle, i vector. And now this part, the top part, the vector is going to be R cosine. Let's put this nicer, go cosine, theta. And now the next one's going to be, well, we're going to write this here. R prime, sine, theta, plus R cosine theta, D theta over DT, and then move some space there. This is going to be our J. Now this is going to be, well, R -sign theta. Yeah, R-sign theta like that. And then lastly, well, we also need a K component, because the number is 3D, the K is going to be of cross-products in 3D, so it's going to be zero, has zero components, because it's, the position vector is just in a, in a 2D plane, well, it's just in a plane, like on a flat plane, but again, it cross-products 3D, so the Z components are zero. So we have it like that. All right here, just move this over here. So now the next setup is, well, we're using a basic cross-product rule. We'll cross this out. So multiply this by this by this, and then subtract by this by that. That's just basically you get a bunch of zeros there. So this equals two. This is going to equal to, well, I'm going to put the first one, it's going to be, well, zero. Yeah, zero I. Now the next one is we cross this. out, and now we multiply this by this part, and then subtract by this one here, actually, but it's the reverse direction, but, but anyways, it's going to be, and there's zeros everywhere, so you're going to have a zero, so it's going to be minus i0, the reason is because you minus it in the middle one, and then lastly, the next one is going to be the only one that matters, is you cross out this part, so again, this is why it's perpendicular to both vectors, so now this multiplied by that, is going to get, R, now let's just multiply that inside R, cosine, so we're going to have a R, multiplied inside, we're going to have an R, R prime, that we're going to have a cosine, theta, sine theta, and then multiply that, yeah, it's still inside here, now we're going to have a plus R squared, cosine, squared, theta, and then there's a D theta over DT, like that. Alright, so we have this, and now continuing further, I'm going to make it all in one line. Now the next one is going to be subtracted now. Now you subtracted by these two multiplied by each other. So minus, so now minus r times r prime, so that's going to be minus r times r prime, and now we have a sine theta, but now we have a cosine theta, sine theta. All right, and then just move it over. Now the next one is multiplied by this negative R-sign, so we're multiplying by negative, but we're already doing a negative there, so it's going to be, well, positive. Just put a positive there. Yeah, so positive, and now it's going to be R-squared, sine theta. So, R-squared, theta, and then times it by D-theta over D-T, and this is our K component. Like that, all right, and here, just quickly just pause video and move it over here. It's better to run it over there. So now what we're going to get is, well, this zero just goes away. And now we're going to be left with is, well, this right here, R, and then R cosine, sine theta, same as this one here. So this just cancels with this, so these two are the same. And now we're just left with this setup here. So this equals to, well, we could factor this R squared out, so we'd be R squared. Yeah, so r squared, now we're going to have a cosine squared theta plus sine squared theta, and then we have a d theta over d t, and this is going to be then multiplied by the k vector, standard base vector. So what we end up having as well, and then this one here's just a Pythagorean identity, this is equals to 1. Yeah, so what we get is H, you remember this whole thing was H, the cross product of this, so we're going to get H vector equals to, well, R squared, D Theta, over DT, times by the K vector. So we could block this out. All right, and here, just fix that up. But now if we look at above, that's this one right here, that's what we were asked to show. So we're asked to show that we want H equals R squared D theta over DT, and then K vector there. So that's what we got. So we got the exact solution. And now also just quickly recall here from my earlier video on the Pythagorean identity, or the Pythagorean trigonometric identity is in my link to my earlier notes there. And there's the, I got to say, I didn't make a proper thumbnail, this was a decade, like I think 10 years ago. So I just found one online right here. So A squared plus B squared equals C squared. This is just a Pythagrin theorem of the triangle, so A squared plus B squared equals C squared. And then here you just replace it with the sine theta, the definition number, sine theta is equal to adjacent over hypotenuse. But in this case, it's 1, so multiplied by hypodonus is 1, just become sine theta. Same thing, it's cosine, it's adjacent of iPodonis, this one's opposite over iPodonis. And then where C is a unit circle equals 1, the radius 1. So then basically using Pythagorean theorem, oops, yeah, using Pythagorean theorem, we get the Pythagorean, I mean, yeah, Pythagorean identity. So co-squared plus sine square equals to 1. So that's this part right here. So again, just want to include that. All right, and now let's take a look at the solution to part B. And if we go back up to B, basically, so we have... We show it A is this, this is the H vector, and now we have to deduce that this equals to H, so R squared D Theta over DT equals H, basically the magnitude equals this. Essentially the magnitude is that. All right, so what I'll do is, I'm just going to copy this. All right, so let's go back to the solution. So it's pretty straightforward, because we know that, well, since H is a constant vector, we could write it as, well, before I write that, I'm going to go, we want this. All right, let's put it here. Okay, so we want, all right, we want to show that R squared D theta over DT equals H. All right, so we know, we know that, so H is a constant vector, so the H vector is equal to, I'm going to put this, the calculus book, just raises this gamma, it's gamma, and then k vector like that. Yeah, so basically it's a constant, so it's going to be just a constant vector, I'm just a constant scalar multiplied by k, it's in that direction, that unit vector, and where, where, I'll put this, whoops, where this one here, where alpha is greater than zero, and is a constant, and actually move it down here, to have some room, is a constant. All right, so thus, yeah, we know that, so thus, if it's a constant, well, it's just going to be the magnitude of it. So that's if it's a constant, and the magnitude of it is also a constant. So thus, what we have is, thus, this alpha is equal to, this the magnitude of this H, and it's a scalar value. All right, so in other words, what we have is, we have alpha, I mean, H vector equals to, equals to basically the magnitude H, and this one is equal to lowercase H. Like that equals to the multiply by the K vector, again, it's all constant, equals to HK vector, and then, well, we know that from, so we have that, but we also know that, well, we also know this part right here. So we also know H is that, so basically, in other words, this equals to, this is going to be, R squared, I mean, it's R squared like that, D theta over DT, K vector. And this is from part A, this is from, from A, like that. All right, so thus what we have is well, H is equal to this part right here. So that's what age is. So thus, or I'll say so. Yeah, just because we already have a thus there. So the, so we'll have H, the magnitude of the H vector is equal to R squared D Theta, and it's a, yeah, just a constant, D Theta over DT, like that. And yeah, basically this solves Part B there. So this is what we wanted, and that's what we got. So now let's take a look at the solution to Part C. So let's scroll back up to Part C, what is asked. So if A equals A of T, so the function of T, is the area swept out by the radius vector R equals R of T in the time interval T not to T as in the figure below. So it goes from T0 all the way to T there. and then show that the rate change of the area d a over d t is equal to one half r squared d theta over d t in other words well this part here is just h and that's concept so in other words it's that's the uh Kepler's second law so it's the equal area that's being swept with time. So the rate change of the area is always constant as you sweep around, as the planet goes around the sun there. All right, so let's take a look at that. What I'm going to do is I'm going to just copy this and paste it just because it's easier to be drawing it myself. So let's draw out a thin sector of the circle to obtain an integral. So we have, this is the integral, I mean, this is the area, A of T, and this is R-T-Nut. And actually, before you even get to that, I want to quickly recall from my earlier videos videos. Yeah, this first one, there's a second link to the one screenshot below this. on proving the area of a sector of a circle, because we're going to need this equation. So we know that the area of a circle is pi r squared or pi d squared over 2, but for a sector, so like this segment there, it's actually a equals 1 half r squared theta, like that. That's the area of a wedge like that. And again, it's from my earlier video on area of a circle proof, and I proved it like this setup here using. this circle equation and then using a section of it, and then area of a circle revisited, and this one, I use different integrals, this setup, but again, it's the same equation, you can go and, yeah, watch those videos for the derivations, but basically the area is equal to one-half r squared theta. So now what I'm going to do is I'm going to draw out a thin sector of the circle to obtain an integral. So I'm going to draw out from here to here. So let's say we have up to T there, and I'm going to draw a... slightly thin here, I'm going to call this, so let's say we have a theta here, from here to here is theta, and I mean, yeah, theta, this is theta, yeah, I'll write theta not, and then the other angles from here to here is theta, yeah, just theta, all right, and now let's say we go slightly further, and this is going to be an infinitely small wedge. like this, and this is going to be D-theta. So it's like infinitely small. Let's make this a bit better. D-theta, all right? And so what that means is, well, and now this length right here, from here, all the way over across here, so this length is going to be... Yeah, this is going to be D, yeah, this one's going to be R equals to the magnitude of the position vector, R like that. Like that, all right, so now we have a wedge, so we have a wedge like this. So what that means is, so this wedge, we could, we know that the equation of it is, we're going to call this DA. So this is going to be this area. I'll call this DA. So DA, yeah, this DA is equal to one half, I'll move it all the way across here, actually, yeah, I'll just put DA like that, all right, so that's DA. So infinitely small, yeah, change in area there, just a small segment there. So now what we'll write is, well, we know that D of A and its function of T, this equals to using this equation, a, of the, uh, the slice there, so one half r squared, and the radius is, again, this can be r, so this equals two, one half, we'll just write magnitude for completeness for now, it's a magnitude of r squared, and then…