3Speak - YouTube - Telegram - Summary - Notes - Sections playlist - Vector Functions playlist - MES Links In this video, I derive Kepler's 3rd Law of planetary motion and obtain a formula for a pla...
Transcript
the solution to part two, and this one we're asked prove Kepler's third law. All right, so let's read this out again, then go over parts A, B, and C. So, let T be the period of a planet about the sun. that is, T is the time required for it to travel once around its elliptical orbit. Now suppose that the lengths of the major and minor axes of the ellipse are 2A and 2B, and it says part A, use part D of problem 1 to show that T equals 2 pi AB divided by H. And then part B says show that we have this H squared over GM equals ED. equals b squared over a then part c says use parts a and b to show that t squared equals 4 pi squared over gm times a cubed and also notice that the proportional uh the proportional constant right here And actually, yeah, it's pronounced proportionality constantly, I mean, spelled this one here. So proportionality constant 4 pi squared over GM, it's independent of the planet. Basically, this is the mass of the sun, and this is the gravitational constant, so it doesn't matter about the sun there. So it could be considered as a constant there. All right, so since it's independent of the planet, this proves Kepler's third law, basically the square of the period of revolution of a planet. So right here is a square of the period. The planet is proportional to the cube of the length of the major axis of its orbit. And there's a cube right there. So it is basically, yeah, a cube, so proportional. So t squared proportional to a cube there. So fascinating stuff. Alright, so now let's take a look at the solution to part A, all right, and then A, we're asked, use part D of problem 1 to show that T equals 2 pi A, B over H, and then part D of problem 1 is basically show that this is constant, D,A, over DT, equals to this setup, and that's equals to constant. And before we get to that, we're given that the supposed the lengths of the major and minor axis of the ellipse are 2A and 2B. So let's draw that out first. So if we have an ellipse, let's say we have an ellipse and we just centered about the origin. I'll just make it. Actually, I'll just draw this out using the built-in thing here on one note. So let's say we have an ellipse like this. So it's an ellipse-ish. It's somewhat like an ellipse. So let's say you have an ellipse like that. There's the origin. and then basically the minor or the major axis is going to be length, well, the radius A, I mean the point from the origin to the end is A, so then the length of major axis is 2A, so that's A and A. And this right here is the minor axis, so that's going to be B, and then this bottom one is going to be B. So in other words, it is 2B, minor axis, major axis is 2A. Alright, so now we're going to be given use Part D of problem 1 to show that the period equals 2 pi AB over H. So we have this right here. All right, I'm just going to move it here. I'm just going to copy that and paste it right here. So let's say we paste it. All right, so we have this, and again, it just equals to 1.5H. It's all the matters. So like that, and this is constant. And now let's write this out again. So this is constant. All right, so DA over DT is constant equals to one half H. All right, so what happens is, so I'll just write since, all right, since, what we have is now the area, a of T, is going to equal to, well, the integral of the area, D a over DT. You know, basically we move with this DT over to this other side, so then basically integral of D.A. and this equals to, again, this proper notation, one half, and now it's going to be h d t. And now this equals two, so what is this equal to? This equals two, it's going to be, well, one half, it's going to be just a constant, and then multiplied by, well, integral of H is a constant, so it's a constant, so we just add a T there. So the integral is just going to be HT. and then plus, well, we're going to have to have a constant of integration. I'm going to call this C. I'm called C1, so this is just a constant of integration. So this right here is just a constant of integration. So we integrate its constant, just because when you take a derivative, it just, constant goes to zero. So we can have any constant we want, like that. All right, so we know that, but we also know, but... We know that at basically A of 0, A at, well, T equals 0, this just equals to 0, because if you look at the area, the setup over here, let's go all the way back here, so this area, if you have the angle 0, and it hasn't gone anywhere, it's just going to be a thin, zero line, yeah, just a line of zero area. if it hasn't traversed anything. So that area swept is just zero at the beginning. So A at equal 0 is equal to 0. So what we have is A of 0 equals to 0, and this equals to plug this inside 1 half, well, 0, this is going to be 0 plus C1. So in other words, the constant of integration is equal to C1. So that means and a of t is just equals to one half two h t. So that's what this is. And that is what the area is with a time t. Alright, so going further, so note that the area A at T equals to capital T, in other words, equals to period, as it does a full revolution, is the area of an ellipse, whose formula is given for my earlier video, and there's a link to the notes, and there's a thumbnail, it's basically the area of ellipse is equal to pi times b times a, where A is... this axis there, the major axis point there, and this is B is the minor axis there. So that's the area there, and I proved it over there, and I compare that with a circle pi r squared. Again, ellipse is just a special kind of circle where it's the A and B are equals to radius. Like that, so that's why you have this pi r squared. It's actually ellips is just pi a times b. So a circle is actually a special case of ellipse. I can consider that. So that's pretty fascinating. So what that means is, well, we know that a ellipse, I'm just going to put this in completeness ellipse, equals to pi times a times b, all right? Thus, that the area at t equals to capital T, where this is the period, All right, so that's a period. So in other words, it makes a full revolution. So it goes basically a full revolution, so we got a full lips. That is when T is equals to the period, and so on. So what that means is this is going to equal to, well, we know that, well, first I want to write this out, we know that it's going to be this A over T, so one-half H-T, but then we put a capital T there. So this equals to one-half-h capital T, and now this equals to pi times a times b and again where a and b are this length right here this is a at a zero and then this is at b at zero b like that all right so if you're going back over here i just moved it i just moved it in the center there but it was going back here so what we happen uh what we have you now is we move this over there and solve for t we get uh so the yeah so then we'll get uh right here t is equal to, this is going to be the two is on top, so two, and then divided by, this is going to be divided by H there. Yes, we're going to get a 2 pi, AB over H. And I'll move this a bit in here, like that. I'll move this better, and move it right here. All right, so there's a period, and that's what we wanted to solve. All right, now if we go scroll all the way back up, well, we want it to show, It equals to 2 pi AB over H, and that's what we just solved. So there it is, we just solved that. Let's check mark. All right, and yeah, now we have that. Let's take a look at the solution to part B, and this one states, well, go all the way back up here. So part B show that H over GM equals ED equals B squared over A. And again, number ED is from the equation of the ellipse from part one. I am not part one, from the first law derivation. So a recall from the earlier proof of Keper's first law for the value of E .D, which are the eccentric. and directrix of an ellipse in polar coordinates. I'll write that in here. In polar coordinates. Yeah, all right. So that's using the unified conic theorem setup. So this is what we want. I'm going to write this here. So we want here, so we want. So we want to show that we have this setup. Basically, H squared over GM equals ED equals B squared over So what I'm going to do now, I'm going to scroll all the way back up and then find the equation of the setup there. I'm going to go all the way back to the first law derivation. So if we go all the way back, and that was before the applied project. So we had this setup. All right, so we have this. I'm going to just copy this. All right, so we have that, and we already know those values, ED. So we're going to go all the way back down here and just paste it straight into here. All right, so we have that, so I want to paste this here. All right, this is, I'll put it like this. All right, so from, this is from Law 1. I'll put that, like that law, 1. So from law 1, we have this R equals ED plus a divided by 1 plus E cosine theta, where E equals C over GM and D equals H squared. Well, we could easily solve for ED. That's just, we already know what E and D are. So that means, so thus, E, D just equals 2, well, C over GM, times it by this, we just throw these inside, that's all it is. So, yeah, that's what we have. So, obviously, this like that, times it by H squared over C. In other words, this cancels, and we're just left with an H squared over GM. All right, so we have this first part. So we got this. Now we got to show that it equals to B over A. Alright, and now what's needed, the next step, to prove that it equals to B squared over A, the major and minor axis, we have to basically convert this back into the normal ellipse setup. Instead of using the polar ones, we need to use the XY set up like this. And this is from earlier video, so again, it's a recap on that. So the ellipse we are dealing with is just horizontally shifted from the origin. as in the, yeah, as I illustrated in my earlier videos, so there's a link to that, as well as obtaining the values of A and B in terms of E and D. So for my earlier video, screenshots off from my calculus book, I just need her than my writing there. So recall that if we had this equation, R equals E, D divided by 1 plus E cosine theta, and again, that's what we derive for. the planet around the sun, and then where f is the focal point focus, and that's where the sun is. So I'll just draw this, the sun there. All right, so that's the sun, that's a bright spot. And then the directrix is, is on the, on the side here, X equals D directrix, and then basically that is a, yeah, that D part right here. So that's a D there. And then the eccentricity is, remember, eccentricity is going to be, well, point on there, and then this is going to be, well, this is PeakD there, yeah, or PL or PD. I think it's just, I'll write this as, yeah, PD, or PL, PeakD, this matter, I'm going to scroll back up, just to recap quickly on ellipses, and that's right over, it's going to be way above. Yeah, it's right here, so I chose the L there, or you could also put D, it doesn't matter, D, or PL, etc. So again, that's, and the E-centric is just the, it has to be constant there, so the point from P-F, so this point P-F, divided by PL. All right here, I just quickly teleport it back here, and again, this could be L, and so it doesn't matter. Anyways, that's just what the E is. All right, so we already have that covered, and then I showed, I basically derived this, made it back into the, the normal ellipse equation set up, and what we end up having as if E is less than zero, we recognize this equation, 3 here as the equation of an ellipse, in fact, it is the form of, this right here is a horizontally shifted one, and this is horizontally shifted there, all right, so basically x minus h over a squared plus y squared over b squared equals 1, where, and then now you saw this h equals to negative e squared d, total by 1 over 1 minus e squared, and then a squared equals to e squared, all squared and then b squared is e squared d squared divided by one minus e squared so again i'm not going to derive these again so make sure to watch my earlier video on that and we're going to just use these results namely this one and this one here all right so now that i've recap this so yeah what we want to do is again we want to get ed equals to this setup here b squared over a squared and we already know that uh ed is equal to h squared over gm so basically relating these elliptical orbit of the Earth around the Sun or the planet and then with these major and minor axis. And then that's from over here. So what I'll do is, so thus, let's write down these two equations, the ones we want, because there's an ED in both them. But notice that they're squared. So what we have is A squared is equal to E squared D squared over 1 minus E squared So keep an eye on the squared, and then squared. Whereas B squared only squares the top part, E squared over D squared, and then the bottom is just 1 minus E squared. So you don't square this one here. Notice this one is squared, and then this one is not squared, the 1 minus E squared. Squared. Alright, so that's what we have. So what I'm going to do now is, well, if I take the square root of that, this is going to be squared both sides, same thing over on the other side. like that and like that. So what we get is, let's put this like this. So what we get is, we're going to get a A squared. I mean, A is just going to be like that, and then we have the top E, D over, this one squared, this is going to be 1 minus e power of 2, whereas B is going to be, well, B equals to E d over square root 1 minus E squared. All right, and again, number is all plus or minus, we just take the plus, as always, because you could have plus or minus for some reason. Anyways, so that's what we have. All right, but what we're, but what we are actually interested in is, is actually the, the ratio. So if we, remember here, we want B squared over A squared, so that's what we want. But we know that, well, A squared, I'll just, I'll just write this here, A squared, I mean, A divided by B, I mean, A divided by, yeah, A squared, divided by B squared. Yeah, so what this equals 2 is, well, we're going to have, this would cancel with this, and then this, only the square is canceled, this one is canceled with this. Because you're just dividing it by it, so we'll just erase all this, and erase, and erase, all right? That's what we have, so this is going to be, well, just EDs all cancel, we're going to have a 1 over, 1 minus, E squared, yeah, so basically the only thing it has more is just this, this is an extra 2 squared there. So that's what we have there. All right, so what I'm going to do now is I'm going to take this right here and then multiply it inside there. So what we get is an ED equals to a times by 1 minus E squared. And again, what is that equal to? Well, let's just, well, we know this E squared right here. This right here. 1 minus e squared, well, you could just move it over to this side and then make, move this, and then flip these around. So then if you flip these around, this would be on top. So we just flip those around. So in other words, we could just have it like that. This is just going to be equals to, and then basically throw that in here. Like that. So what we're going to get is, and now because it's the a minus e squared on the numerator, this is a denominator, we just flip these, this equals to times it by b squared, so we flip these over a squared. Alright, so that's just this part right here, yeah, let's put it like that. And this equals to, well, this cancels with this, and we're just left with a b squared over a. Alright, so we solved it. There it is. Alright, if I go back up here, we have B squared over A, so now we basically solved everything, H squared over GM equals this right here, H squared equals ED, and then which equals to B squared over A. So let's just write that down, so thus, so thus we solved, we have H squared over gravitational constant times mass of the sun equals to e-centricity times by directrix equals to the minor axis squared over the major axis. not squared. All right, so that's what we have. So solves all this. All right, yeah, so now we have this and put a checkmark. Now let's take a look at the next part, and that's solution to part C. So let's just scroll all the way back up to Part C, and that's right here. So use Parts A and B to show that T squared equals 4 pi squared over GM. times a cube. So use parts A and B, and basically use this equation and these equations, and then basically solve for t squared in terms of this A there. All right, so what we'll do is, let's just put that in there. So from part A, yeah, from A, we get, well, T, remember this one is equal to 2 pi, That's this border here, 2 pi AB over H. 2 pi AB over H. So 2, oops, 2 pi A, b over H. And then from B, we just have this part, top there, is we have an H squared over GM over, I'll fix this up. GM equals to E, D equals to B squared over A. All right, so though we have that, so then what we want to do is get the t squared. So thus, if we square t squared, we're going to get, well, this just becomes all squared, so it becomes 4 pi squared, and then a squared, b squared, like that, over h squared. All right, so over h squared. Alright, so what that means is, well, look what we have. We have a B squared over an H squared there. So yeah, we have this right here. Yes, we have a B squared over H squared. What that means is, well, we could just, yeah, take this H squared, move it over to the bottom, and then move the A over to the top. In other words, What we get is, I'm going to, I'm going to just, yeah, I'll just leave that here, but what I'm just going to write here, it's for completeness, just for ease of you. So this is A squared, times it by B squared over H squared. And also we have this part right here. So that's what we have, and then we also have this, is going to be now, well, we have H squared, I'll just put end. So we have that, and, h squared over gm equals to b squared over a squared i mean over a so then basically this move this around move this down and move this one up what we get is a b squared over h squared equals to a over gm again we just move it around like that basic algebra Alright, and then we can throw this inside there. So, so we get t squared equals to 4 pi squared, a squared, and then multiplied by A over GM. Yes, so the B is gone, yeah, A over GM. In other words, we get 4 pi squared over GM. and then this a squared times a is just a cubed, right, so we solved it. So there it is, so that's quite fascinating. So, t squared, yeah, so I'll put this all together, yeah, so we just have, t squared is equal to, so it's proportional to, and these constants right here are independent of the planet, that's just gravitational constant, that's mass of the sun, and pi is just pi, and that's four is four squared, times by a. cubed. And yeah, so the period of the square of the period of the planet is proportional to the cube of its major axis of its elliptical orbit. It's fascinating stuff. All right, yeah, so this basically proves Kepler's third law, notice that the proportionality constant is independent of the planet. And again, now we could just go all the way back up here, and then just double check it. there is the problem right here all the way here where's the problem and that is uh yeah this right here so t squared equals 4 pi squared over GM times by a cube. Again, just double check down. Let's go back here. Yeah, so that proves Koupler's third law. This, yeah, this proves Koubler's third law. Notice that the proportionality concept is independent of the planet, and the square, basically, here's the third law, 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. Again, fascinating, fascinating stuff.