## Philosophy 167: Class 13 - Part 13 - De Motu Corporum in Gyrum: Theorem 4, Problem 4, and Two Key Scholia.

Smith, George E. (George Edwin), 1938-2014-12-02

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Synopsis: Further mathematical analysis of De Motu.

- Subjects
- Astronomy--Philosophy.
- Astronomy--History.
- Philosophy and science.
- Celestial mechanics.
- Newton, Isaac, 1642-1727.
- Genre
- Curricula.
- Streaming video.
- Permanent URL
- http://hdl.handle.net/10427/012724
- Original publication

ID: | tufts:gc.phil167.157 |

To Cite: | DCA Citation Guide |

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Theorem for supposing that centripetal force be reciprocally proportional to the square root of the distance from the center. The square is the periodic times an ellipses, r as the cubes from their transverse axes. And you notice, he got messed up with the figure. I'll do the proof in just a moment but I wanna pause for a second to look at this.

What's he saying? By my account, well, inverse square at least is the dominance force. Ha, we get the ellipse. If it's an ellipse, we don't need anything but inverse square, that's the way I wanna phrase it. And what about the three halves power rule? We only approximated it before because of the changes in motion.

Lo and behold, the inverse square gives you the three halves power rule all over again. Not in an if and only if now, just one way. If it's confocal ellipses under inverse square, then the three halves power rule is going to hold across it. So it look like all three of Kepler's principles come right together with inverse square of centripetal forces.

And the only issue is if versus if and only ifs. And that's principally on that. You have to be careful. And I will at the end, be careful by showing you what's there. This is a complicated proof and I'm running out of time. I'm gonna just describe it briefly.

What he does is take the ellipse and take the equivalent circle passing through the point s going to point p. And he requires the two to have the same periods. Okay? Then he sets up an equal times, the force at this point and the force on the circle, this sp being the radius of circle, he sets up and compares the force in the circle to the force in the ellipse.

And the upshot of this is, and I'm saving time so that it's not that hard to work through, and when I get to the, well the Principia doesn't use this proof so you won't see this again. What it comes down to is the little area, they are doing two equal times.

One in the circle and one in the ellipse. And it turns out, the ratio of those two areas is the same fraction of the total area of the ellipse to the total area of the circle. But that means the periods have to be the same in the two.

That's the way I should say it. And therefore, if I have confocal ellipses, I can compare them to the concentric circles of this sort and since I know the three has power holds for the concentric circles, it has to hold for the confocal ellipses. In other words, I reduce the problem to the circle case and draw the inference from the confocal ellipses then from the fact that it holds for the circles.

It's a nice proof, actually. The proof here makes more clear what's going on, than the proof you'll see in the Principia, which has no diagram like this. It makes clear the equivalent circle is a circle that passes through the end points of the minor axis. That has to be the equivalent circle.

Okay, so I more or less, I hope I've given you enough that you can figure that out. I wanna move ahead because there's other problems, at least two more problems. A scholium to that, hereby in the heavenly system from the periodic times of the planets ascertained the proportions of the transverse axis of their orbits.

That's Horrock's principle, if we know the periods, we can use the three as power rule to get the transverse axis, he's just shown that that's totally legitimate. Therefore it's law-like, therefore we can infer the axis. It will be permissible to assume one axis from that the rest will be given.

Once their axis are given, however the orbits will be determined in this manner. Now, I'm not gonna describe the manner, it's actually something Halley had published while he was an undergraduate. And it's pulled out of Astronomia Nova, either Halley discovered it independently of Kepler or Halley learned it and decided he maybe didn't appreciate Kepler had it entirely, and published it on his own.

It's his first published paper for Halley. And the striking thing to notice here is the number of different lines that are gonna intersect to get the other focus, that's where that other point is, where's the other focus lie. Okay, you know the axis, you know where the Sun's in one focus, you now have to get the other focus.

What he's allowing for here is not imprecision of observation, though it could be that. What he's allowing for is, you're taking observations from a moving Earth and you do not necessarily know the Earth's orbit. So rather than depend on knowing the Earth's orbit, use more than enough different observations and take the average of them.

From the average of the intersection points. This is first place we know of where anybody uses averaging to compensate for uncertainty. Okay? It doesn't show up in the Principia for reasons you'll see next week. But it's striking that Newton is the first person to start using averaging to compensate for uncertainty.

There is a parallel result, second half for the inner planets, that uses maximum elongation instead, but again it's taking an average of locations for the empty focus. Which is very, very clever. Once, of course, you have the length of the major axis and you have the two foci, you've got the others.

Cuz remember, it's the points on the circumference are the sum of the distances from the two foci, that's a constant, so you have that automatically. Okay, problem four, supposing the centripetal force be reciprocally proportional to the square root of the distance from it's center, and with the quantity of the force known, there is required the ellipse with the body shown, described when released from a given position with a given speed following a given straight line.

Two points about this that are important. The first is, he's assuming it's an ellipse or at least a conic section. He is not taking the general question. Let the force be inverse square, what's the trajectory. He is rather saying, which ellipse is it, given the inverse square rule and a velocity at some point p in a direction and a magnitude.

Okay, that's the first point. Second, how does he specify the strength of the centripetal force? By giving a referent circle in effect. Now, this is a key point that I'll be using all the rest of the year, and we still use it today. The strength of the centripetal force, the center is always a cubed over p squared, that's a constant.

Then the centripetal force just varies with the square of the distant from that center. So a cubed over p squared gives you a measure of what we call the absolute field strength now. That's how we do it still, by the way. That's how we get absolute field using that.

And that he portrays in the form of a reference circle out here. That's how you represent that. I'll come to the solution in a moment, what I wanna, of course, point to is the scholia, a bonus indeed of this problem, once it is solved is that we are now allowed to define the orbits of comets and thereby their periods of revolution.

Whoever said anything about comets having a period of revolution? And then to ascertain from a comparison of their orbital magnitude eccentricities, inclinations to the ecliptic plane. And there are nodes ,whether the same comet returns with some frequency to us. That's a remarkable statement and using my prior comment about a surprising striking consequence of the theory.

That's gonna be it. Cuz it's one thing to say six planets go in particular orbits in a manner that has a ratio among them. Comets sweep completely across that. So, comets are showing you the inverse square force works at all, it fills the space. Comets come in at all angles.

The inverse square force fills the space. Newton actually uses that expression in the Principia, filling the space around the body. We call that a field, but that's a modern, that's a post Maxwell term. That's certain that scholium is certainly a candidate for breaking Halley back to Cambridge, but I'll leave that up to you.

It turns out this method doesn't work for comets. In fact, it turns out that he was still in the middle of 1686, trying to find a method that works. The problem is observations of comets, very small imprecision, given the way they move, makes it extremely hard to determine the trajectory.

The reason is, they're so highly eccentric, ellipses are, they're so eccentric you can approximate them by parabolas. But they're so highly eccentric they're very hard to catch. I'm gonna push on for the moment. This is the full solution of that. And I wish I had the time to go through it.

The important thing, I've given you a sketch of how to do that. It should not be that bad to get down to this point. When you get to this point, by the series of steps, you get this relation between sp, which is known, and ph, which is unknown, and this relation, we've got alatis rectim etc.

So once you get the length of ph that determines the other focus and everything falls into place. But this is a funny quantity because when you determine this l, there are three possibilities. It can be less than this quantity, in other words, this whole thing can be, what do I want to be?

If it's less than l, less than that, then this is a number greater than one and you get an ellipse. If this number becomes one, l becomes to equal to that, you get a parabola. If l becomes greater than that, this number is less than one, then you get a hyperbola.

And Newton recognized that. In other words, Newton recognizes that, given the initial speed of a body, at some point relative to a force center, and the strength of the force center. That speed and its location and that strength will tell you whether you have an ellipse. And if so, which one?

Whether you have a parabola, there's only one, or whether you have a hyperbola, and if so which one. And it all depends on the combination of the velocity, the speed, the angle, the distance from the center and the strength of the center. Okay, so we've got a full solution of what's called the initial value problem under the premise that we can only get a conic section.

And it's this solution only works, presupposing throughout your working with a conic section, but you're leaving which conic section?