Philosophy 167: Class 3 - Part 11 - Astronomia Nova, Part 4: Using Triangulation to Show that Mars' Trajectory is An Oval.

Smith, George E. (George Edwin), 1938-


  • Synopsis: Shows how Kepler proves the orbit of Mars is an oval.

    Opening line: "And then, he does now the typical Kepler thing to do. He starts looking at other ways of considering this whole manner."

    Duration: 6:22 minutes.

    Segment: Class 3, Part 11.
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And then, he does now the typical Kepler thing to do. He starts looking at other ways of considering this whole manner. And the best way is to do triangulation. Because what? He knows the following. He knows the heliocentric longitude of the Earth at every moment. That's from his theory of the Earth.
He knows the distance between the sun and the Earth, which is variable. That's from this theory of the Earth, the modified with bisected eccentricity of the Tycho Theory. He also knows the geocentric position of Mars at every moment from observations of Tycho. He may have to interpolate among those observations, but he knows them.
And finally from the vicarious theory he knows the heliocentric longitudes of Mars. So he can do triangulation. And that's what he proceeds to do. He starts triangulating distances on either side of Mars. Either side of the aphelion. And comes to conclusions that the aphelial distance is 166,510. Notice six significant figures.
Perihelia distance 138,173. Which means the semi-diameter is 152, 342, and the eccentricity is, on those units, 14.169. Or, if he does the eccentricity in the classic way, it comes out 0.093, or 9,301 out of 100,000. That's very close to our orbit for Mars. So, what does he say, having gotten that?
And, by the way, he notices here, I'll stress it in the next slide. He notices when he does the triangulations, it's symmetric on either side of the line of apsides. I'll come back to that. Nevertheless, because our observations, especially at perigee, do not bear out that great a difference.
And since it can happen that the vicarious hypothesis since it is false, also might introduce some falsity into the eccentricity. All the votes should be counted before the result is announced. That's a literal translation of what he says. All the votes should be counted and so we shall adapt.
I think that's right. The applehelia distance found here. 166, 510 to the eccentricity of Chapter 42. When he comes up with a temporary orbit. I'm not gonna go on and read the rest of this. You can read it on your own. But it's an orbit in which you're uncertain about what the aphelia distance is.
You're uncertain about where the perihelia distance is. You're somewhat uncertain about what the eccentricity is, but you're reasonable close on all of them. Now what's the problem here? Why can't triangulation pick out the ellipse? Observations are imprecise. You're too sensitive to them. You need much better than two minutes of arc to settle on the specific ellipse.
A further set of triangulations where he wants to see, I'll just run the argument. It's a beautiful argument. So let me just read it. This is the figure that goes with it. I'm not going to bother to explain the figure. But you can see the two orbits and a general idea of what's going on.
For the peculiar strength of the method I have used is this. But it shows that, at whatever point at the plane of the Earth's circle is chosen, it has a determinant position with respect to the Sun's body. Both things are dialogical longitude and in distance from the Sun described to a number of observations, also shows the distance of Earth and Mars from that chosen point.
And it does these things without any knowledge of the equated anomaly of the eccentric corresponding to that point. What he's just said is, he's got the distance of Mars without assuming the motion. That's in effect saying, I've got something Copernicus didn't have. In fact the only reason why I use that knowledge in Chapter 26 was that it was a short cut.
But in addition there is another way to argue the point. It was proved in Chapter 44 that the planet's orbit is not a circle, but an oval, such that the diameter on it, which is called the line of apsides is the longest. Just now, in chapter 51, it was proved that regions that are equally removed from the point of the aphelion G, also make an equal incursion at the side.
That is it's symmetric about the line of apsides. There is thus a real oval situated about the line AC, and therefore it is not situated about the line FH. And one who would compute the various distances of Mars from the point H by the method just recommended, will find a great irregularity in the distances incapable of being included by any means in a circle, or in any other possible figure set up about at FH..
So again, the faith that was pledged in Chapter 6 and in many other places in this work. I have redeemed from all the tincture of self justification and have shown that the asytric of Mars cannot be referred to anything but the sun itself. And that in addition it is not only reason that stands with me, but the observations themselves in my releasing the observations of Mars from the Sun's mean motion In measuring them out by the apparent motion of the Sun.
So he's using triangulation to reach a lot of conclusions. But he can't use it to say what the specific trajectory is, other than it's one of a number of possible ovals very close to one another. Fair enough? That's where he stood and he stood there for a long while doing lots of calculations like this.
I'm showing you only a subset of them that are in the book and they are subset of the ones in the notebook.