I'm trying to understand the degree to which I'm at sea, anchored, or on firm ground, and how to firm up my understanding as needed.
I think Kepler's equation reduces to a task (call it $t$, which I'll describe better shortly) that is equivalent to finding a circle with equal area as a square. So:
1) Is it appropriate and correct to say that solving Kepler's equation reduces to squaring the circle?
I suppose you could tell that I'm not a real geometer, or algebraist, or:
2) What area of mathematics am I grasping at here? And am I using appropriate language?
3) Is all of this a reasonable way to explain why Kepler's equation is said to be transcendental? Is it also a reasonable way to explain why "how to square the circle" is an interesting question even now that we know there's no method using the compass and unruled straightedge (or, what I read is equivalent, the "quadratic closure of the rationals")?
Where I mentioned a task $t$, what I have in mind is finding a region partly bounded by a circular arc with equal area as a triangle. I got this notion when I read "Computing the position as a function of time". It presents a figure that it calls a geometric construction:
So when I ask in (1) whether I'm correct, I guess what I actually want to be correct about is that point $x$ (or equivalently $P$ or $d$) in the so-called construction is not "constructible" from points $c$ the center of the ellipse, point $S$ a focus of the ellipse, and point $y$ on the circle around $c$ of radius as large as the semi-major axis of the ellipse and with angle $Scy$ proportional to the time since periapsis, that is, the nearest approach of $P$ to $S$.
It is claimed that Kepler's equation provides that the area of region $zSx$, bounded by two line segments and arc $zcx$, equals the area of circular sector $zcy$. Removing the region in common and giving name $k$ to the point where $cy$ intersects $Sx$, we have equal areas of triangle $Sck$ and region $ykx$ that is bounded by two line segments and arc $ycx$. Then task $t$ is to find $x$ such that we get the equal areas that we have just mentioned where $k$ is determined as we have just defined.
As to squaring the circle, well, it is trivial to find a square with the same area as triangle $Sck$ and not difficult, I imagine, to find a circle with the same area as the region $ykx$.
After a brief chat with @OrangeHarvester I think it's this way: I was overreaching for a connection between Kepler's laws of planetary motion and the non-constructible.
Yes, Kepler's equation is in the nature of squaring the circle, in that it is transcendental, due to the use of the sine function.
I just learned, as I did not previously know, that the constructible numbers are a proper subset of the algebraic numbers, which are the complement of the transcendental numbers. The "impossibility" of squaring the circle, we remember, is that it requires using non-constructible numbers.
But no, Kepler's equation is not so special in that way. There were transcendental equations long before Kepler.
However, if we prefer, then Kepler's equation is a particularly lovely example of a transcendental equation, in the sense that de gustibus non est disputandum: in matters of taste there is no proof.