Counter-Earth

[Мастер]

Can this be simplified?

[Мастер]

Actually, I like this one a little better, because cosines (unlike tangents) don't blast off to infinity.



I'd like to simplify where it says <arccos(stuff*cos(stuff))> though.

The idea here is, we have a description of an orbit around a large object located at x=0 and y=0. In polar coordinates, I can work out a closed-form solution for an elliptical orbit, in a certain sense - you can choose the angle your are interested, and the formulae tell you how far the orbit is from the centre at that angle, and also at what time that point should be reached. I can't invert the formula and get position as a function of time, though.

The arccos funcion is badly behaved. Every time the cos function included in its argument passes through zero, the arccos jumps between 0 and pi. This formula tells you what time it is - so, at time is this object at -1 degrees? At one time is it at -0.01 degrees? At what time is it +0.01 degrees? As soon as you cross zero, the formula jumps backward in time, and tells you the *last* time the object was at +0.01 degrees. I'm wondering whether there is a way to do better.

[Мастер]

This one is better.



I still wonder whether it is possible to simplify the arccos term.

[Мастер]

I like this one best of all, so far at least.

[Arneb]

Mactep, not that I could follow you (that's far above my level of math) - but your first equations come out as code on my screen, while all the others come out as sleek-looking Tex equations style.

But maybe I get your argument verbally this time: Since Earth and counter-Earth have non-negligeable masses, they would have to have the same distance from the Sun in order to be at eacho other's L3 . Therefore, they would not share the same orbit but would be in two otherwise identical orbits with their perihelia and aphelia oriented at 180° from each other. Therefore, they would NOT necessarily deviate from their exactly opposite positions wrt the Sun, as they would be in perihelion, aphelion, or any other equivalent points of their respective orbits all the time - and therefore, the Wikipedia argument (and therefore, mine) of visibility close to the Sun is spurious. Do I get it this time?

[Lance]

Sometimes I am amazed by the discussions we have on this little, bullshit forum of ours.

[Enzo]

I see them the same way Arneb does, all look good but the first.

As to the math, well..... way over my head. I was a physics major 50 some years ago, but it didn't stick.

[Lance]

I like this one best of all, so far at least.

=42

[Мастер]

Huh. I wrote a long, detailed response to Arneb's post, but I don't see it now.

[Мастер]

I see them the same way Arneb does, all look good but the first.

The first one is a self-contained LaTeX document, to be cut-and-paste into a text file, then compiled with pdflatex (probably some other LaTeX compilers will do). The others were short snippets of LaTeX code, displayed with the help of an on-line service that Llance found a while back.

As to the math, well..... way over my head. I was a physics major 50 some years ago, but it didn't stick.

The expressions are complicated (except for a=0, in which there is an incredible amount of simplification), but the basic idea isn't that bad. The sun is at (0,0), and exerts gravity according to Newton's inverse-square law. So the trajectory of the planet, given by x(t) and y(t), must always accelerate towards the sun, with the magnitude of the acceleration given by the inverse square of the distance to the sun.

It turns out to be easier to use polar coordinates, r(t) and theta(t). However, I can't find explicit solutions in terms of r(t) and theta(t), so I had to choose theta as the independent variable. So I found r(theta) and t(theta). Those are the expressions given.

So, the concept isn't too bad. The devil is in the details.

[Lance]

Huh. I wrote a long, detailed response to Arneb's post, but I don't see it now.

Perhaps you saved it as a draft?

[Мастер]

Huh. I wrote a long, detailed response to Arneb's post, but I don't see it now.

Perhaps you saved it as a draft?

I can see a "Save as draft" option at the bottom when posting, but I can't find where they go if you choose it. Where are the drafts?

[Lance]

I can see a "Save as draft" option at the bottom when posting, but I can't find where they go if you choose it. Where are the drafts?

User Control Panel -> Manage Drafts

[Мастер]

I can see a "Save as draft" option at the bottom when posting, but I can't find where they go if you choose it. Where are the drafts?

User Control Panel -> Manage Drafts

Nothing there

[Lance]

[Мастер]

I don't think I imagined writing the response

[Мастер]

Trying to recreate the lost reply . . .

Mactep, not that I could follow you (that's far above my level of math) - but your first equations come out as code on my screen, while all the others come out as sleek-looking Tex equations style.

The first is a stand-alone document that will create a PDF file if you copy it into a text file and compile it with pdflatex. The others are much shorter snippets, created with an on-line tool that Llance found. This on-line tool will not handle a full document.

But maybe I get your argument verbally this time: Since Earth and counter-Earth have non-negligeable masses, they would have to have the same distance from the Sun in order to be at eacho other's L3 .

I believe that would be the case if they had equal mass.

Therefore, they would not share the same orbit but would be in two otherwise identical orbits with their perihelia and aphelia oriented at 180° from each other.

That's what I'm getting - their orbits would be "mirror images" of each other, and the two would reach perihelion at the same time, and aphelion at the same time. This would be the case, even if they had unequal masses - in that case, the orbit of one would be a scaled version of the other. But they would still be mirror images, reaching aphelion simultaneously, and reach perihelion simultaneously.

Therefore, they would NOT necessarily deviate from their exactly opposite positions wrt the Sun, as they would be in perihelion, aphelion, or any other equivalent points of their respective orbits all the time

That's what I'm getting. The "mirror image" solution I propose would be an unstable equilibrium, i.e., the two objects would orbit in perfect synchronicity forever, except that the tiniest perturbation of the orbit would break the system. Much like a pencil balanced perfectly on its point - it's an equilibrium, until someone sneezes. By contrast, the situation described in the article does not seem even to be an unstable equilibrium, unless the two objects each have negligible mass. (And in that case, they could have just about any set of orbits.)

I'm pretty confident of these results, but I want to get it down to rock-solid mathematical results. I need to work a little bit more on that. My formulae are OK for a single object orbiting a much larger object at the origin, except that my time-as-a-function-of-angle formula is really only correct for half the orbit. The problem is the "arcsin" function, where it is necessary to choose a branch of this multi-valued function that is different than the standard branch to describe the other half of the orbit.

[Heid the Ba]

All well beyond me, but entertaining none the less.

[Arneb]

Thanks, Mactep, for your patience and effort. I feel a lot better know. I think I understand why you called the articles' argument "specious" instead of just "wrong" or whatever: by dictcc, "specious" is superficially convincing, but not holding up to scrutiny.

I'd be happy to see more equations, even if I'll need someone like you to walk me through it.

[Lance]

I don't think I imagined writing the response

The only other thing I can think of, because I also continue to do it once in a while, is hit [Preview] then, believing I hit [Submit], navigate away and lose the post.