Counter-Earth

[Мастер]

https://en.wikipedia.org/wiki/Counter-Earth

Last paragraph under "Scientific Analysis" - it seems wrong to me. Does anyone have the same problem with it that I do?

[Lance]

This paragraph? It sounds reasonable to me. What exactly is troubling to you?

For a Counter-Earth orbiting the same path as Earth to always stay 180 degrees from Earth, the two planets would have to have circular orbits, but Earth's orbit is elliptical. According to Kepler's second law, a planet revolves faster when it is close to the star, so a Counter-Earth following the Earth on the same orbit with half a year of delay would sometimes not be exactly 180 degrees from Earth. To remain hidden from Earth, the Counter-Earth would require an orbit symmetrical to Earth's, not sharing the second focus or orbit path.

[tubeswell]

Its kind of like juggling I guess. They could be in the same orbit and the juggler-star is tossing each one onward more forcefully when it is in the near-sun part of the orbit, and each one is slowing down as they get to the far-sun part of the orbit, and they each take the same time to complete the orbit.

[Arneb]

I find no fault with it, but I'd be happy to think about anything you throw at me.

[Lianachan]

Wouldn't it be detectable due to the influence of its gravity on things like comets and Earth-crossing asteroids?

[Lance]

If they are in the same, identical orbit; one would be at its fastest while the other would be at its slowest. Q: Would the variation be enough to have the faster poke out from behind the Sun long enough to be observed?

[Arneb]

At least in radar observations, yes it would. I seem to recall that this kind of irregularities would make the whole setup unstable.

[Мастер]

If they are in the same, identical orbit; one would be at its fastest while the other would be at its slowest. Q: Would the variation be enough to have the faster poke out from behind the Sun long enough to be observed?

This would be the case if "identical" means the perihelion were at the same point relative to the sun. So the two planets would pass through the exact same points (measured by a fixed inertial coordinate system that always has the sun at the centre), just not at the same time. Then the criticism would apply; one is moving its fastest when the other is moving its slowest, so they wouldn't stay 180 degrees apart, but would "wobble" around the 180-degree separation point.

But, a counter-earth hypothesis would work better if the two planets had perihelions on opposite sides of the sun, so that they reached them at the same time. They would both be at their closest point at the same time, and at the farthest point at the same time, and would always move at the same speed.

Wouldn't it be detectable due to the influence of its gravity on things like comets and Earth-crossing asteroids?

I have no criticisms of these other methods of detection, just the argument that the two planets couldn't maintain constant 180-degree separation. They can (although I'm not sure whether it's a stable orbit).

[Halcyon Dayz, FCD]

Wouldn't it be detectable due to the influence of its gravity on things like comets and Earth-crossing asteroids?

Yes. That's how we know there isn't anything of significant size there.
Also, the STEREO spacecraft would have directly observed it on their way to their final destinations.

I have no criticisms of these other methods of detection, just the argument that the two planets couldn't maintain constant 180-degree separation. They can (although I'm not sure whether it's a stable orbit).

Such a set-up would be so improbable to occur naturally that discovering it would lead to some wild speculation.

[Arneb]

But, a counter-earth hypothesis would work better if the two planets had perihelions on opposite sides of the sun, so that they reached them at the same time. They would both be at their closest point at the same time, and at the farthest point at the same time, and would always move at the same speed.

No, that wouldn't work at all. It was discovered shortly after the publication of the Principia that there are almost no stable solutions to the gravitational equations if more than two bodies with significant masses are involved. ("three-body problem"). There seemed to be none at all, until Euler and Lagrangre discovered five special situations in which the setup can be stable. Thus, if counter-Earth had an orbit with a perihelion at Earth's aphelion, it wouldn't be in a Lagrange point, and the setup would be unstable right from the beginning.

As the article says, the counter-Earth idea only works for almost perfectly circular orbits. That probably has to do with the fact that L1,2 and 3 are metastable, and any small perturbation quickly leads to deterioration of the equilibrium. That is also why satellites at, say, the Sun-Earth Lagrange points (SOHO, for instance, is in L1), need fuel in order to maintain their position. It's a different story with L4 and 5, which are truly stable and tend to attract and collect things around them, acting almost as if there were a body with proper gravitation - witness the Trojan asteroids in the Sun-Jupiter system, and the single Trojan in the Earth-Sun system, 2010 TK7.

I read the speculation somewhere that the Earth-Theia collision that created the moon was not the result of a blind-luck, on-off orbital crossing but rather a Lagrange situation gone sour - the colliding body would have been in Earth's orbital plane all along, and the collision would hve been unavoidable from the beginning, rather than a freak event.

[Мастер]

But, a counter-earth hypothesis would work better if the two planets had perihelions on opposite sides of the sun, so that they reached them at the same time. They would both be at their closest point at the same time, and at the farthest point at the same time, and would always move at the same speed.

No, that wouldn't work at all. It was discovered shortly after the publication of the Principia that there are almost no stable solutions to the gravitational equations if more than two bodies with significant masses are involved. ("three-body problem"). There seemed to be none at all, until Euler and Lagrangre discovered five special situations in which the setup can be stable. Thus, if counter-Earth had an orbit with a perihelion at Earth's aphelion, it wouldn't be in a Lagrange point, and the setup would be unstable right from the beginning.

The sources I am finding (I don't know how to do a stability analysis, but I thought about it a bit today, and have some ideas I will try out later) suggest that L3 is unstable.

My orbital solution is an equilibrium - earth and counter-earth reaching perihelion at the same time (on opposite sides of the sun), and also aphelion at the same time, is a solution to Newton's second law and the inverse-square gravitational law. It may be an unstable equilibrium, although the sources I am finding suggest that it would be unstable even if the orbit were circular.

As the article says, the counter-Earth idea only works for almost perfectly circular orbits. That probably has to do with the fact that L1,2 and 3 are metastable, and any small perturbation quickly leads to deterioration of the equilibrium. That is also why satellites at, say, the Sun-Earth Lagrange points (SOHO, for instance, is in L1), need fuel in order to maintain their position. It's a different story with L4 and 5, which are truly stable and tend to attract and collect things around them, acting almost as if there were a body with proper gravitation - witness the Trojan asteroids in the Sun-Jupiter system, and the single Trojan in the Earth-Sun system, 2010 TK7.

The L3 point may be unstable; I'll work on that one in a bit. But the situation described in the article (earth at perihelion when counter-earth is is at aphelion) is not even an unstable equilibrium.

The paragraph I cited earlier strikes me as being of the form, "if we assume that counter-earth is not at the L3 point, then it doesn't maintain a constant position relative to earth." I agree with that, but I see no need for the assumption.

Stability is a separate issue. From what I can tell, my solution is most likely unstable. But it's better than the non-Lagrange point solution proposed in the article.

[Мастер]

Such a set-up would be so improbable to occur naturally that discovering it would lead to some wild speculation.

I agree that it is unlikely to occur, especially if (as suggested by the sources I am finding) the L3 point is unstable. My point is only, the argument in the article is specious.

[tubeswell]

TFT Dr Arneb. You get Lagrangian Points! (instead of Brownie Points)

[Arneb]

tft?

[tubeswell]

The effing truth

[Lianachan]

New one on me, though it makes perfect sense and I'm definitely going to start using it. Up until now, for me TFT was to do with TV/monitor displays and stood for "Thin Film Transistor".

[Lance]

Tough Fucking Titty = Too Bad So Sad = Sucks to Be You = Oh, Gee, I am So Sorry for your Misfortune = Life's a Bitch and Then You Die

[Arneb]

The reason why Mactep finds the explanation on Wikipedia (and by extension, mine) specious may may be that the article and I treat it as a fact or a given that no three-body combination is ever going to be stable except in the five situations described by Euler and Lagrange - and of these, the first three are only metaastable (that is, stable as long as nothing happens, and somehing always happens). So IF there was counter Earth, it COULD ONLY to be at L3, and since that conformation would steer it away from behind the Sun twice every year, ergo no counter-Earth. A case of two orbits whose long axes are at 180 degrees wrt to each other but have otherwise identical orbital parameters needn't be considereed because it wouldn't be an Euler/Lagrange situation, and simply couldn't exist for any legth of time begin with.

That sounds totally convincing if you accept the stable-only-at-Lagrange dogma as true, but pretty forced if you don't.

Mactep, did I get you correctly here?

[Lance]

So IF there was counter Earth, it COULD ONLY to be at L3, and since that conformation would steer it away from behind the Sun twice every year

Do we know for sure the offset would be enough to make it visible to Earth 1?

[Мастер]

The reason why Mactep finds the explanation on Wikipedia (and by extension, mine) specious may may be that the article and I treat it as a fact or a given that no three-body combination is ever going to be stable except in the five situations described by Euler and Lagrange - and of these, the first three are only metaastable (that is, stable as long as nothing happens, and somehing always happens). So IF there was counter Earth, it COULD ONLY to be at L3, and since that conformation would steer it away from behind the Sun twice every year, ergo no counter-Earth. A case of two orbits whose long axes are at 180 degrees wrt to each other but have otherwise identical orbital parameters needn't be considereed because it wouldn't be an Euler/Lagrange situation, and simply couldn't exist for any legth of time begin with.

That sounds totally convincing if you accept the stable-only-at-Lagrange dogma as true, but pretty forced if you don't.

Mactep, did I get you correctly here?

Stability is something else. Three of the Lagrange points are not stable, apparently (haven't done the analysis myself), including the L3 point.

The reason I disagree with the article is because it describes a counter-Earth which is not in the L3 point. In my scenario, it is in the L3 point.

If we have two objects in an elliptical orbit, they're in each other's L3 points if the reach perihelion at the same time, not half an orbit out of synch with each other. Newton's gravitational law is satisfied perfectly if the orbits are mirror images, as I propose; it isn't satisfied if the orbit is literally the same (reaching perihelion at the same point, not the same time), as the article proposes, unless earth and counter-earth are so small that their gravitational effects on each other can be ignored - and if that's the case, then any orbits would be OK.

[Мастер]

So IF there was counter Earth, it COULD ONLY to be at L3, and since that conformation would steer it away from behind the Sun twice every year

Do we know for sure the offset would be enough to make it visible to Earth 1?

If it were at L3, it would always be behind the sun. If it were not at L3, as described in the article, I'm not sure how much "wobble" there would be, but I'll try to work it out later.

[Мастер]

The reason why Mactep finds the explanation on Wikipedia (and by extension, mine) specious may may be that the article and I treat it as a fact or a given that no three-body combination is ever going to be stable except in the five situations described by Euler and Lagrange

I have not looked at the original work of Euler and Lagrange on this, and do not know whether they considered elliptical orbits. However, if they did, I am confident they would have come up with the correct analogue to the L3 point for a circular orbit, not the wrong one that appears in the article.

Again, this is not about stability - the L3 point is unstable even in a circular orbit (or so I've been told - I haven't worked out this part myself).

[Enzo]

WOuld their toilets flush in the opposite direction? I mean clockwise/counterclockwise, not upward.

Lianachan, you surprise me, I wouldn;t have guessed a digger in the dirt sort would be up on electronics terms. You probably know what OLEDs are too then.

[Lianachan]

Lianachan, you surprise me, I wouldn;t have guessed a digger in the dirt sort would be up on electronics terms. You probably know what OLEDs are too then.

I do indeed! Alas, I am not a professional digger in the dirt sort - I have a day job. Not one in electronics, admittedly, but definitely technology related.

[Мастер]

Here is what I have so far. I am trying to evaluate the integral in the last step, I think it can be done, but I haven't figured out how yet.

Next step is the two-body problem with both objects having non-negligible mass. Once that is done, I should be able to find the five Lagrange points with elliptical, rather than circular, orbits.
---------------------------------------------------------
\documentclass[a4paper]{article}

\usepackage{amsmath}

\begin{document}
\section*{Two Body Problem, One Large, One Small}
There is a large point mass at the origin of a plane with Cartesian coordinates. Another, much smaller point mass has position $x$ and $y$, which are functions of time. The law of motion is
\begin{equation*}
x^{\prime\prime}=-\frac{c x}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}}\qquad\text{and}\qquad y^{\prime\prime}=-\frac{c y}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}},
\end{equation*}
where the value of $c>0$ depends on the size of the massive object. Switching to polar coordinates,
\begin{equation*}
x=r\cos\theta\qquad\text{and}\qquad y=r\sin\theta,
\end{equation*}
the law of motion is
\begin{equation}\label{PolarCoordinatesConditionX}
r^{\prime\prime}\cos\theta-2r^{\prime}\theta^{\prime}\sin\theta-r\theta^{\prime\prime}\sin\theta-r\left(\theta^{\prime}\right)^{2}\cos\theta=-\frac{c\cos\theta}{r^{2}}
\end{equation}
and
\begin{equation}\label{PolarCoordinatesConditionY}
r^{\prime\prime}\sin\theta+2r^{\prime}\theta^{\prime}\cos\theta+r\theta^{\prime\prime}\cos\theta-r\left(\theta^{\prime}\right)^{2}\sin\theta=-\frac{c\sin\theta}{r^{2}}.
\end{equation}
Multiplying \eqref{PolarCoordinatesConditionX} by $\cos\theta$ and \eqref{PolarCoordinatesConditionY} by $\sin\theta$ and adding the two, the result is
\begin{equation*}
r^{\prime\prime}-r\left(\theta^{\prime}\right)^{2}=-\frac{c}{r^{2}}.
\end{equation*}
Multiplying \eqref{PolarCoordinatesConditionX} by $-\sin\theta$ and \eqref{PolarCoordinatesConditionY} by $\cos\theta$ and adding the two, the result is
\begin{equation*}
2r^{\prime}\theta^{\prime}+r\theta^{\prime\prime}=0.
\end{equation*}
The solution, using $\theta$ as the independent variable instead of $t$, is
\begin{equation*}
r=\frac{k^{2}}{c\left[1-a\cos\left(\theta-\theta_{0}\right)\right]}
\qquad\text{and}\qquad
t=\int^{\theta}\frac{k^{3}d\theta}{c^{2}\left[1-a\cos\left(\theta-\theta_{0}\right)\right]^{2}}.
\end{equation*}
\end{document}