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I am designing a new exoplanet. Do these physical parameters seem plausible?

Sigma Implexis- b is one of three planets orbiting the star called Sigma implexis (S-intertwined), designated a red dwarf star.

Sigma Implexis-b has an annual orbit takes 200 earth days; its daily cycle varies depending on where you are on the planet, the day is slightly longer at the equator because of its ellipsoidal shape. It has three small moons.

Sigma implexis is actually part of a binary star system. The 3 planets and their red dwarf are in a very long elliptical orbit around a blue sun- also a dwarf.

When Sigma Implexis-b and its sun reach the long end of their orbit around the Blue dwarf star, and Sigma Implexis-b gets between the two stars, the planet undergoes “the Bake”.

The Bake can last as long as two Earth weeks. Only in the deepest oceans can the inhabitants survive during the Bake. The Bake only happens once every 450 earth years.

When the Bake happens, most of the polar caps melt, and the seas rise significantly. The continental landmasses are mostly around the equator, and some of the peaks rise as high as the Exosphere on Earth. The air is very thin up there, almost a vacuum.

The volcanism on the planet at the equator affects the magnetic field, and causes some of the atmosphere to be stripped on the upper ranges of the mountains.

The atmosphere is similar to Earth's during its pre-industrial stage, in terms or O2 and CO2 content.

Am I missing anything?

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  • $\begingroup$ ...and without life, you have problems getting to that type of atmosphere. worldbuilding.stackexchange.com/questions/9857/… $\endgroup$ – anonymouse Oct 23 '16 at 18:14
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    $\begingroup$ "the day is slightly longer at the equator because of its ellipsoidal shape." - no. If the planet is terrestrial (or anything predominantly solid), this just doesn't happen. $\endgroup$ – John Dvorak Oct 24 '16 at 0:44
  • $\begingroup$ I’ve never heard of a constellation named Implexis. Is that even the posessive form (in Latin)? It sounds like the common way of naming stars but is nonsense. $\endgroup$ – JDługosz Oct 24 '16 at 1:20
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    $\begingroup$ @JDługosz - Does that matter? We're World-Building. he/she could name his/her constellation Bob and he/she'd still be in their right as a writer to do so. (Not that I would. I'd want my constellations to be at least latin-sounding) $\endgroup$ – Raisus Oct 24 '16 at 7:48
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No this system would not work as you think it might.

Firstly, the planet:

Sigma Implexis-b has an annual orbit takes 200 earth days

EDIT: In light of some comments, I re-did the calculations and, based on a star 0.25 M☉; with an earthlike planet (1.0 M⊕) and a 200-day orbital period, the actual orbit would be 0.421683 AU, based using this calculator, or just further than the orbit of Mercury (0.387 AU), still too far from the parent star to be in the habitable zone, but not nearly as much as before.

Next the stars:

Sigma implexis is actually part of a binary star system. The 3 planets and their red dwarf are in a very long elliptical orbit around a blue sun- also a dwarf.

Firstly, by Blue Dwarf, do you mean something like Vega or Sirius A? If so; then this would mean a very short lived star, only enough for some tens of millions of years, not nearly enough for life as we know it (approx 4 billion years - History of life)

EDIT: Okay, so according to Wiki, a Blue Dwarf can also mean the theoretical and not acknowledged "Blue Dwarf (Red Dwarf Stage)", which tells you all you need to know.

Next, the "The Bake" as you called it:

When Sigma Implexis-b and its sun reach the long end of their orbit around the Blue dwarf star, and Sigma Implexis-b gets between the two stars, the planet undergoes “the Bake”.

The Bake can last as long as two Earth weeks. Only in the deepest oceans can the inhabitants survive during the Bake. The Bake only happens once every 450 earth years.

This would kill all life trying to exist on the planet, if any had made it at all. As the planet and its parent star get really close to the binary, the influence of that other binary's gravitational well, electromagnetic pulse emissions and ultra-violet radiation would completely irradiate any life on the planet's surface.

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    $\begingroup$ The relative masses of a binary pair doesn't determine the eccentricity of the orbits. It's perfectly possible to have two stars of equal mass in highly eccentric orbits around each other. See this animation for example. $\endgroup$ – Kyle Oct 24 '16 at 0:01
  • $\begingroup$ Re: "This puts the orbit approximately at the orbit of Venus (225 days)": That's not quite right. Orbital period is not just a function of distance, but also of the mass of the central body; and a red dwarf will be much less massive than the Sun. (Your conclusion -- that the planet would be outside the supposed habitable zone -- is still correct AFAICT, but not as drastically so.) $\endgroup$ – ruakh Oct 24 '16 at 7:14
  • $\begingroup$ @Kyle - If that's how the OP intended it, that wasn't made clear. From what I understood, only the Red Dwarf was on the eccentric orbit $\endgroup$ – Raisus Oct 24 '16 at 7:32
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    $\begingroup$ @Raisus It doesn't matter what their masses are, the eccentricity of the orbits must be the same. Like I said, the semi-major axes will be different, so the more massive body will trace out a small ellipse while the less massive body will trace out a larger ellipse. The ellipses will both have the same eccentricity, they'll just be different sizes. If this weren't the case, the barycenter of the system would wobble which, in the absence of an external force, violates conservation of momentum. $\endgroup$ – Kyle Oct 24 '16 at 13:34
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    $\begingroup$ @Kyle - Right. I'm getting mixed up between Semi-Majoral Axis, then. (because it is possible that the common barycentre is inside one star if that star is large enough) $\endgroup$ – Raisus Oct 24 '16 at 13:40
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its daily cycle varies depending on where you are on the planet, the day is slightly longer at the equator because of its ellipsoidal shape.

Except for polar night, it's not really feasible.

Sigma implexis is actually part of a binary star system. The 3 planets and their red dwarf are in a very long elliptical orbit around a blue sun- also a dwarf.

This does not sound like a stable system. Read this Wikipedia article and, if you can, its sources to know why. As far as I understand, it's not proven totally impossible, but expectations aren't high.

When Sigma Implexis-b and its sun reach the long end of their orbit around the Blue dwarf star, and Sigma Implexis-b gets between the two stars, the planet undergoes “the Bake”.

The Bake can last as long as two Earth weeks. Only in the deepest oceans can the inhabitants survive during the Bake. The Bake only happens once every 450 earth years.

This event would not start abruptly and would not end abruptly. Without seeing actual masses and orbits it's hard to tell, but I think it would last longer than that.

The volcanism on the planet at the equator affects the magnetic field,

Umm how would it?

and causes some of the atmosphere to be stripped on the upper ranges of the mountains.

What mechanism restores lost atmosphere?

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If we only look at it with classical mechanics, this system would be highly chaotic. Each time the two stars get close, all the planets will be thrown everywhere. We haven't yet found planets orbiting closely a single star in close binary systems. Most of the systems where we do find plausible planets are those where the stars in the binary system are very close and the planets are orbiting from far away. Example: Proxima Centauri b.

If we ignore the color of the blue 'dwarf' (you can explain the color using some kind of future tech or yet-to-be-found astronomical body, since blue stars, which fuse hydrogen at a faster rate, and thus hotter, are massive.) and assume both stars are the same mass, you would have an orbit like this.

Two binary stars of the same size

Now for planets to orbit a single star in that binary system, you would need to solve the n-body problem, a problem that is currently unsolved, because it is chaotic. If you do a simulation of three bodies for million of years, the result would be highly different than what we would see in real life. The errors of a numerical solution are just too great in cosmic timescales.
https://en.wikipedia.org/wiki/N-body_problem

Three bodies of the same mass

Here you can see the problem, for three objects orbiting each other, the movement becomes chaotic and unpredictable. This would also happen for the planets orbiting one of the binary stars.

If we were to try and find out if such a system were possible, we would need infinite time to compute the numerical solution, since we do not, we cannot know which initial parameters would make such a system possible.

Here I made a crude representation of what I mean. And in this case, the star collided with the planet...
Crude representation of the Sigma implexis system

I would suggest the classic sci-fi trope where the planet is just highly elliptical. But don't let this plunder your hopes, this is only based on the current understanding of orbital mechanics.

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    $\begingroup$ The n-body problem isn't solved analytically, as in there is no general proof. But it is still solved in the sense that there are numerical solutions to multi-body gravitational interactions. Also, just because the n-body isn't solved for a particular configuration, doesn't mean it can't exist. This is proven because there ARE planets that orbit just one star in a binary system, most notably the one just found around Proxima Centauri. $\endgroup$ – kingledion Oct 23 '16 at 20:17
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    $\begingroup$ In general cases, yes. Proxima centauri does have a planet, but remember the OP says the blue star passes close enough to the planet so it gets heated up considerably. Proxima b is incredibly close to proxima centauri, and Proxima centauri is very very far from the binary system. The orbit of proxima centauri lies so far that we can approximate alpha centauri with its barycenter. Exceptions don't make the rule. $\endgroup$ – Bloc97 Oct 23 '16 at 23:22
  • $\begingroup$ Returning to the n-body problem, since there are no analytical solution, no numerical solution can be considered 'correct', since any variation of the step length in computing trajectories will significantly affect the end result after long periods of time. Numerical solutions are what they are: approximations. Only as the step length approaches 0 is the solution remotely 'correct', and we know that the compute-cost is absurd. $\endgroup$ – Bloc97 Oct 23 '16 at 23:25
  • $\begingroup$ To clarify any misunderstandings, numerical solutions can be considered as 'good enough' solutions in non-chaotic systems. But unfortunately, it is not the case for chaotic systems. Numerical solutions are only very rough approximations in this case. $\endgroup$ – Bloc97 Oct 23 '16 at 23:30
  • $\begingroup$ If numerical solutions are only 'good enough' for non-chaotic systems, then global warming isn't real because the models are only numerical. This is a bad argument. $\endgroup$ – kingledion Oct 23 '16 at 23:57
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Since Jupiter, Earth, Mars, can all be describe starting with being relatively similar in many ways, and have not change in the same way has you might expect. Looking at the way that the atmosphere work initially in Mars and Earth and how they have changed from one another will give you things that really should be included if you are looking at what is going to actually happen in a planet that way. It is not the complete answer, but it gives you an idea of why things don't go the way and the time you might expect.

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  • $\begingroup$ Thank you all, for your insightful comments. I got a copy of Universe Sandbox2- and have been working out these mechanics in that space. Great fun! Thanks again! $\endgroup$ – AC_Latham Oct 25 '16 at 23:32

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