Planetary tidal locking of Close Binary Red Dwarf Stars

[Althistorybuff]

Hi, I'm new to the forum and have an idea for a novel. I can't find anything on this question so I thought I'd sign up for a forum to ask. I'm not a scientist and lack much of the higher technical knowledge.

Here's my question:

Most planets in close orbits (they'd have to be to support liquid water) of Red Dwarf Stars are "tidally locked" due to proximity. I'd like to have a planet that isn't.

There is also the problem that Red Dwarf Stars give off varying levels of heat and I think having two stars would "even this out" and perhaps the additional heat may push back the habitable zone.

If I had a pair of very close Red Dwarf Stars in a binary system (probably close enough to mutually orbit over hours or days), size isn't overly important. Would having not one but two stars affect whether or not the theorized life-supporting planet is "tidally locked"?

Picture a planet about .25 AU's orbiting a pair of Close Red Dwarf Stars of relatively equal size.

Would the planet still be tidally locked, perhaps to the common barycenter between the stars?

Or would the varying gravity of the nearby stars at different positions of the planet's orbit "pull" the planet in some manner that allowed for continued spin/rotation or even encourage it?

Not sure how it works.

Also, in a related question on tidal locking, would the lack of rotation not hamper the magnetic field, thus damaging the prospects of life on any tidally locked planet?

Answers like this are surprisingly tough to find. I don't need the math, as I'd probably not understand it. Just a dumbed-down explanation.

Thanks for any feedback.

 

[John Park]

Would the planet still be tidally locked, perhaps to the common barycenter between the stars?

Reference https://www.physicsforums.com/threa...cking-of-close-binary-red-dwarf-stars.908224/

I suspect that at 0.25 AU you'd be fairly safe from tidal locking. I'm basing this on the fact that Mercury isn't locked, at about 0.4 AU and with a stellar mass probably at least an order of magnitude greater than you are considering.

Edit. Some quite uniformed thoughts about magnetic fields and atmospheres. (1) If the planet was tide-locked, its day would be equal to its year, so in a fairly close orbit it could still have a significant amount of spin. (2) Venus with a rotation period of several months has managed to hang on to a very thick atmosphere.

 

[Althistorybuff]

I suspect that at 0.25 AU you'd be fairly safe from tidal locking. I'm basing this on the fact that Mercury isn't locked, at about 0.4 AU and with a stellar mass probably at least an order of magnitude greater than you are considering.

Edit. Some quite uniformed thoughts about magnetic fields and atmospheres. (1) If the planet was tide-locked, its day would be equal to its year, so in a fairly close orbit it could still have a significant amount of spin. (2) Venus with a rotation period of several months has managed to hang on to a very thick atmosphere.

Thanks for your feedback.

Some good points, though I think Venus still has a weak magnetic field and I assume continues to receive replenishing inflows of material (water, carbon dioxide, etc) from its 3000 volcanos. I pretty sure the slow rotation and weak magnetic field is losing at least the free hydrogen to space as the solar wind hits the upper atmosphere.

While Mercury is not completely tidally locked, it is pretty close. There is a bit of wobble, to what I understand but you could call it "virtually tidally locked". Certainly, the "dark side" does not see much of the way of light. I don't see that as likely for a good ecosystem for life to thrive.

I thought that, based on recent discoveries of terrestrial planets around Red Dwarves, that entire planetary line ups are tidally locked, from planet 1 to 7. I'm curious if the Close Binary Red Dwarves would alter this given that there is no one direct gravitational point to which an orbiting planet may Lock To.

Or would the planet be tidally locked to the common barycenter (let's just say right in the middle of the two even-sized Red Dwarves). I'm going to make up a number and say there is 1/100th of an AU between the two co-orbiting suns.

Also, on another note, would there be any reason why this planet could NOT have a large moon being so close to the star(s)?

If so, would this large moon have any affect on the planet being or NOT being tidally locked? Our moon is exchanging momentum with the Earth and slowing down the rotation while the moon spirals out. Is there any way for a moon to help PRESERVE a planet's rotation?

 

[John Park]

Some good points, though I think Venus still has a weak magnetic field and I assume continues to receive replenishing inflows of material (water, carbon dioxide, etc) from its 3000 volcanos. I pretty sure the slow rotation and weak magnetic field is losing at least the free hydrogen to space as the solar wind hits the upper atmosphere.

According to Wikipedia, Venus' weak magnetic field is caused by interactions between the ionosphere and the solar wind--it's not produced by a planetary dynamo, and it has a "negligible" effect on protecting the atmosphere (which contains virtually no hydrogen) from cosmic rays.

While Mercury is not completely tidally locked, it is pretty close.

It has been known for decades that Mercury makes three rotations about its axis for every two revolutions about the sun, so there is no dark side. Mercury presents the same face to the Earth when observing conditions are good here, which gave the illusion of tide-locking, but observations from spacecraft showed the true situation.

I thought that, based on recent discoveries of terrestrial planets around Red Dwarves, that entire planetary line ups are tidally locked, from planet 1 to 7.

If you're thinking of the recent Trappist-1 discoveries, note that those planets are all much closer than 0.25 AU to their star--within 0.063 AU in fact--which increases tidal forces by about a factor of 60 for a given stellar mass.

As to the effect of a binary star at the centre, I think the biggest question would be whether orbits in the habitable zone could be gravitational stable. I'm inclined to doubt it, but that's only a guess. If the system was stable, the two stars would have to be close together and their period of revolution would be appreciably less than any of the planetary years. (For you 0.01 AU example, applied to the outmost Trappist planet, I think there would be an order of magnitude difference tween the two periods.) It seems likely, therefore that two stars would behave roughly like a single body of their combined mass acting at their centre of mass. Probably there would be some wobble in the direction of "tide-locking", but I imagine that direction would stay close to the direction to the barycentre.

would there be any reason why this planet could NOT have a large moon being so close to the star(s)? If so, would this large moon have any affect on the planet being or NOT being tidally locked?

I don't know if anyone could answer the first part definitively. It is a fact, though, that the relative size of Earth's Moon is unique among the planets of our solar system, and apart from Luna, the four rocky planets have a total of two tiny moons. I think a large moon would affect tide-locking. The planet would be in a competition between being locked to the star and locked to the moon; the moon would have the advantage of closer distance (tidal forces go as 1/r³), but of course would have much lower mass. What ultimately happened would--I think--depend how the spin angular momentum (or rotational energy?) of the planet compared to the corresponding orbital quantity for the moon, and how effective tidal forces were in dissipating energy. I suspect this is an answerable question but I'm not sure I can provide the answer myself.

 

[Althistorybuff]

Yes, you are right. I've looked up some data on Mercury and the 3/2 ratio is correct.

I'm still uncertain as to the probability of avoiding tidal locking. I thought this binary star might be the solution to my problems.

For argument's sake, let's assume a pair of identically sized Red Dwarf Stars, 150,000 miles in diameter each, separated by say, six diameters for 900,000 miles. I won't bother with the math but this is probably less than 1/1000th of an AU.

The theoretical planet would be out at about, say .15 to .2 AU's, which is about 15 to 20 million miles, in a circumbinary system. \ Given that close confines of the stars, I would think that, seen from the perspective of someone standing on the planet, the two stars would be virtually one unit for much of the day.

The planet would not be bound ( or less bound) by the 3 body system (correct me if I'm wrong) as a P-Type Orbit, see below for a more complex one. In this diagram, the stars are much further apart than my assumptions relative to the orbit of the planet (planet orbit is 15 to 20 times further out than gap between starts). In this diagram, the planets orbit would almost equal the gap between starts. I would think my planet's orbit would be exceptionally circular around both stars provided something didn't nudge the planet early in its history and the early nebulae couldn't straighten out its orbit.

Thoughts?

Thanks.

 

[Althistorybuff]

Note that I'm actually trying to put together a quadruple or quintuple system. Here is generally what it looks like:

Two Sol-type stars in circular mutual orbit at about 74 to 78 AU's. I chose this range as the rule of thumb for the Hill Sphere for planets circling one of these suns (I read somewhere) is about a 5 to 1 ratio to the point where the other star risks pulling on the planets of the first.

Two red dwarf (in mutual close orbit of about 1/1000 of an AU) stars cast out of the system early into an generally circular orbit about 450 AU's plus around the two larger stars.

I may even add a brown dwarf at about 4000 AU's, for fun.

 

[Vanadium 50]

If you have one planet orbit two stars, you will have an enormous quadrupole moment of the gravitational source and thus enormous tides. Your planet may find itself in a resonance not with its own orbit but with the orbit of the two stars.

 

[John Park]

My immediate, gut-feeling, reaction is that the final planet is doomed. It's going to fall into one of the suns or be kicked out of the system entirely.* Realistically its only change of finding a stable orbit is to get into a Trojan configuration (the L4 or L5 point) , where it and the two stars rotate together at the vertices of an equilateral triangle--and that configuration places limits on the relative masses of the two stars. Otherwise, for any planet to have a reasonably stable orbit, either one star has to dominate gravitationally (your '"other kind") or the two stars have behave approximately like one body. But even then, getting such systems to be stable for billions of years is I gather a delicate business, particularly if there is more than one planet.

* The average field of the two stars might act through the centre of mass, but the planet doesn't feel the average field, it feels the gravitational force from the stars where they are at that moment. At some point it's going to pass very close to the primary, and be strongly attracted by it--and this is the sort of situation that could lead to the planet being ejected from the system.

 

[John Park]

This might be amusing to experiment with:
https://phet.colorado.edu/sims/my-solar-system/my-solar-system_en.html

 

[Althistorybuff]

If you have one planet orbit two stars, you will have an enormous quadrupole moment of the gravitational source and thus enormous tides. Your planet may find itself in a resonance not with its own orbit but with the orbit of the two stars.

Just to clarify, the planet would be orbiting at .15 to .2 AU's the two Red Dwarf Stars which would orbit around .001 to .002 AU's from another. I would think the planet would not be pulled inward but am not sure of the gravitational stresses.

The two - Sol-type planet would be co-orbiting over 400 to 500 AU's away and would not have any gravitational effects on a planet orbiting the Red Dwarf at .15 to .2.

 

[Althistorybuff]

My immediate, gut-feeling, reaction is that the final planet is doomed. It's going to fall into one of the suns or be kicked out of the system entirely.* Realistically its only change of finding a stable orbit is to get into a Trojan configuration (the L4 or L5 point) , where it and the two stars rotate together at the vertices of an equilateral triangle--and that configuration places limits on the relative masses of the two stars. Otherwise, for any planet to have a reasonably stable orbit, either one star has to dominate gravitationally (your '"other kind") or the two stars have behave approximately like one body. But even then, getting such systems to be stable for billions of years is I gather a delicate business, particularly if there is more than one planet.

* The average field of the two stars might act through the centre of mass, but the planet doesn't feel the average field, it feels the gravitational force from the stars where they are at that moment. At some point it's going to pass very close to the primary, and be strongly attracted by it--and this is the sort of situation that could lead to the planet being ejected from the system.

I believe that Circumbinary orbits are quite common, are they not, at least for planets around stars? I've seen many artist representations but I don't know of the specifics of the system.

From the planet's point of view, it has already been kicked out (or formed out there) and orbit the two others.

 

[Althistorybuff]

Here is the Caster System:

 

[John Park]

I believe that Circumbinary orbits are quite common, are they not, at least for planets around stars? I've seen many artist representations but I don't know of the specifics of the system.

Reference https://www.physicsforums.com/threa...e-binary-red-dwarf-stars.908224/#post-5720592

Yes they are quite common, but they don't work if the all the bodies are separated by comparable distances that are also changing. In your large diagram showing a planet in a circular orbit about two stars in eccentric orbits, in fact the gravitational forces on the planet would be changing dramatically as it drew near one star or the other. So I can't see any hope of a circular or even stable orbit. Similarly (using Wikipedia as source), I think your picture of the Castor system is misleadingly out of scale. The A and B pairs orbit each other with a period of about 450 years. The C pair is in a 14,000-year orbit about A or the AB pair. But the components of the pairs themselves have orbital periods of a few days or less than a day. In other words--three very tight pairs orbiting at considerable distances from each other. There would be no chance of any two pairs closing to a distance comparable to the individual pair-separation.

Remember your planetary system has to be stable for probably billions of years. The periods of the Trappist planets were all less than three weeks--meaning around twenty orbits per year; think at least fifty billion orbits. If there's any likelihood that the planet will get too close a star it will happen over those time scales. In one likely consequence the planet would be ejected--meaning it would be given escape velocity and kicked out of the system completely. I believe this is pretty common in young star clusters for example.