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This question has been rewritten to incorporate all clarifications.

On Earth, half the planet is illuminated at any time (let's ignore eclipses). Axial tilt lets day lengths vary, but over the course of a year, every location is illuminated half the time.

It's easy to make a planet where, over a year, everywhere is illuminated more than half the time. Use a binary star.

But is there a naturally occurring, stable solar system that satisfies the more restrictive requirement that the planet is always more than half illuminated?

In the general case, if it orbits one star of a binary, there will be a point in its orbit where the other star passes behind the one the planet orbits. If it orbits both stars, there will likewise be a point where all three are in a line. And note that, even if the planet's orbit is inclined relative to the plane containing the stars' orbits, a collinear situation is still possible... barring some resonance that prevents it.

Constraints:

Note that I'm only talking about solar system geometry. Cloud cover means you can't see the sun all the time (though light gets through). Atmospheric refraction and diffraction extend visible light onto the 'night' side; this gets extreme with a dense atmosphere like Venus. I know this, so I'm not asking for answers involving that. All solutions must work for a vacuum world. My purpose is to explore the geometry of solar systems.

The planet must satisfy both of "At any time, >50% of the surface is illuminated" and "At any location, illuminated >50% of the year."

Approximate scale of the effect: Let's say that "more than half" means at least 195/360 of the surface (IE, an extra hour in an Earth day). It must also be light providing meaningful illumination, not just technically visible. Let's say that said area is illuminated to a level at least 1/40 of (should it be "the brightest illumination it receives" or "the brightest illumination Earth receives"?).

Before asking this question, I thought of a Trojan planet of a binary star. I then saw a figure of a minimum mass ratio of 25 for two bodies to generate stable L4/L5 points. With stars, luminosity is roughly proportional to mass to the 3.5 power. This means one star must be at least 78000 times brighter than the other, and the planet is equidistant from them. Given that full moonlight on Earth is about 1/400000 of full sunlight, this is hardly better, nowhere near enough to count as "day". That's why I asked the question.

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  • $\begingroup$ Wont that depend on your definition of "day"? I think Alpha Centauri has 3 stars but one is so far outwards it doesnt illuminate much anymore. A similar system with the third star closer to the center (perhaps a "small" red dwarf?) would illuminate enough to give that "day" cycle. $\endgroup$ – Demigan Dec 30 '18 at 20:57
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    $\begingroup$ Demigan: Yeah, definition is vague. But, for example, moonlight on Earth isn't bright enough to be thought of as 'day'. At what intensity does light become considered 'day' on Earth -- a few % of its full intensity? Karl: Yeah, I know those. But like eclipses, they're minor effects. $\endgroup$ – Tristan Klassen Dec 30 '18 at 21:14
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    $\begingroup$ Your title asks a different question from the text. One asks for a day that's longer than night. The other asks that the planet always be more than half illuminated. That is not the same question at all. $\endgroup$ – chasly from UK Dec 30 '18 at 22:03
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    $\begingroup$ For a planet with no axial tilt and turning on its axis at a constant rate, they are the same AFAICT. $\endgroup$ – Tristan Klassen Dec 31 '18 at 1:04
  • $\begingroup$ Regarding "If it orbits one star of a binary, there will be a point in its orbit where the other star passes behind the one the planet orbits." and "If it orbits both stars, there will likewise be a point where all three are in a line" - these are true only if both stars share the same plane as the orbit of the planet (consider e.g. a state where the secondary star is "above" the orbital plane of the planet and the primary star...) $\endgroup$ – G0BLiN Jan 1 at 11:35

13 Answers 13

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You can have a binary star, and the planet in a Trojan position in the same orbit as the smaller of the two stars.

Basically, the two stars describe one side of an equilateral triangle, and the planet occupies the third vertex, in either L4 or L5 position.

One such configuration is presented here (figure 2, on the right).

Wikipedia gives stability requirements as "m1 > 100 m2 > 10,000 m3" so you'd need a large F-star as m1, and a small red-yellow dwarf as m2. This also requires a large orbital radius for habitability.

This configuration is not long lived enough for life to originate on the world, though. For that you'd need a smaller, colder, longer-lived main star (for example a Sun-type G star), and then you'd need a very small brown dwarf which, at a distance of some 8-10 light-minutes, wouldn't probably supply much of a daylight.

But if you don't require habitability, this would get you four hours of main daylight, eight hours of "reinforced" daylight, four hours of "secondary" daylight and eight hours of "night" every 24-hour day (thanks to @ltmauve for pointing this out).

A simpler setup

Again a large star, and a secondary smaller star. This time, though, the planet orbits around the smaller star, inside its gravitational well. The limits on the stars' sizes and luminosities are now more relaxed.

We need two additional constraints: the secondary star's ecliptic is not coplanar with its orbit. Ideally they are perpendicular (so there are only two syzygy points where the daylight period might be 50% of the rotation period); and the secondary star's revolution period is an odd multiple of the planetary revolution period, so that at syzygies, when the three bodies are aligned, the planet is always in the middle and actually gets a 24-"hours" day (the two "daylights" are not overlapping).

I'll try and run some simulation after New Year's Eve ;-D to check whether this setup does indeed work - I might well have missed something obvious.

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    $\begingroup$ I thought about a Lagrange point setup before. However, luminosity goes as mass to the 3.5 power, and the minimum mass ratio to make a stable Lagrange point is 25, meaning that the minimum luminosity ratio is 78000. That's hardly better than a full moon on Earth (1/400000 of full daylight). I want a day that would be recognized as 'daylight', which means an extended time period with at least a few % of full daylight. $\endgroup$ – Tristan Klassen Dec 31 '18 at 1:08
  • $\begingroup$ The answer that saved the question from a downvote. $\endgroup$ – Joshua Jan 1 at 17:26
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    $\begingroup$ Shouldn't the daylight levels change at 4 and 8 hour intervals, not 6? The stars should stay at about 60 degrees apart in the sky, so you should have 60 degrees of just one star, 120 of them sharing, 60 of the other star, and 120 degrees of night. $\endgroup$ – ltmauve Jan 2 at 3:48
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    $\begingroup$ This is somewhat a dangerous place to put a planet: the key thing about L4 and L5 is that they accumulate space rocks, so any planet living in this space is kind of being hit by a constant meteor shower of varying intensities. $\endgroup$ – CR Drost Jan 2 at 23:30
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When you have a large geostationary moon that reflects enough sunlight to extend the day it should be possible. To increase the day time significantly.

Earth Example:

Formulas:

  • $\alpha_z$ = $\frac{v^2}{r}$ which is the same as: $\alpha_z = \frac{4\pi^2}{T^2}$
  • $F = G\frac{m_1 m_2}{r^2}$

$\alpha_z$ is the centripetal acceleration it has to be the same as $F$ from the second formula which is the gravitational force.

T is the time it takes for your object to complete one orbit in the case of earth about 24 hours or 86400 seconds.

$r$ is the distance between the two objects.

$m_1$ and $m_2$ are the masses of your moon and planet.

$G$ is the gravitational constant.

$v$ is the speed of your orbiting object relative to in this case earth.

In our case we get about 0,0000000052885 for $\alpha_z$

For the second formula you leave the mass of the moon out because it cancels out with the first formula where you originally also have to adjust for the mass of the moon but you use it in both formulas so you can ignore it completely.

If we plug this value for F in the second formula and switch the equation around to give us r we get $r^2 = G\frac{m_1 m_2}{F}$ and a value of 7.537137 × 10^22 we have to take the square root of it as it is $r^2$ and so we land at 2.74538468 × 10^11 meters which tells us that in our case it would not work because the moon would be to far away to reflect enough light onto your planet. So you would have to tweak your system if you want this solution to work.

The things you could tweak to make it possible are:

  1. You could make the moon lighter
  2. Let the Planet rotate much faster around it self
  3. decrease the mass of the planet

Also you could place the planet between two stars in such a way that both stars pull on the planet with the same force which leads to a cancellation of gravitational forces but you would have constant day and no seasons. Hope this helps.

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    $\begingroup$ The formulas looked funny because you were using separate TeX blocks over multiple lines, I've suggested an edit to fix the issue. $\endgroup$ – Οurous Dec 30 '18 at 23:41
  • $\begingroup$ Thanks I hope I can recreate this next time. $\endgroup$ – Soan Dec 31 '18 at 0:03
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    $\begingroup$ "Also you could place the planet between two stars in such a way that both stars pull on the planet with the same force which leads to a cancellation of gravitational forces but you would have constant day and no seasons." Isn't that unstable? $\endgroup$ – Tristan Klassen Dec 31 '18 at 1:11
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    $\begingroup$ depends on how consistent the stars orbit each other. But yeah in most cases it is. Which does not stop him from creating the planet with the perfect double star solar system where it is stable. $\endgroup$ – Soan Dec 31 '18 at 1:16
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The star is usually much larger than the planet, and there is diffraction, so any planet is always more than half illuminated. ;-) Significantly more than half, no, imo.

Except ... if you put the orbit of the planet perpendicular to the plane in which the two binaries circumvent each other. It'll probably be tricky to stabilise such an orbit, which needs to be synced up to make sure the three objects really never line up.

An astrophysicist could probably tell us if there is a "natural" resonance which would drive a planet into such an orbit, and keep it there.

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    $\begingroup$ "a "natural" resonance which would drive a planet into such an orbit, and keep it there." You'll notice the same could also happen in a coplanar system! Can a resonance be set up so the three bodies never line up that way? $\endgroup$ – Tristan Klassen Dec 31 '18 at 1:17
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    $\begingroup$ Yes, except it's not the absolute size of the star, but its angular diameter as seen from the planet. The same geometry applies: from Earth, both moon and sun are about 1/2 degree, IIRC, so on average 181 degrees would experience some sun/moon light during the day. (Plus, as you say, some extra because of refraction.) $\endgroup$ – jamesqf Dec 31 '18 at 4:01
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    $\begingroup$ @TristanKlassen No, in a coplanar system, there will always (once per year+a bit) be a moment when the three bodies line up. In a perpendicular system, you can (possibly) make sure the outer body always passes through the plane when the inner bodies are furthest from each other. $\endgroup$ – Karl Dec 31 '18 at 8:46
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    $\begingroup$ @jamesqf Of course it's also about the angular diameter, but the moon is smaller than the earth. You will always see the sun from the north or south pole, but twice per month, you don't see the moon from either! $\endgroup$ – Karl Dec 31 '18 at 8:50
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Imagine a planet with an advanced civilization. It has launched many mirrors into space, to illuminate (part) of the planets dark side. You can tweak the amount of mirrors and increase the average length of the day to fit your story.

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Pyramid planet.

pyramid planet

With one light source and a spherical planet, I could not think of a way to illuminate more than half. It's a sphere thing. But if you can use shapes other than a sphere it is easy. The (tidally locked) pyramid planet keeps its apex point at its sun, and each of the triangular faces stays in the light. You could have it rotate around the axis down through the apex. The square side stays in the dark.

Other tidally locked elongated shapes would also keep their elongated faces in the light and the base in the dark.


OK. Tidal locking not allowed. I will borrow my answer from

Why is my Dark World so dark?

disc world

This world is a disc, turning on its axis. It stays with its edge facing its sun. On the ground, the sun is always moving along the horizon, never setting, never rising, never stopping. Sunlight is always redshifted and oblique. Shadows are long.

There is sunrise and sunset on the edges of the disc. The edge is a minuscule fraction of the disc.

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    $\begingroup$ "Other tidally locked elongated shapes would also keep their elongated faces in the light and the base in the dark." I guess I forgot to specify that tidal lock is ruled out, because everywhere on the planet must receive day at least some of the time. $\endgroup$ – Tristan Klassen Dec 31 '18 at 1:14
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    $\begingroup$ Also, a planet is generally understood as a body which becomes spherical under the own gravity. $\endgroup$ – o.m. Dec 31 '18 at 7:23
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    $\begingroup$ Yer killin me, @o.m. Yer killin me. Here is some light reading on a fictional disc planet. It spins fast partly counteracting its gravity and flattening it out. That would be neat here too because the sun would appear to race around the perimeter and the long shadows would wheel around wildly. en.wikipedia.org/wiki/Mesklin $\endgroup$ – Willk Dec 31 '18 at 13:21
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If the planet is not required to be habitable for humans or to have advanced lifeforms or any lifeforms, the answer is simple.

Make the star a star which has left the main sequence and swelled up to a red giant stage.

Such a star could have expanded until it almost reached the orbit of the planet. If it reached the planet's orbit the drag of the star's gases would cause the planet to spiral down into the star.

In such a situation light emitted from the edges of the star as viewed from the planet could reach the planet far into the side facing away from the star.

But the increase in stellar a radiation and thus in the planetary temperature as the star became a red dwarf would have wiped out any preexisting native life on the planet. Of course a sufficiently advanced civilization could have terraformed the planet and introduced lifeforms from other worlds and/or made it habitable for humans.

If the planet has to be naturally habitable for humans and/or have advanced native lifeforms the problem is more complex.

The star could be made a red dwarf or main sequence star instead of a red giant. All red stars, giants or dwarfs, have low surface temperatures and thus emit less energy per unit of surface.

So in order to have a surface temperature equal to that of Earth, a planet would have to orbit the red star close enough that the star appears several times as large in the planet's sky as the Sun does in Earth's sky. And that will help the light from the sun to reach more than half of the planet's surface.

Of course the fainter the star, the closer the planet would have to be to it in order to have an Earth-like surface temperature, and the greater the proportion of the Planet's surface that would be illuminated by the star at any one moment. Thus it is desirable for the star to be a very, very faint red dwarf for as much as possible of the planet to illuminated at any one time.

But for that the happen the planet would have to obit so close to the red dwarf star that the planet would probably become tidally locked to the star so that one side always faced the star and the other side always faced away from the star.

But that would fail the original question.

Thus the planet would have to be saved from being tidally locked to its star by being tidally locked to some other astronomical body. If the planet was actually a giant, Earth sized moon of a gas giant planet orbiting close to a red dwarf star, the planet/giant moon would become tidally locked to the gas giant planet instead of to the star.

And the gas giant planet could appear several times larger in the sky of its planetary-sized moon than the red dwarf star appears. Meaning that the light from the gas giant planet planet could cover even more of the planetary sized moon than the light from the star.

What light from the gas giant planet? Possibly light from countless lightning strikes in its atmosphere ever second.

And certainly light from the red dwarf star reflected from the planet, just as sunlight is reflected from the Moon onto Earth. But probably many times as bright as a full moon on Earth.

So if the planet sized moon orbits the gas giant planet in the same plane that the gas giant planet orbits the red dwarf star, there will be a moment in the orbit of the moon when it is directly between the red dwarf star and the gas giant planet and will be casting a shadow on a tiny part of the gas giant planet. And the rest of the gas giant planet will be reflecting light in all directions, and some of that light will illuminate the side of the moon facing away from the star.

In that moment every part of the moon will be illuminated by the red dwarf star or the gas giant planet, and some parts will be illuminated by both.

The closer the moon is to that part of its orbit, the greater the proportion of its surface that will be illuminated, and the farther the moon if from that part of its orbit, the smaller the proportion of it is surface that will be illuminated.

When the planet sized moon is about 90 degrees from the line between the star and the gas giant planet, somewhat more than half of the moon will be illuminated by the the star and somewhat more than half of the moon will be illuminated by the gas giant planet. About one quarter of the moon will get light from both the star and the planet, one quarter will get light from only the star, one quarter will get light only from the planet, and one quarter or less will get no light.

And when the moon is more than 90 degrees from the line between the star and gas giant planet the proportion of the moon's surface that is illuminated will get smaller and smaller. When the moon is exactly 180 degrees from the line between the star and the planet it will receive light only from the star. But since the star is assumed to be a red main sequence star that the moon and planet have to be very close to in order to have Earth like surface temperature, it should have several times the angular diameter of the sun as seen from Earth and thus should illuminate a bit more than half the surface of the moon.

Would the moon be eclipsed by the planet one time every orbit, when it is 180 degrees from the star? Yes, if the moon orbits the planet in exactly the same orbital plane as the planet orbits the star.

The moon should orbit the gas giant planet in the equatorial plane of the gas giant planet. The equatorial plane of the gas giant planet should be tilted to a greater or lesser degree relative to the orbital plane of the gas giant planet around the star. It is perfectly possible for a moon of a gas giant planet to never be eclipsed by its planet, if its planet has a high enough axial tilt.

And I am not sure if brief eclipses, lasting hours at most, would count as violating the original question.

If the moon is tidally locked to the gas giant planet, one half of that moon would always face the gas giant planet and would always be illuminated by light reflected from the planet, as well as being illuminated by the star for more than half the time.

One half of the moon would always face away from the gas giant planet, and except for the section closest to the gas giant planet, would never be illuminated by the gas giant planet. And that half would be illuminated by the star somewhat more than half the time.

There have been a lot of other questions about habitable moons of gas giant planets in the habitable zones of stars, and it is a good idea to refer to those questions and answers to see if they have any useful information, as I state in my answer to this question:

How long will it take to discover they live on a moon and not on a planet?1

And I gave links to two earlier questions about habitable exomoons.

The article "Exomoon Habitability Constrained by Illumination and Tidal heating" by Rene Heller and Roy Barnes Astrobiology, January 2013, discusses factors affecting the habitability of exomoons.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3549631/

Note that it states that a moon can not have a stable orbit unless the orbital period or year of the planet is more than 9 times the orbital period or month of the moon.

So if your moon has an orbital period of 0.75 to 15.0 earth days, for example, the planet must have an orbital period of at least 6.75 to 135 days.

The planets of the star TRAPPIST-1 that orbit in the habitable zone hare years of 4.05 to 12.4 days, so it is certailnly posible for a planet and it s moon in the habitable zone of red dwarf to have orbital periods of the necessary length.

http://How%20long%20will%20it%20take%20to%20discover%20they%20live%20on%20a%20moon%20and%20not%20on%20a%20planet?

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Yes. Admittedly an engineered system answer:

Central object: A black hole. This must be rotating in the plane of the system so the deadly jets never get anywhere near the planet.

Around it, a ring of 6 (or more) stellar objects.

Case A: The planet orbits between the black hole and the stars. In this case if all the stars are burning you have perpetual sun, if you want night most of them must be dead (white dwarf or neutron star.)

Case B: The planet orbits outside the ring of stars. At this point you have more than half light but it's not perpetual.

So long as the black hole is sufficiently more massive than the stars (I don't know the required ratio) this is stable.

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  • $\begingroup$ Won't the jets be illuminating more than half the planet all the time anyway? Maybe in x-rays though. :-) $\endgroup$ – brendan Jan 2 at 4:41
  • $\begingroup$ @brendan Depends on how much the black hole gets to eat. I figure there will be some light from the accretion disk and jets but the question was about "day". If it's not eating too much those will be luminous nighttime objects. $\endgroup$ – Loren Pechtel Jan 3 at 0:55
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Not in the ways you expect, but there is a loophole.

First let's consider the issue of it being day on more than half of the planet always. This is impossible with a single star system and an ellipsoidal planet. If you elongate the planet enough perhaps you could get more than half the planet facing the star at some point in its rotation. However, if you consider an oval in a 2D view then when the short side is facing the star at most only half of the planet is facing the star and likely much less than half if you elongate in any meaningful way (enough to make the long side significantly greater than 50% of the surface area).

Now let's try a different approach with a single star system. Maybe it just needs to be day for more than half of a rotation cycle. Ok well then let's consider a point A. When A is on the side facing the sun the planet slows down. Then when it facing away the planet speeds up in rotation. This works, right? No, this still fails because now point A' which is the point on the opposite side of the planet will have long nights and short days.

This means that short of making the sun send light in a larger band and somehow curve around the planet this is impossible. However, light travels along the geodesics of the surface of space-time which bends according to mass. This means that light can only curve according to mass. Furthermore according to one of Einstein's most famous thought experiments light will bend under gravity as if you were in a box accelerating upwards at the rate of gravitational acceleration. What this means is that light can only bend further around the planet if something either reflects it to the other side of the planet (however this would be side dependent) or something lifts the light around the planet. This might be possible if your planet has significantly large rings. Another possibility is that your planet has naturally low enough gravity such that the light actually curves to the other side of the planet. These are doable but it would a simple cop out, imo and pretty lackluster.

I'm done bashing other potential methods so I'll now I'll just move on to how I think you can do it your scenario. Just before I continue I should explain why I'm considering only one planet and one star. It's because in certain orientations a binary star system is equivalent to a single star system in terms of the propagation of light. I mean technically one star will cause the other stars light to curve which might in turn allow it to bend around the planet, but that would be impossible for me to claim one way or the other, and it wouldn't be satisfying.

Now here's a simple loophole.

Your planet has literal oceans of glass.

It is literally that simple. The sunlight can't get around the planet, so just make the planet transparent periodically. Enough locations and in size so that the light bleeds through past the one half of the planet and other to the other side. Crystals may also work better. If done properly it shouldn't affect life significantly. After all, sand is pretty much the same as glass. Sure it's not livable in those areas for the most part but there could be small rivers and islands or things that are adapted to such extreme conditions. Plus, the requirement here wasn't that things be able to live on this thing in particular. It's a bonus I imagine but the large glass sections are simply assumed to be unlivable.

Now you might ask me what will happen when people or things damage the surface thereby lowering it's transparency. Sure, by all means let's go drain the entire atlantic ocean such that water can no longer evaporate and cause hurricanes. Do you get the point? The sunlight shining through it is on a grand scale. It's not going to be prevented by chipping away a volume equivalent to a square block surrounding the entirety of Manhattan island all the way from bedrock to the top of the empire state building let alone whatever humans or animals might damage. Perhaps if someone strikes the glass with a nuclear bomb it might have an effect, but at that point we're dealing with planetary level changes here and in that situation I'd say that anyone doing that would be really stupid as causing the glass to crack to that large of a degree would risk magma leeching out from under the surface and if the damage were significant enough might expect a super volcano to form. In fact, I'm going to go ask a question about that.

tl;dr if the planet has large regions of glass all the way to the mantle or seafloor then it's likely you'll get the desired result.

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    $\begingroup$ Instead of glass you could use diamond. Or ice! $\endgroup$ – Willk Dec 31 '18 at 13:42
  • $\begingroup$ I'm not so sold on the glass ocean idea. Glass, in small quantities, can be quite transparent. If you take a pane of glass and look at its face, you'll see right through it—sure. But if you take that same pane and look through it edge-on, you'll likely get a different view, one that isn't wholly transparent. Any small, recurring artifacts in your ocean of glass will render it opaque at some distance away, not to mention that light itself won't propagate through glass indefinitely. Light loses energy as it passes through glass and eventually is absorbed by it. $\endgroup$ – B.fox Dec 31 '18 at 13:45
  • $\begingroup$ Also, I would suspect that an ocean of glass would be massive, and have varying levels of densities. Light passing through these densities would owe itself to refraction, and there's hardly any telling in what way it'll become distorted. $\endgroup$ – B.fox Dec 31 '18 at 13:50
  • $\begingroup$ Given that I said "I'm not talking about atmospheric diffraction (Venus has some weird stuff going on there). Solutions must work for a vacuum world." I don't think making the planet transparent should count. $\endgroup$ – Tristan Klassen Dec 31 '18 at 18:28
  • $\begingroup$ @TristanKlassen as I said. It is a loophole. However, I doubt you can get the result you want from geometry alone. Any time the three elements of your system align along a line you will have roughly 50% of the planet in daylight. So to have a significantly longer day period always is likely unfeasible. The only way to avoid that would be to never allow the three bodies to align but I'm pretty sure even that is not possible. However if you simply want longer days 90% of the time then I have a sneaking suspicion that binary systems will naturally trend towards have mostly longer days. $\endgroup$ – The Great Duck Jan 2 at 2:48
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Does a "forever" day count?

The moon has a few peaks of eternal light, and it has been theorized that Mercury may have them too. While the rest of the planet experiences solar days (lasting 176 Earth days each), the north pole has areas in perpetual shadow, so it would stand to reason that the south pole experiences the opposite phenomenon.

For the lunar examples given in the linked article, the peaks don't all experience daylight 100% of the time. However, some experience daylight for more than 50% of their year, which could be simplified to longer days than nights, on average, in perpetuity. While this is specific to a satellite rather than a planet, I don't see why similar effects could not occur on a planetary scale, given the right combination of axis, orbit, and craters.

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  • $\begingroup$ Yeah, you can get it locally, but I don't think this can be extrapolated to an entire planet. It only works near the poles. $\endgroup$ – Tristan Klassen Dec 31 '18 at 18:30
  • $\begingroup$ For spheres orbiting a single light source, that the best you're going to get. For the whole planet to experience longer days than nights, then more than 50% of the surface has to be illuminated at any given time. @LSerni's answer is the most viable solution. $\endgroup$ – CactusCake Dec 31 '18 at 18:41
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A possibility (also at the end of LSami's answer) might be a binary system with a suitable 3D configuration of the orbits.

enter image description here

Imagine a 3D coordinate system with the $x$-axis pointing right, the $y$-axis pointing up, and the $z$-axis pointing at the viewer. We have a central yellow star (sitting at the origin in the animation, but it really should orbit the center of mass also), a red dwarf slowly orbiting the bigger component of the binary system in the $xy$-plane. And, finally, a planet (the blue dot), orbiting about the red dwarf in a plane parallel to $xz$-plane (i.e. one that has the $y$-axis as its normal). The animation tries to give a top-down view of the motion

The points:

  • Unless the smaller star is close to the $x$-axis, it will not eclipse the bigger star, because the bigger star is not on the plane of the orbit of the planet.
  • When the smaller star is very close to the $x$-axis, it may eclipse the bigger star, but if we synchronize the periods we can arrange the planet to always be either above or below the $xy$-plane at those instants, when the small star crosses the $x$-axis.
  • When the dwarf is near the $y$-axis, 3/4 of the planet bathes in starlight. The ratio goes close to 1/2 in those years, where the bigger star is closer to the plane of the planet's orbit, and even then only for a single season (one season the planet will be nearly between the stars and be fully lit).

The caveats:

  • I suck at celestial mechanics, but I suspect the long term stability of this set up may be in doubt. At least the ratio of orbital periods likely needs to be quite high, may be something like one hundred (if not thousands) of "planet years" per a single orbit of the red dwarf about the bigger star.
  • Also, gravity of the bigger star may make the plane of the planet's orbit rotate over time.
  • Also, if the ratio of periods is 1000:1, then the above synchronization idea doesn't help very much. The planet will reach the $xy$-plane, at points when the red dwarf has moved only very little off the $x$-axis. At those points the dwarf may almost eclipse the bigges star, resulting in something like only $50.001$ per cent of the panet having a semblance of a day. (in the animation the ratio of those periods is 10:1)
  • But, those close to 50-50 days are few and far between. It might make for an occasion for the culture living on the planet!
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  • $\begingroup$ Oops, the same idea is in LSami's last paragraph. I only read the part about using Lagrange points. Sorry. $\endgroup$ – Jyrki Lahtonen Dec 31 '18 at 22:40
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A synopsis of a paper

So the standard reference here would be Siegfried Eggl's Habitability of Planets in Binary Star Systems and his general conclusions are that binary star systems

  1. Support predominantly two kinds of stable reasonable orbits: ones where a planet orbits both stars as if they were one, and ones where a planet orbits one star but is perturbed by the other, and
  2. Can indeed have habitable zones in either configuration, but
  3. The presence of the second star perturbing a planet's orbit can draw it temporarily out of the habitable zone, requiring atmospheric inertia to keep it habitable for that part of the year until it comes back to proper temperatures.

On a positive note based on the calculations here it does seem like two Sol-mass stars co-orbiting a common center at a distance of 10 AU (ten times the distance the Earth is from the Sun) can indeed sustain a planet orbiting one of them somewhere between 0.95-1.55 AU away from the one star, as long as the two stars do not co-orbit with an eccentricity greater than 0.2-0.3 or so.

A binary star system with some inclination

So you are gonna need a brighter star, probably something off of the main sequence.

The issue is that you want some orbit of some Sun-like star at something like 1AU or so, but you want that Sun to be part of a binary system where the other star is maybe 10AU or more away. Since the brightness of a light bulb decreases with the square of its distance from you, if you want both stars to be approximately equal in the sky, this one star needs to be maybe 100+ times brighter than the Sun. Looking at the Hertzsprung-Russell diagram makes this very easy to see, but there is a noticeable "gap" between the very blue and the yellow-orange-red side: the main sequence stars would be really really blue and that's really really bad, because the more blue light you have the more ionizing radiation you get from the star. So you would have to go off the main sequence, to a red giant star. These do not have to be too dramatically red and can have surface temperatures (hence colors) similar to an incandescent light bulb; maybe a good (not-too-massive not-too-bright) giant to model the giant around would be Arcturus, 25 times larger in radius than our Sun, 170 times as bright if you're at the same distance from it, roughly the color of an incandescent light bulb. At 10 times further away it would only visually appear to have 2.5 times the radius of the main star, and though we might have to shrink both stars a little to get the temperatures down a bit, it shouldn't be too bad. Importantly, Arcturus is still about one solar mass in size.

Each star illuminates half of a planet but as another answer points out, putting the orbit of the planet off-axis with some past collision might mean that the suns never eclipse each other, and so during the most "day/night" part of the year they might still be separated by, say, 10 degrees of the sky, causing 190 degrees of planet to be illuminated at once. But the more dramatic feature is that there would be a season where the sun rises just around when the giant sets, and vice versa, meaning that you have a very constant half-illumination for the entire day across nearly the whole planet, with some "South Pole" still having a day/night cycle but the "North Pole" not seeing either star set. So the seasons themselves would be quite interesting to work out here, and also the weather (our weather is dominated by the fact that our equator is approximately lined up with our orbit leading to a warm equator and cold poles; not clear how that transfers over and it depends in part on how the tilt lines up with the two other orbits). Since there is also an orbit of the stars about each other, there would also be a regular exchange between which pole had the day/night cycles and which one did not; this could change over a 100-year cycle or so, maybe.

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If the planes's self rotation period is equal to the around-the-star period then always one side of the planet faces the star constantly, like the moon and the earth. Additionally if the planet is an irregular shape, for example, like a cone shape that one side is "flatter" than the other side yields one side area is larger than the other side. When the large-area side faces to the star, the planet have longer daylight than the night.

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    $\begingroup$ A planet is a body in hydro-static equilibrium. This means it has to be a spheroid. $\endgroup$ – L.Dutch Dec 31 '18 at 11:07
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Firstly, given a single light source at sufficient distance that light can be treated as parallel beams, together with a spherical planet and no other reflecting object, the answer is clearly 'No' since only half the planet is illuminated at any time.

Given more than two sources of light, or reflections, we are into three-body problem so long term predictions are generally out which means that you need to decide how long you want the system to remain stable for.

I can't give all conditions you want at a single point in time ('significantly' over 50% illuminated and day longer than night, but I can give each on a single planet.

Consider a non-rotating planet with respect to fixed stars (implausible in the long term due to tidal effects but it could be 'stable' for many human lifetimes) which was in a highly eccentric orbit around a large star. Then, during periapsis (closest approach), over one face would be illuminated but day length would be comparatively short because the orbital speed would be high.

During apoapsis, the opposite face of the planet would be illuminated although because of the distance, only slightly over one hemisphere in total. However the orbital speed is slower at apoapsis so the day on that face would be longer than the night.

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  • $\begingroup$ In trying to explain sunrise and sunset times to someone yesterday, I happen to have worked out that a similar effect exists for a planet in an eccentric orbit which has two sidereal days, and thus only one solar day, per year. One side gets a longer day, the other side a longer night. $\endgroup$ – Tristan Klassen Dec 31 '18 at 20:03

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