0
$\begingroup$

Working off this question and its suggestion of binary star systems to increase the length of 'day' on a planet without increasing the length of 'night'

I have a gas giant (1.2 Mass/Jupiter, 1 Radius/Jupiter, Obliquity around 20°, Rotational Period (right now) 6.242 days but up in the air, Orbital Period 1.76 yrs, Semi-major Axis 1.46 AU) with rings around which orbits a habitable, tidally locked moon on which the actual worldbuilding takes place (0.6 Mass/Earth, 0.926 Radius/Earth, Rotational/Orbital Period 6 days (72h night, approx. 25h day, approx. 22h eclipse, approx. 25h day), Semi-major axis 0.00677 AU).

At the moment, the system has a single sun (Mass 1/Sun, Radius 1.01/Sun, Surface Temperature 5752K, Luminosity 1/Sun, Speed 24.9 m/s, Rotational Period 26.2 day, Rotational Period 4.66 gigayear) but in order to achieve the longer period of brightness of the gas giant (as seen from the moon), I think turning the system into a binary star system would be my best bet. I think it would be neat if the rotation of the planet (as seen from the moon) would be 2/3 bright and 1/3 dark. I'm trying to figure out how big, bright, and how far away the second sun should be to 1) have that dark-bright ratio I just mentioned 2) not affect the planet-facing side of the moon too much.

By 2 I mean that I'd really like to keep the six days with technically 3 nights and 3 days, with a large portion of that day period being eclipse (right now it's a little less than 1/3 of the day period). I have no plans for the side facing away from the planet, so technically it could be bright the whole time, if it, for example, faces the second sun.

It can't be a Trojan position, I think, because then the smaller star in the same orbit as the planet would illuminate the moon when it faces in that direction. S-type would seem the most plausible to me, angled in such a way as to illuminate half of the planet when it 'rotates away' from the sun that it orbits. I've tried doing that in Universe Sandbox. What I have right now is a second star with 0.123 Mass/Sun, 1.44 Radius/Jupiter, temperature 3331K, luminosity 0.158 L/Sun, rotational period 3.76 days, orbital period 11.8 years. It's 5.41 AU away from the sun and positioned 5.06 AU away from the planet. Watching the rotations of the planet and the moon, it looks to me like the second star would automatically turn part of the night-portion of the moon into day, which is something I want to avoid.

If it's not possible, then it's not possible. I'm just wondering whether there's a reasonably logical mathematical solution.

$\endgroup$
8
  • 1
    $\begingroup$ Basic geometry says that the dark side of the primary and the dark side of the satellite face in the same direction. If a light source is able to illuminate the dark side of the primary then it is also able to illuminate the dark side of the satellite. $\endgroup$
    – AlexP
    Dec 29, 2023 at 15:20
  • $\begingroup$ @AlexP That's what I thought would be the case and saw in US. Right now, and my diagrams might be wrong, night happens on the satellite when its planet-facing side faces away from the sun and is only illuminated by the planet it orbits. If a second sun is there on the opposite side, then the night period would become bright and thus day. I guess that's just something I have to figure out with the planet's rotational period and how that fits best into what I want without fiddling around with binary suns. $\endgroup$ Dec 30, 2023 at 10:47
  • $\begingroup$ @Alex P I think I had a major thinking error. It seems to me that with the planet's original rotational period of 10-ish hours, the satellite faces the sun while the planet is dark and when M has night, P is illuminated. Watching US, P rotates very slowly, several rotations of my satellite. So my 'problem' only becomes a problem once P's orbit makes it so that the 'shadow' over it moves from the minuscule differences before. I think that's more correct than my prior thoughts so I wouldn't have to change the 10-something hours I had originally? Unless I misunderstood even more now. $\endgroup$ Dec 30, 2023 at 11:01
  • $\begingroup$ I simply do not understand what is the relevance of the rotation period of the primary. The primary may rotate fast or it may rotate slow, so what? For the purpose of estimating the illumination of the satellite, the primary is a featureless diffuse reflector; its rotation period is immaterial. But then anyway, I don't understand why you would think that "night happens on the satellite when its planet-facing side faces away from the sun". What happens when its planet-facing side faces towards from the sun? Isn't it night on the other side of the satellite? $\endgroup$
    – AlexP
    Dec 30, 2023 at 11:55
  • $\begingroup$ @AlexP The more than unscientific method of drawing a diagram (imgur.com/a/IfTKVSc) and dividing it according to whether it faces the sun or not. My worldbuilding takes place on the planet-facing side of the satellite only, so I haven't looked at the other side, where it would be the other way around. Though I have the feeling I'm going to go back to the drawing board anyway $\endgroup$ Dec 30, 2023 at 12:12

1 Answer 1

1
$\begingroup$

The two stars will illuminate different regions of the planet and the moon. But those different regions will will be overlapping.

As the two stars orbit around each other and as the planet and the moon orbit around one or both of the stars, the angles between the stars and the planet & moon will constantly change, thus changing the degree of overlap of the illuminated regions on the surfaces of the planet and the moon.

If you ever have difficulty visualizing the relationships between different objects in space, you should draw diagrams of the spatial relationship I will mention.

There are two types of orbits a planet can have in a binary star system. A circumbinary or P-type orbit or a non-circumbinary or S-type orbit.

This article has a diagram of a binary star system, with one planet in P-type orbit and one planet in an S-type orbit.

https://en.wikipedia.org/wiki/Habitability_of_binary_star_systems#:~:text=Planets%20that%20orbit%20just%20one,planets%20within%20stable%20orbital%20ranges.

Part One: A P-Type Orbit Around Both Stars.

Suppose that the planet orbits around both of the stars, in what is called a circumbinary or P-type orbit.

In that case there will be times when the two stars are lined up in a line with the planet and thus will be seen side by side from the planet. If the orbit of the planet is in the same plane as the orbit of the two stars around each other, at that time the nearer star will eclipse the farther star. If the two stars have the same luminosity, their combined light on the planet will drop down to one half of its normal amount during an eclipse. If the two stars have different luminosities, then alternate eclipses will have greater and lesser total luminosity.

And as the two stars orbit around each other, and the planet orbits farther out at a slower speed, the angular separation between the two stars as seen from the planet will gradually increase to its limit and then gradually decrease back down to zero.

How far apart can the two stars orbit compared to the orbit of the planet orbiting both of them?

For a circumbinary planet, orbital stability is guaranteed only if the planet's distance from the stars is significantly greater than star-to-star distance.

The minimum stable star-to-circumbinary-planet separation is about 2–4 times the binary star separation, or orbital period about 3–8 times the binary period. The innermost planets in all the Kepler circumbinary systems have been found orbiting close to this radius. The planets have semi-major axes that lie between 1.09 and 1.46 times this critical radius. The reason could be that migration might become inefficient near the critical radius, leaving planets just outside this radius.[9]

https://en.wikipedia.org/wiki/Habitability_of_binary_star_systems#:~:text=Planets%20that%20orbit%20just%20one,planets%20within%20stable%20orbital%20ranges.

So if your planet orbits both stars the minimum possible distance of the planet's orbit around the barycenter of the two stars will be 2 to 4 times the separation between the 2 stars. And the planet could orbit much farther out from the stars.

So you could draw a diagram of 2 stars orbiting around their common center of gravity. And draw orbits of planets orbiting around the center of gravity at 2, 3,, 4, 5, 6....times the diameter of the orbit of the two stars. And you could draw cones from the two sides of the orbit of the two stars to each of the planetary orbits and then use a protractor to measure the angles. Or you could calculate the maximum angles for various ratios of orbits.

According to my rough calculations, if the planet orbits at 2 times the separation between the 2 stars, the maximum possible angular separation between the two stars as seen from the planet will be about 28.648 degrees of arc, if the planet orbits at 3 times the separation between the 2 stars, the maximum possible angular separation between the two stars as seen from the planet will be about 19.0986 degrees, if the planet orbits at 4 times the separation between the 2 stars, the maximum possible angular separation between the two stars as seen from the planet will be about 14.3239 degrees, and so on.

And the degree of overlap between the hemispheres of the planet illuminated by the two stars will decrease the farther apart the two stars are. It will equal 180degress minus the degrees of angular separation between the 2 stars as seen from the planet.

And if the planet rotates in the same plane as its orbit around the stars and their orbit around each other, the number of degrees of longitude on the planet illuminated by one or both of the stars will equal 180 degrees plus the angular separation between the two stars.

So if the planet orbits at twice the orbital separation of the two stars, the absolute minimum possible for a stable orbit, the angular separation between the two stars as seen from the planet will be between zero and 28.648 degrees. And thus the degree of overlap between the longitudes illuminated by the 2 stars will vary between 180 degrees and about 151.352 degrees, and the percent of the planets surface illuminated by the two stars will vary between 180 degrees and 208.648 degrees of longitude.

And of course the two stars will usually appear much closer together as seen from the planet.

As the habitable moon orbits around the planet, the angle from the moon to the stars will be slightly different than the angle from the planet to the stars. But it should be a slight difference. The orbit of a moon around its planet has to be a tiny fraction of the orbit of the planet around its star for the moon's orbit to have long term stability. And long term orbital stability over billions of years is necessary for a planet or moon to become and stay habitable and thus interesting for the purposes of most science fiction stories.

Part Two: A S-type Orbit Around One of the Stars.

In that situation the two 2 stars will orbit around their common center of gravity and the planet will orbit around one of those stars, and much closer to that star. It will orbit much faster over a shorter distance than the stars orbits around their center of gravity, and so will have a shorter orbital period.

And sometimes the orbit of the planet will take it where the farther star is lined up behind the nearer star and possibly eclipsed by it as seen from the planet. Then the angular separation between the 2 stars as seen from the planet will be about zero. And there will be times when the orbit of the planet takes it between the two stars and they appear on opposite sides of its sky 180 degrees apart.

And most of the time the angular separation between the two stars in the sky of the planet will be between zero and 180 degrees. The angular separation will gradually grow from zero to 180 degrees and then gradually diminish to zero degrees, over and over again. And thus the percentage of the planet which is illuminated will gradually grow from 50 to 100 percent and then gradually shirk from 100 percent to 50 percent, over and over again.

Since the orbital period of the Moon around the planet will be much shorter than the orbital period of the planet around one of the stars, the angle between the two stars will not change much as seen from the Moon during one orbit of the moon around the planet. But that angle will gradually change as the planet orbits the star.

Part Three: A S-type or a P type orbit?

In either a P-type or an S-type orbit, the two stars will sometimes appear at least a few degrees apart. Which means that the planet might be able to eclipse them at different times. So the side of the moon facing the planet could have periods lit by both stars, eclipse of one star periods, eclipse of the other star periods, possibly overlapping eclipse of both stars periods, and both stars beneath the horizon but some light reflected from the planet periods, all during one orbit around the planet. But that depends on how separate the two stars are in the sky of the planet, which will be always changing.

In an S-type orbit the maximum angular separation between the two stars will be much greater, and their average angular separation will be be greater. Thus separate eclipses of the stars by the planet will happen a much larger percentage of the time.

So that is one reason why you might want to consider making your system a S-type system.

But on the other hand, consider the masses and luminosities of stars. There is a great range in the masses of stars, which causes a many times greater range in their luminosities. But there is a much narrower range in the possible masses and luminosities of stars which can be found in star systems with habitable planets or moons.

In a P-type system, both stars will be about the same distance from the planets, the more so the farther away the planet orbits relative to the distance between the two stars. So if the planet orbits at five or ten times the separation between the two stars, the distances to the two stars will not change very much as seen from the planet.

So if you give both stars the same mass and luminosity, the eclipse of star 1 by the planet with change the illumination of the moon approximately as much as the eclipse of star 2 by the planet will.

In an S-type orbit, the planet and its moon will orbit much closer to star 1 than the separation between Star 1 and Star 2. So Star 2 will always be at least a few times farther away from the planet and moon than Star 1 is.

So if Star 1 and Star 2 have the same mass and luminosity, if Star 2 is three times as far from the planet and moon as Star 1 the planet and moon will receive only 1/9 or 0.11111 as much radiation from Star 2.

If Star 2 is 4 times as far the planet and moon will receive only 1/16 or 0.0625 times as much radiation from star 2.

If Star 2 is 5 times as far the planet and moon will receive only 1/25 or 0.04 times as much radiation from star 2.

If Star 2 is 6 times as far the planet and moon will receive only 1/36 or0.27777 times as much radiation from star 2.

If Star 2 is 10 times as far the planet and moon will receive only 1/100 or 0.01 times as much radiation from star 2.

If Star 2 is 20 times as far the planet and moon will receive only 1/400 or 0.0025 times as much radiation from star 2.

If Star 2 is 31.62278 times as far the planet and moon will receive only 1/1,000 or 0.001 times as much radiation from star 2.

So Star 2 is far enough away for the orbits of the planet and moon to be long time stable, it will provide much less heat to the planet and moon and thus its eclipses by the planet will not change the temperature very much.

But the picture with respect to illumination is much brighter.

On Earth, the brightest sunlight has an apparent magnitude of -26.832. which is about 400,000 times the brightness of a full moon at an apparent magnitude of -12.90. And the full moon is bright enough to see what you are doing and where you are walking. If you want humans to be able to see colors by the light of star 2, it will have to be many times brighter than the full moon, but can still be only a tiny fraction as bright as full sunlight.

And of course you can compensate for the greater distance of Star 2 in an S-type system by making it more luminous than Star 1. But as I wrote above there are is a comparatively limited range in the masses and luminosities of stars which can be found in systems with habitable worlds.

You want your moon to have an orbital period around its planet of 6 Earth days. That means your planet will have to have an orbital period around its star or stars of at least about 54 days, because the planet's orbit needs to be at least 9 times as long as the moon's orbit around the planet for the moon to have orbital stability. You can fudge that a little, because the moon's sidereal orbital period is relevant for orbital stability, while the moon's synodic orbital period is relevant for the lengths of its day/night cycle.

According to the table in this article:

https://en.wikipedia.org/wiki/Red_dwarf

An M0V type star will have a mass of 0.57 Sun and a luminosity of 0.069 Sun. The square root of 0.069 is about 0.26268, so a planet orbiting a M0V type star at a distance of 0.26268 AU will receive as much radiation from its star as Earth gets from the Sun. it will be at the Earth Equivalent Distance or EED of that star.

According to this orbital velocity calculator, a planet with 381.36 times the mass of Earth at an semi-major axis of 0.26268 AU from a star with a mass of 0.57 Sun will have an orbital period of 65.06 days.

A M1V type star should have a mass of Sun and a luminosity of 0.041 Sun. Its EED will be at about 0.20248 AU. The calculator says a planet at its EED should have an orbital period of 47 days.

So an M0V class star should have the minimum mass and luminosity for your purpose.

A G0V type star would have a mass of 0.06 Sun and a luminosity of 1.35 Sun.

https://en.wikipedia.org/wiki/G-type_main-sequence_star#:~:text=A%20G%2Dtype%20main%2Dsequence,about%205%2C300%20and%206%2C000%20K.

Its EED would be at a distance of 1.1619 AU. That is 4.4232 times the EED of a M0V star.

An F5V type star would have a mass of 1.33 sun and a luminosity of 3.63 Sun.

https://en.wikipedia.org/wiki/F-type_main-sequence_star#:~:text=F%2Dtype%20stars%20have%20a,they%20begin%20to%20fuse%20carbon.

It would have an EED at about 1.9 AU, about 7.233 times the EED of a M0V type star.

An F0V type star would have a mass of 1.61 Sun and a luminosity of 7.24 Sun.

It would have an EED at about 2.69 AU, 10.243 times the EED of a M0V type star.

And that is as massive and luminous a star as most scientists in the field think could be in a system with a habitable planet. Probably much more luminous and massive than they would accept.

So if your planet has an S-Type orbit, the correct selection of the luminosities and orbits of the stars and the orbit of the planet can make the planet and the moon receive as much radiation from the more distant Star 2 as from the closer Star 1, but it would be a rather tight squeeze.

If you desire the planet and moon to receive a total radiation equal to that which Earth gets from the Sun, and to get equal amounts from each of the stars, then than the planet will have to be at a distance of 0.7071 of each star's EED from each of the stars to receive 0.50 of Earth's radiation from each of the stars.

These are some factors to consider when deciding between an S-type orbit and a P-type orbit in a binary star system.

Part Four: The Moon's orbit.

What did you write about the orbital parameters of the moon's orbit around the planet?

I have a gas giant (1.2 Mass/Jupiter, 1 Radius/Jupiter, Obliquity around 20°, Rotational Period (right now) 6.242 days but up in the air, Orbital Period 1.76 yrs, Semi-major Axis 1.46 AU) with rings around which orbits a habitable, tidally locked moon on which the actual worldbuilding takes place (0.6 Mass/Earth, 0.926 Radius/Earth, Rotational/Orbital Period 6 days (72h night, approx. 25h day, approx. 22h eclipse, approx. 25h day), Semi-major axis 0.00677 AU).

Naturally I used the orbital period calculator to check your orbital period. I got an orbital period of 6.007 days or 144.168 hours. The orbital period calculator gives an orbital period of 6.0 days or 144 hours with a semi-major axis of 0.006765 AU.

I don't know which calculation is more accurate.

0.00677 AU is 1,012,777.585 kilometers, and 0.006765 AU is 1,012,029.5953 kilometers.

If the Moon's orbit is perfectly circular, it will have a circumference of about 6,363,463 or 6,359,391.569 kilometers. Each degree along its orbit will be 17,676 or 17,664 kilometers.

If the planet has exactly Jupiter's equatorial radius of 71,492 kilometers, the shadow it casts upon the orbit of the moon should be about twice that, or 142,984 kilometers long, or about 8.09 degrees of the orbit.

If the eclipse of the moon by the planet takes 22 hours out of a total 144 hours, or 0.152777 of the orbital period, the moon should travel 55 degrees along its orbit during the eclipse.

Those calculations don't exactly agree.

If the Moon has an orbit around the planet with a circumference of about 6,361,000 kilometers and a radius of about 1,012,385 kilometers, a circle around the the moon which goes through the center the planet would have exactly the same circumference and radius. According to my rough calculations, the planet with a diameter of 142,984 kilometers would have an angular diameter of about 8.09 degrees as seen from the moon.

The eclipse time of 22 hours would seem to contradict the other two calculations.

With an orbital circumference of about 6,361,000 kilometers, and an orbital period of 144 hours, the moon would travel at about 44,173.611 kilometers per hour, and would take about 3.236 hours to pass through the shadow of the planet which is 142,984 kilometers wide.

To make the eclipse last longer, you have to increase the radius and diameter of the planet, or else the make the moon orbit closer to the planet.

Unfortunately giant planets can't get much larger than Jupiter. After a certain point adding more mass just makes them denser and denser, not larger. The one way to get around that is to swell the atmosphere of the giant planet to several times the normal size, by heating it up a lot. Heating it to thousands of degrees F, C, or K. A process to do that would also heat up the moon to such hellish temperatures.

If the moon orbits closer to the planet, the shadow of the planet will be large relative to the orbit of the moon, and the moon will spend a larger proportion of its orbit in the shadow. Unfortunately, as the moon gets closer to the planet it necessary orbital speed will increase, decreasing the total orbital speed and the total orbital period and the period spent in the shadow of the planet.

But you can have a six day orbital period with your moon orbiting much closer to your planet, if your planet is much less massive than Jupiter, thus lowering the orbital speed and lengthening the orbital period. But a planet much less massive than Jupiter will be smaller than Jupiter.

Probably a lot of calculations would be needed to find the situation where the orbital periods and eclipse periods come closest to what you desire.

And maybe the planet's rings, with the right diameter, thickness, and opaqueness, might extend the apparent diameter of the planet and the size of its shadow.

I also note that it has been calculated that a habitable moon should orbit a giant planet at a distance of at 5 least planetary radii to avoid tidal heating causing a runaway greenhouse effect. According to my rough calculations that means that a giant planet can appear no wider than about 20 degrees in the sky of a habitable moon. And thus if the orbital period is 144 hours, the eclipse should last no more that about 8 hours.

And maybe the rings can extend the eclipse a bit. But the wider they are compared to the planet, and the closer the moon is to the planet, the more likely the moon is to rapidly destroy the rings.

Part Five: The Planetary Day.

You write:

Rotational Period (right now) 6.242 days but up in the air,

And the Moon has a rotational period of:

...tidally locked moon...Orbital Period 6 days

You might need to adjust the orbital period of your moon, making it longer or shorter.

And it is a good thing that your planet's rotational period of 6.242 is up in the air and subject to change.

How did your moon get tidally locked? Earth's moon and many other moons in the solar system got tidally locked to their planets by tidal interactions. Those tidal actions slowed down the rotation periods of the planets and moons while also pushing the moons farther from their planets. Once the moons became tidally locked to their planets, the tidal interactions continued to push the moons farther out and to slow down the rotation of the planets. If the planets and moons last long enough eventually the planets will also be tidally locked, to the moons.

However, there are exceptions.

Some moons experience tidal deceleration.

Moons orbiting in retrograde directions, opposite to the directions their planets rotate in, will slow the rotation of their planets, but slowly move closer to their planets. Triton, the big moon of Neptune, is slowly getting closer to Neptune and is expected to crash into it in a few billion years.

Moons orbiting in prograde directions, the same directions as their planets rotate, move outward from their planets if they orbit above the synchronous orbital level where their orbital periods are longer than their planets' rotation periods.

Moons orbiting in prograde directions, the same directions as their planets rotate, suffer tidal deceleration if they orbit lower than the synchronous orbital level and had orbital periods shorter than the rotation periods of their planets. The tidal interactions speed up the rotation of their planets and pull the moons closer to their planets. Thus Phobos, the inner moon of Mars, is expected to crash into Mars in just tens of millions of years.

You choose a rotation period of 6.242 Earth days for your planet and an orbital period of 6 Earth days for your habitable moon. That means that your moon is orbiting below the synchronous orbit and is slowly descending toward the planet, doomed to be destroyed in millions or billions of years. That may or may not be a reason to change the rotational speed of your planet.

I note that the giant planets in our solar system have rotational periods of 0.41354, 0.44002, 0.71833, and 0.67125 days. The giant exoplanet Beta Pictoris B has a rotation period of 8 hours or 0.33333 days.

So your planet's rotational period of 6.242 Earth days could be made much shorter without anyone thinking it was odd.

$\endgroup$
2
  • $\begingroup$ Thank you for the very detailed explanations. Reading everything you said, there doesn't seem to be a way to keep the day-night-eclipse cycle of my moon the way it is while simultaneously having only the planet be illuminated more by either of the two suns since "The two stars will illuminate different regions of the planet and the moon. But those different regions will be overlapping." When I looked at satellites through Stellarium, I found that most often when the planet is dark, the satellite had daylight. Is that always the case or was it just a coincidence? $\endgroup$ Dec 30, 2023 at 10:50
  • $\begingroup$ Regarding the day-night-eclipse length, I put everything I mentioned into Universe Sandbox (and ran it for gigayears, everything looked stable), plus looked at questions in this forum that asked the same thing, and used the formulas there. It gave me 72h each during the approx. 6 day rotation/orbital period, of which I assumed half to be night (since the planet-facing side would face away from the sun during that period). Then I calculated the eclipse period and got the approx. 22 hours I mentioned, which divides the day portion into more or less three periods of day, eclipse, day. $\endgroup$ Dec 30, 2023 at 10:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .