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Is this physically possible?

A non-technological phenomenon visible in the same position of the sky, for 18 hours of a day using the other 6 to do whatever, rise and set, just be impossible to see, it doesn't matter.

What matters is the 18 hours of constant position in the sky, on a repeating cycle.

For an observer that is assumed to be watching from the same place, each cycle.

The viewer's planet is not Earth, just has enough similarities for humans to live on it.

If it is possible, how complex a system would I need to make such a thing happen, and how stable would that system be?

The cause can be in the atmosphere, as long as it cycles, and has the same visibility.

I'd like planets, but if that's not possible, then use whatever is possible.

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    $\begingroup$ Does it have to be visible from everywhere on earth at some point (or at least, all around some great circle of the earth's surface), or is having it only visible from one part of the globe acceptable? $\endgroup$
    – Starfish Prime
    Commented Apr 28, 2019 at 19:08
  • $\begingroup$ @StarfishPrime - Observer of single position. Can you help me think of how to add that into the title, while not going over 150 characters? $\endgroup$
    – Malady
    Commented Apr 28, 2019 at 19:13
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    $\begingroup$ What does "int the same position for 18 hours a day" mean? Is it in the same position with respect to the fixed stars? Or is it immobile, that is, it does not rise and does not set, and in this case what happens during the other 6 hours? Must it be visible in daylight? What does "in the sky" mean? In outer space, or is an object flying in the atmosphere acceptable? $\endgroup$
    – AlexP
    Commented Apr 28, 2019 at 19:21
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    $\begingroup$ This is an interesting question, but so far all the comments are asking for clarifications which indicates to me you're not exactly clear what it is you're asking. I've voted to put your question on hold until you edit it. If you clarify before it actually gets closed, I'll happily retract the VTC! $\endgroup$
    – elemtilas
    Commented Apr 28, 2019 at 19:38
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    $\begingroup$ Keep the title short; put all the "stuff" in the query body. $\endgroup$
    – elemtilas
    Commented Apr 28, 2019 at 19:45

13 Answers 13

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It's physically possible.

Put a spherical object in a geostationary orbit, make it rotate about its own axis at a rate suited to your own visibility/non-visibility requirements, and make a portion of it have very low albedo.

Staying at a single point in the sky, the object will only be visible while the higher albedo portion is facing the planet and become invisible while the low albedo portion rotates into view.

It could technically happen by chance but would only be stable for as long as the orbit is stable, which really depends on your planetary system.

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    $\begingroup$ Difficult to imagine another possible answer. +1 $\endgroup$
    – Escaped dental patient.
    Commented Apr 28, 2019 at 20:16
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    $\begingroup$ @user10915156 Well, if the creator of this world is a god then those terms suddenly mean natural phenomenon again! Amazing how that terminology problem got solved. $\endgroup$
    – Samuel
    Commented Apr 28, 2019 at 20:26
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    $\begingroup$ No intention of taking away from this answer, but in thinking about providing a similar answer I realized the precession of a geostationary orbit would render the position not fixed. The largest factor here would be a slow cycling of the inclination of the orbit (satellites use thrusters to counter this effect). You do note that the stability depends on the planetary system so the answer still works on shorter timescales but for long term stability a planet with less J2 (equatorial bulge) than Earth would help reduce the precession and keep the effect stable for longer. $\endgroup$
    – ben
    Commented Apr 28, 2019 at 20:40
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    $\begingroup$ @ben It's entirely possible that the planet in question has no precession. I was thinking about a way for the orbital mechanics to work such that solar pressure/off-gassing corrected the orbit, but that's unlikely to add much more time without violating the other precepts of the question. $\endgroup$
    – Samuel
    Commented Apr 28, 2019 at 20:42
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    $\begingroup$ @ben: Would the planet's own equatorial bulge make a geostationary satellite precess? By symmetry that ought to be unlikely. What nudges Earth's geostationary orbits out of the equatorial plane must be the righting moment of solar/lunar tides, causing gyroscopic precession -- but since the planet is not Earth, we can posit that it has negligible axial tilt and do away with the moon. Then a geostationary orbit should be stable, right? $\endgroup$
    – hmakholm left over Monica
    Commented Apr 29, 2019 at 21:56
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You didn't say how far up in the sky you need your object and what type of object you want, so I suggest the plume of a volcano.

Some volcanos and geysers are quite regular in their eruptions. Your volcano will erupt not long after midnight every 24 hours and emit only a short burst of gaseous matter and fine dust-like particles that will drift upward in the still air and remain visible for 18 hours, until the evening wind scatters the cloud and it disappears.

If the eruption is just a short puff, the plume will be a small near-spherical cloud, as this one over Popocatepetl:

enter image description here

Etna even does smoke rings:

enter image description here

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  • $\begingroup$ pretty clever answer! $\endgroup$
    – Fattie
    Commented May 1, 2019 at 12:44
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A geostationary satellite follows an orbit which keeps it over the same point on the Earth.

https://www.skyandtelescope.com/observing/how-to-see-and-photograph-geosynchronous-satellites/

satellites

The streaks are stars which are elongated by the rotation of the earth and the long exposure. The satellites are rotating with the earth and so they look like dots. I was surprised that the satellites this blogger photographed did not track out an analemma like the sun, but he says they stay put.

Unlike the ISS and the many objects in low Earth object, geostationary satellites are visible all night long every night of the year.

Satellites are technological objects but a thing can be in orbit and not be technological.

If something were bright and in orbit you might be able to see it all the time. You could have it get bright alternately. With satellites these are called satellite flares.

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

Satellite flare, also known as satellite glint, is the visible phenomenon caused by the reflective surfaces of passing satellites (such as antennas, SAR or solar panels), reflecting sunlight toward the Earth below and appearing as a brief, bright "flare".

The satellites that are famous for this apparently rotate so as to present their reflective surfaces. Something in orbit could be slowly rotating, and when the non reflective side was presented it would seem to disappear to the viewer on the ground.

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    $\begingroup$ Great minds... But you did include a picture, so I suspect you'll be the winner by votes. $\endgroup$
    – Samuel
    Commented Apr 28, 2019 at 20:16
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    $\begingroup$ @user10915156 Our satellites are technological, but a satellite does not have to be technological. It can show up and get caught by a planets gravity. $\endgroup$
    – Willk
    Commented Apr 28, 2019 at 22:54
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    $\begingroup$ Also it is worth pointing out that the term satellite just seems to mean something orbiting. The moon is considered a satellite. $\endgroup$
    – Captain Man
    Commented Apr 29, 2019 at 14:36
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The Sun of a tidally locked planet, eclipsed by its Moon.

Your people live on a tidally locked planet. The Sun is always in the same spot in the sky.

(Traditionally such planets' habitable zones form a ring with the Sun near the horizon, for practical reasons)

Six hours per day, the Moon passes over, eclipsing it.

Caveats:

  • The eclips's start and end are not instantaneous

  • The satellite would have to be huge and/or very close for a 25% cover. You'd have to crunch the numbers to see if the system is feasible gravitationally.

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  • $\begingroup$ I couldn't find a reference for the habitable ring concept, probably because I'm missing the standard term for it. If anyone knows, please do edit in a link. $\endgroup$
    – Emilio M Bumachar
    Commented Apr 29, 2019 at 7:39
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    $\begingroup$ If you don't want the object to be the sun, it could be a trojan asteroid at the planet's L4 or L5 point if the planet is tidally locked to the sun. The asteroid could be eclipsed, or have a variable visibility due to its own rotation and shape / albedo. $\endgroup$
    – patstew
    Commented Apr 29, 2019 at 11:04
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    $\begingroup$ Not possible. In order to get 25% coverage from a contact binary, you'd need the moon to be about 40% bigger than the planet; if you want a stable situation, you'd need to be further away (and have a bigger "moon"). In such a situation, the planet would tidally lock to the "moon", not the Sun. $\endgroup$
    – Mark
    Commented Apr 29, 2019 at 20:58
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This is the equation you are looking for:

$$T^2GM=4𝜋^2R^3$$

This is Kepler's third law, and it correlates mass, semi-major axis length and orbital period.

For a geostationary orbit, you have a circle with a radius of approximatelly 40,000 km. Notice, however, that what the law actually states is that:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Starting from a perfect circular orbit, you can make it elliptical. As long as you keep a semi major axis as long as the radius of a geostationary orbit, your satellite's orbital period will be 24h - but it will have a periapsis much closer to Earth, and an apoapsis much farther. It will look like this:

Elliptical orbit

Bodies always spend more time closer to the apoapsis than closer to the periapsis. That's because their orbital speed is at its maximum at the periapsis and at its lowest in the apoapsis.

Just fine tune the eccentricity of the satellite to spend a quarter of its time closer to tje Earth on the day side and you're all set.

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  • $\begingroup$ Could you explain how this fulfills the most important criteria, "18 hours of constant position in the sky"? $\endgroup$
    – pipe
    Commented Apr 29, 2019 at 7:39
  • $\begingroup$ @pipe by having a 24h orbital period, it will stay pretyy much in the same position. It will move back and forth a little, but not more than a few degrees. That's how a lot of communication satellites work. $\endgroup$
    – The Square-Cube Law
    Commented Apr 29, 2019 at 9:49
  • $\begingroup$ You need to have a circular, or very- near-circular orbit to stay within a few degrees of geostationary (and near-equatorial on top of that). The more elliptical the orbit, the lower your angular velocity near apoapsis, and the higher your angular velocity near periapsis. See Kepler's Second Law. $\endgroup$
    – notovny
    Commented Apr 29, 2019 at 10:13
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    $\begingroup$ @Renan If you play KSP, you can see how much Ike moves around over a Dunar day, and its orbital eccentricity is only 0.03. The issue with an elliptical synchronous orbit is that the planet rotates with a constant angular velocity, but an elliptical orbit does not, so an object that's in a synchronous orbit but not a near-circular, near-equatorial orbit will change its sky position noticeably in a repeating 24-hour pattern. celestrak.com/columns/v04n07 $\endgroup$
    – notovny
    Commented Apr 29, 2019 at 11:43
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    $\begingroup$ If you want the position to be "effectively constant" for 18 hours, your eccentricity needs to be so high you're passing through the planet at periapsis. $\endgroup$
    – Mark
    Commented Apr 29, 2019 at 21:02
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The Coandă effect (Wiki)

It's the effect that allows a ping-pong ball to float in a hair dryer. The air clings to the rounded surface of the ball and air pressure magic keeps it in the jet, while the force of the jet itself keeps the ball afloat.

Replace the ball with a sufficiently shaped object (a smooth rock or something) and the hair dryer with a gaseous vent of sufficient strength and you can plausibly get yourself a rock floating (mostly) stationary in the sky for as long as the vent spews. If you want the rock to be higher in the sky, you can put the whole construct on top of a hill and view it from the foot of the hill or some distance away.

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    $\begingroup$ I wonder if topologically speaking a tidally locked planet could take a continuous solar flare and cause this effect...without rendering the entire planet a blasted wasteland (and/or pushing it out of the solar system) $\endgroup$
    – Ruadhan
    Commented Apr 29, 2019 at 15:09
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While geostationary satellites are the ideal, and obviously-correct answer if orbital altitude is permissible, they don't work so well in higher latitudes.

For these there's a less-stationary but still viable option: a highly elliptical inclined orbit such as a Molniya Orbit or Tundra Orbit, which are designed to give a high dwell time over the area of interest.

This will appear to trace a "γ" gamma-shape in the air, slowing down to essentially stationary and then reversing in the loop:

enter image description here

It spends 2/3 of its time in the small eye of that tail - for the geostationary Tundra orbit, that's the 18 hours you asked for.

With two or more satellites following this same orbit (a "constellation"), you get essentially constant coverage.

If, because of angling of solar panels or something, the satellites are only visible at certain times, such as at the apogee (the very tip of the gamma tail) they can then essentially look like a single stationary object, that periodically blinks out briefly and then turns back on (slightly to one side of where it turned off, but you'd have to be very accurately monitoring it to notice that).

The requested gap of a few hours could either be due to a gap in the constellation, or because to be visible they require the sun to be shining on them, and they are in the earth's shadow at that time.

However, for these to be non-technological would be a stretch. A highly elliptical orbit is feasible though unlikely for a single object, but multiple objects in a constellation, not so much. So, the tricks to make it seem extremely stationary won't work.

Against the sun or stars, though, a single object in a Tundra orbit would appear essentially stationary, rising, hanging there, and setting at the same horizontal position.

Without solar panels, it'd need to be very high albedo - clean white or perhaps crystal?

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    $\begingroup$ In many ways this is the best answer. $\endgroup$
    – Fattie
    Commented May 1, 2019 at 12:46
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The planet of a tidally locked satellite.

Your people live in a satellite tidally locked to a gas giant, around a red dwarf.

Similar to, and inspired by (hover to show spoiler)

https://en.wikipedia.org/wiki/Nemesis_(Asimov_novel)

The gas giant looms huge and fixed in the sky. It completely eclipses the star for six hours a day, the "day" being a revolution of the satellite around the gas giant. Having no inner light, it disappears to the naked eye while not illuminated by the star.

When the satellite is between the planet and the star, the planet is still illuminated, as the satellite is too tiny to eclipse anything.

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  • $\begingroup$ pretty clever, Emilio ! $\endgroup$
    – Fattie
    Commented May 1, 2019 at 12:46
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Polaris already does this, at certain places on the Earth and at certain times of the year.

The title asks for an object while the body asks for a [sic] 'phenomena'; to lean towards the latter we might also entertain:

A rainbow -- you might need to fiddle with the atmosphere a bit, but I think this could be arranged; I might guess something like this already occurs on Earth near waterfalls or that sort of thing.

The auroras -- by which I mean the aurora borealis and the aurora australis -- it seems likely to me that you could fiddle enough with a planet, its magnetic field, and its sun to make these visible 18 hours a day, at least on some parts of the planet. They do tend to take up quite a large segment of the sky.

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  • $\begingroup$ Okay, well, good job on not being an object, I guess. But you need Hard Science evidence to back up such things lasting 18 hours, cyclic, in a human-hospitable atmosphere. $\endgroup$
    – Malady
    Commented Apr 29, 2019 at 23:55
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Lenticular wave clouds stay in place, relative to the mountain, ridge or other topographic feature that creates them, and they can persist as long as the conditions are favorable. On Earth, around the summer solstice at the right latitude (e.g. London, 51.5 degrees North), you can have around 18 hours of combined daylight and civil twilight for almost a month (depending on how precise it needs to be -- note that the length of daylight does not change rapidly from day to day around the solstices), so the visibility requirement seems feasible. You could even posit a diurnal weather pattern in which, for example, the wind dies down overnight, causing the cloud to dissipate, only to re-form as the wind picks up in the morning. With this approach, you can separate the duration issue from the hours of daylight, if you assume the cloud is visible against the stars, or your planet has sufficient moons (or bright enough stars, in a globular cluster or near a galaxy's center) that it is never fully dark at night.

Continuing with the meteorological theme, consider also the Catatumbo lightning - towering clouds by day, lit up by lightning at night, confined to a specific and relatively small geographic area, and on a diurnal cycle. From the Wikipedia article:

Italian geographer Agustin Codazzi described it: "like a continuous lightning, and its position such that, located almost on the meridian of the mouth of the lake, it directs the navigators as a lighthouse."

Lenticular clouds, unfortunately (for your purpose), are not likely to generate lightning, but maybe it's not too much of a stretch...

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Make Mars go a little faster. Instead of ever seeming to move retrograde for a few months, it just seems to stop for about a day. Decrease their orbital periods and it happens as often as you want.

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    $\begingroup$ In order for this to work, Earth's orbital period needs to be about a day, with Mars's period being only slightly longer. That close to the Sun, planets tend to evaporate. $\endgroup$
    – Mark
    Commented Apr 29, 2019 at 21:05
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I can imagine a hot-air balloon type creature (similar to what Sagan imagined a living creature on Jupiter might look) that sits up however high you want in the atmosphere but periodically comes down to feed or rest (maybe it feeds on microbes high up in the atmosphere or has ultra stable DNA which allows it to live in higher radiation environments). If you're worried about it blowing around, just make it have an adaptation where it can track itself relative to the ground and is territorial.

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Since all the other answers are super-large scale, let's go with something a little smaller (as the question doesn't state that the same object must be visible at all places on the planet or even region).

Birds and Heat Vents

Imagine you have a bird (-like) creature, which can fly for extended periods of time, and is pretty communal. These birds have a very long range, but nest in large groups. Specifically, they've learned to build their nesting colonies around natural hot air vents (whether these are caused by volcanic activity, burning seams of coal, gigantic sleeping creatures, whatever). For a significant portion of the day, a constant column of these birds can be seen rising on the thermals produced by the vent at the center of their colony.

After rising on this natural vent (lots of free altitude), they break off formation, gliding away and using their keen eyesight to catch some prey before diving for dinner. They may bring some food back if you want, or something else that causes them to stay in packs (maybe they're pack hunters, something raven-sized that eats adult deer, and carry that back to the nest).

The constant column of birds leaving the nest from before dawn until after dusk would be a fixed sight, they'll always need to eat, and it even opens up some interesting plot (the bird column isn't out today?!?!?).

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