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I've been working on my story planet and have come up with an unlikely but possible scenario to help explain some of my stories origins as well as myths and folklore. Over the last day or so, I've been playing around with a small moon in a geostationary orbit around the planet. Now I just need to clarify how the phases of the moon would appear to my surface-dwellers.

A moon in a geostationary orbit is possible as answered here on astronomy SE. Unlikely but possible. For any particular location the moon would appear fixed and always in the same place in the sky. There would be no rising or setting of such a moon, it's just always there. A constant Big Brother looming over you. Some locations on the planet wouldn't even know of its existence.

Typically a moon takes at least a few days to orbit a planet. The way that the light from a star reflects off the moon surface gives observers the phases of the moon. Here is a helpful image from Wikipedia which shows the lunar phases in relation to their position around Earth over a one month lunar cycle. Drawing of lunar phases and orbital positions

If we now interpret this image as a one day lunar cycle, with the moon always in the same location above the planet. Ignoring how long such an orbit would remain stable, am I correct in figuring that from an surface observer's point of view the moon would appear to wax and wane fully in one day?

OR would it be always in the same phase for a particular location?. eg observer in Location A would always see a crescent moon and Location B always a full moon?

Note: for the purposes of this scenario, it's a very small moon and as such won't have huge tidal forces on the planet. It also won't block out large portions of the sky during lunar eclipses or reflect too much light in the evenings. All of which, while very interesting, are not part of this question.

I also don't mind that it probably would be located within the Roche Limit for its mass and density, as that is all part of the story!

I just need help to figure out the phases. sadly, no werewolves are involved...

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    $\begingroup$ Once more, Kerbal Space Program provides a useful example. (Duna, KSP's Mars-analogue, has a single huge moon called Ike that it's mutually tide-locked with, somewhat like the real-world Pluto-Charon system.) $\endgroup$
    – Ilmari Karonen
    Commented Feb 20, 2017 at 9:13
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    $\begingroup$ Yes, the moon would wax and wane in one day. The phase has little to do with the observers location, it is purely how much of the moon that is facing the sun that the observer can see. Two observers at different longitudes would see different phases. $\endgroup$
    – Paul Smith
    Commented Feb 20, 2017 at 12:05
  • $\begingroup$ @PaulSmith, wait. So different latitudes see different moon positions and different longitudes will see different phases at the same time for the same moon position???. Observers would also see the moon either a little bigger or smaller depending on their longitude??? $\endgroup$
    – EveryBitHelps
    Commented Feb 20, 2017 at 12:44
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    $\begingroup$ @EveryBitHelps Yes. Text moved to an answer. $\endgroup$
    – Paul Smith
    Commented Feb 20, 2017 at 13:43
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    $\begingroup$ @EveryBitHelps Yes, and those effects will be significant (in astronomic scale). For example: geostationary orbit has radius of ~35700 kilometres, while Earth radius is ~6400 kilometres. Observer at equator, where moon is in zenith is 29300 km away, while observer at pole or at any longitude 90 degrees away from moon-zenith spot would be 36300 km away from moon, and there moon will appear 35% smaller. $\endgroup$
    – M i ech
    Commented Feb 20, 2017 at 17:30

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In one day you would see the full cycle (if you can see it)

So if you stand a little distance from a lamp holding a ball at arms length, then turn around, you will see the phases on the ball. This is the same process, you are turning at the same rate as the ball, just as the planets surface and the moon would be.

Your small not-too reflective moon might not always be visible...but for the sake of a story there is obviously no harm in you ignoring that.

Inclination and eccentricity

I appreciate Zxyrra's point about the inclination and eccentricity - they would change the moon's appearance if you had anything but a circular orbit angled normal to the planet's rotational axis. However a geostationary orbit is necessarily circular and angled normal to the planet's rotational axis. Due to this the moon will always orbit around the equator of the planet.

What would affect this would be the planet's tilt and eccentricity. If the planet was tilted 90$^{\circ}$ you could instead observe a moon permanently half-lit, half dark top to bottom rather than left to right. This would vary from one extreme to the other depending on how you wish to tilt your planet.

All in all the best way to play about with how this would look involves you, a lamp, a ball and making yourself dizzy

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  • $\begingroup$ Excellent points. Just clarifying. I didn't mean the small moon would be "not too reflective". I was trying to say that if a moon the size of our Moon was in geostationary orbit that the reflective light would dominate the night sky. As my moon would be smaller, this wouldn't be the case. It's still just as reflective...too scale. So would it still be more constantly visible than you mentioned, right? $\endgroup$
    – EveryBitHelps
    Commented Feb 20, 2017 at 7:17
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    $\begingroup$ @EveryBitHelps Please video yourself running around a lamp while holding a ball and spinning around, and upload it to YouTube. Totally for educational reasons, of course. $\endgroup$
    – David Richerby
    Commented Feb 20, 2017 at 13:44
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    $\begingroup$ If the planet was tilted 90°, the phase of the moon would gradually change over the course of a year. If the moon is perpetually half-lit, then 1/4 of a year later it'll be going through all its phases. Then 1/4 of a year later it'll be back to half-phase again. In between these points you'll get to see the moon oscillate between partial-full and crescent to varying degrees. Also your seasons would be pretty extreme. $\endgroup$
    – Kyle
    Commented Feb 20, 2017 at 16:24
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    $\begingroup$ @Kyle, thank goodness my planet is only tilted the 'normal' amount to provide regular seasons. But good to know if I ever what to give my surface dwellers a roller coaster ride :) $\endgroup$
    – EveryBitHelps
    Commented Feb 20, 2017 at 17:06
  • $\begingroup$ To add on to @Kyle's comment: That's because the planet's rotation axis points in a constant direction- it won't rotate as the plane orbits its star. So at the winter solstice, the north pole will point directly away from the star; at the summer solstice, the pole points directly toward the star; and at the equinoxes, the poles are perpendicular to the star. Thus, at the winter (summer) solstice, the entire southern (northern) hemisphere of the planet and moon will be illuminated, at the equinoxes, they'll have a normal day/night cycle everywhere, and in between for the rest of the year. $\endgroup$
    – Someone Else 37
    Commented Nov 6, 2019 at 6:18
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You are (somewhat) correct

Just as shadows of stationary objects on Earth change direction as the sun passes, so too will the shadow of your moon - kind of like a sundial.

However, the moon's eccentricity (shape of orbit) / apsis (distance along orbit) and inclination are able to change this phenomena. It may be farther away from the planet at some times than others, causing phases with differing duration - or cutting off a completely full moon. Additionally, just as we don't have an eclipse every night here on Earth, an inclination could still change the relative location of the moon - regardless of if it revolves at the same speed the planet rotates.

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    $\begingroup$ Geostationary specifies an eccentricity and inclination of 0. Other geosynchronous orbits can vary. $\endgroup$
    – Mark
    Commented Feb 20, 2017 at 4:15
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The phase of the moon depends on the angle between the line from the sun to the earth and the line from the earth to the moon. When that angle is 0°, the moon is between the earth and the sun, so we see a new moon; when it's 180°, the moon is on the opposite side of the earth to the sun, so we see a full-moon; when the angle is between these two extremes, we see some fraction of the moon's surface in shadow. The variation of phases happens across one full orbit.

The same holds true for a geostationary moon. Since the angle between the two lines varies as the moon follows the earth's rotation, we still see phases. Now, the moon's full orbit takes one day, so you'll see the full range of phases during a single day. If you're close to the point the moon orbits above, you'll see the new moon at solar noon and the full moon at midnight, with waxing in the afternoon and evening, and waning through the night and morning. As you move farther away from that point, the times will change because your angle of view will be different. And, of course, if you're on the other side of the planet, you won't see the moon at all.

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Think of it as a triangle with the sun, the moon and the observer. The face of the moon is always lit by the sun (except during eclipse) and how much of that face is visible to the observer depends on where the observer is in relation to the moon. Closer to the poles, observers are further from the moon so it looks smaller to them. An observer on the ground underneath the moon will see a new moon at noon, and a full moon at midnight. At any given time, an observer on the equator closer to the sun will see more of the lit up face then one further back.

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Just go to the Moon and take a look. From the Moon's point of view, the Earth is a selenostationary$^{1}$ satellite just as you describe. And as others have said, it goes through the full range of phases every lunar day.

$^{1}$ "Seleno" refers to the moon so, just as geostationary means orbiting Earth above the same point, selenostationary means orbiting at a particular point above the moon.

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