I am interested in the idea of a moon orbiting only the southern pole of a planet. Could it theoretically be possible (though unlikely) that a moon would, for instance, orbit a planet parallel to the equator, but at a latitude halfway or closer to the southern pole?

  • $\begingroup$ Thanks everyone for the responses to my question! $\endgroup$
    – mrkv12
    Feb 7, 2020 at 19:45
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    $\begingroup$ Even if you could, I’m fairly certain the difference would not be super noticeable? You’re talking about , on an earth-sized planet, a move “southwards” by -4000 miles. The moon is 238,000 miles away: relatively speaking, it hasn’t ‘moved’ very far. $\endgroup$
    – Daniel B
    Feb 7, 2020 at 20:10
  • $\begingroup$ Just a suggestion, but could you have a moon like Saturn's Iapetus solarsystem.nasa.gov/moons/saturn-moons/iapetus/in-depth that has dark & bright hemispheres? Place in in a near polar orbit (or perhaps a molniya orbit), then if you timed the rotation right, you'd only be able to see much of the bright pars of the moon from the neighborhood of the south pole. $\endgroup$
    – jamesqf
    Feb 8, 2020 at 4:29
  • $\begingroup$ So far no one has seemed to consider a Uranus-like planet whose rotational axis points roughly into the solar plane. $\endgroup$
    – Paul
    Feb 9, 2020 at 4:51
  • $\begingroup$ of possible interest: Traveling in a circle while in orbit $\endgroup$
    – uhoh
    Feb 10, 2020 at 1:02

10 Answers 10


Not quite. Orbits are constrained to have a focus in the system's barycenter, so you cannot have the satellite orbit a planet favoring one hemisphere only.

What you can do is alter the orbit so that the dwell time favors one hemisphere - you can have the moon stay 80% of the time over the Southern hemisphere and 20% of the time north of the Equator. It is called a molniya orbit.

Needless to say, it requires a very small moon, and the orbit is not stable for geologically significant periods. Your moonlet would have to be a captured asteroid of some kind.

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    $\begingroup$ Can you expand on the points which are "needless to say"? Unclear why you'd need a very small moon, or why this simple orbit is unstable. $\endgroup$ Feb 7, 2020 at 19:29
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    $\begingroup$ Seems like a highly elliptical orbit at the same angle as a molniya orbit but higher altitude could be stable, no? $\endgroup$
    – Harabeck
    Feb 7, 2020 at 19:33
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    $\begingroup$ Also need to note the in the north (20% of time) the moon will look considerably bigger than in the south (80% of time). $\endgroup$
    – Alexander
    Feb 7, 2020 at 19:43
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    $\begingroup$ @NuclearWang I don't want to put words in his mouth, but one problem with molniya orbits for large bodies is tidal forces. They will be pretty extreme at periapsis and will, over geologic timescales, circularize the orbit. It won't stay in a molniya orbit forever. $\endgroup$
    – Ryan_L
    Feb 7, 2020 at 20:24
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    $\begingroup$ You could also do something similar but reversed, with it only being bright enough to be relevant near the periapsis. $\endgroup$ Feb 7, 2020 at 21:18

No. A moon has to orbit the center of mass, and it has nothing to do with planet's axle of rotation. The Moon, for example, does not orbit around Earth's equator, but rather around Earth-Moon barycenter.



The other answers are correct about barycenter orbits, but we can adjust that with density.

The average density of Earth’s crust is 2.83 g/cm3.

The element osmium is the densest material known that can exist under merely planetary gravity. That metal has a density of 22.61 g/cm3. It’s neighboring elements of iridium and platinum are similarly dense.

That 10x difference can be used to create a planet where one end is much more dense than the other. What you need is a lopsided planet, where a very large ball of these dense metals rests at the South Pole. That moves the center of mass of the planet away from its core.

Creating such a planet is not trivial. What we know about planetary formation says that the heavy mass will sink toward the core.

Here we get out of my comfort zone... others may have a better way to create this planet. But based on other posts I’ve read on this site and their sources: you could have a planet where a metallic comet hit an existing planet, and knocked a big chunk off the side. The asteroid now sticks off one edge. This planet is probably tectonically unstable (because that osmium mass is always trying to work it’s way to the center), has a large asteroid belt (from the debris). The moon in question may be detritus from the impact.

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    $\begingroup$ There can be no lopsided planet. Planet is, by definition, a body in hydrostatic equilibrium. That means the surface is approximately equipotential in self-gravity and that in turn means it is an ellipsoid with its geometric centre coinciding with its centre of gravity. Any difference from a sphere is caused by centrifugal force due to rotation. $\endgroup$
    – Jan Hudec
    Feb 8, 2020 at 23:42
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    $\begingroup$ @JanHudec technically true, but for the purposes of most stories, “planet” is the big rock or gas bag that moons orbit. $\endgroup$
    – SRM
    Feb 9, 2020 at 0:45
  • $\begingroup$ @JanHudec actually, no. Planet is a body massive enough to pull itself into a sphere. The body in my answer is that scale, but it hasn’t achieved that yet. I think that qualifies as a planet under the technical definition. Particularly if it WAS stable prior to the comet impact. $\endgroup$
    – SRM
    Feb 9, 2020 at 0:48
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    $\begingroup$ Such a planet wouldn't be stable rotationally. en.wikipedia.org/wiki/Tennis_racket_theorem youtube.com/watch?v=1VPfZ_XzisU $\endgroup$ Feb 10, 2020 at 3:53
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    $\begingroup$ @SRM a comet won't change planet's center of mass. The requested setup is proposing to change it significantly. Your north pole would effectively become a mountain thousands of kilometers high. $\endgroup$
    – Alexander
    Feb 10, 2020 at 7:58

No. There are different shapes of orbit, but they all go all the way around the planet. You could have a low orbit over the equator which cannot be seen from the pole, but not the other way around.


The moon could be in an orbit with a period of exactly one day, so it is never seen from the other side of the planet. Then, something could happen, e.g. In the way the moon casts its shadow, that causes an equatorial point to be the coldest zone on the planet, much like conventional poles here.

Note that the orbit doesn't need to be around the equator; it might stay over one hemisphere but wobble up and down. This wobbling might coincide with when that polar=icy" area is facing the sun, causing the cold zone.

If the rotational axis was quite oblique from its orbital plane, it would mean the rotational poles would have seasons, but like I say, this one zone is the coldest since it's usually either in winter or in eclipse.


This may not be the answer you wanted, but this might give you an idea.

Suppose the Moon is in a polar orbit. Suppose it is also extremely eccentric, with a periapsis of, say, 100,000 km and an apoapsis of 1,000,000 km.

It will appear very close from the south pole at its closest approach, and very small everywhere else.

If it has the same size as our Moon, from the South pole it will appear to be 16x larger. When it is over the north pole, it will appear less than a quarter the size of our moon.


A moon that orbits at the L1 Lagrange point of a tidally locked planet, in theory, could do the trick.

the planet is tidally locked-- which means that there is now a distinct gradient of hot and cold that may create "poles" for the inhabitants. For civilizations that lives in the twilight zone, the night side would function as the "north" as it gets colder as you move up the night side; whilst the day side would function as the "south" as it gets hotter the further you move down the day side. The North pole would simply be the point that is directly opposite to the star, and the south pole would be the point that is directly facing the star.

Lagrange points are not truely stable-- a moon that orbits there is instead put into a Halo orbit, which fits the "orbiting around a pole" argument.

A large moon will cast shadow onto the planetary body it was orbiting-- for the Earth, it happen rarely as solar eclipses, but for a planet that have a L1 moon, it would be permanent-- directly above the "south" (star-facing) pole of the planet. Due to the sun being obscured by the L1 moon directly above the "south pole", the temperature at the pole will be very cold from the lack of starlight-- just like the real south pole of the Earth!

A map of the hypothetical world

  • $\begingroup$ Welcome to the site. +1 - This is a clever answer - though I'm not sure how long such a system would be stable for, and it requires a very specific planet layout, this definitely addresses the issue directly. $\endgroup$
    – Zxyrra
    Feb 10, 2020 at 4:01

Not in the way you are thinking for the reasons given. However, don't give up hope just yet. Apart from geosynchronous orbit above a fixed point on the equator, you could have is something at or orbiting a Lagrange point (L1 or L2). This wouldn't be over the South Pole as such (unless the axis of rotation points at or near the sun making the planet nearly uninhabitable) but it might give you what you need.


This is possible if we use another planet, not a moon. (this could be the size of the moon, or appear to be a moon at a distance, but have the gravitational effects of a planet)

Step one, take two planets and put them in a binary orbit.

Step two, make the planets rotate at their angular velocity, so the inhabited planet is tidally locked to be in the same orientation at all times with the other planet.

Step three, orbit that system around a sun.

Step four, have the axis of rotation on the inhabited planet be out of line with the line between the two planets, adding a new rotation component.

Step five place the south end of the planet pointing to the other planet.


Now, day and night are caused by the orbit of the planets.

The orbit of the moon is caused by the rotation of the inhabited planet.

The tilted planet still results in seasons.


A lot of answers here effectively state "no, but ..." and what I read was correct.

I'll give you another "no, but ..." solution: You can adapt various other parameters to make it appear as if a moon was only orbiting in the southern area.

  • Give the plant an axial tilt of about 90°. This exists in the Solar System, specifically Uranus.

  • Have the "moon" orbit the L4 or L5 point, depending on what you consider the south pole. The L4/L5 points are Lagrange points staying in the same relative position to the planet

This way, you'd have a moon in a stable orbit, almost only visible from the southern hemisphere.

However, it would not exactly be a moon as it wouldn't orbit the planet. Instead, it would be considered an asteroid or dwarf planet sharing your planet's orbit, similar to Jupiter's Trojans: https://en.wikipedia.org/wiki/Jupiter_trojan

The relative location to the sun would mean that it'd always be sort of a half-moon, always visible on the day-side in the southern hemisphere. The high distance would also mean that it would appear very small.


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