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Picture a planet identical to Earth, but with rings.

It has a satellite similar to the moon, except smaller; I'm not sure how big it needs to be but distance from the planet is more important so assume that it is whatever size it needs to be to have the same effect on the planet as the moon does for earth at whatever distance it lands at.

The key attribute of this satellite is that it orbits the planet along its equator (parallel to the rings, of course) at an extremely low range so that it appears from the planet's surface to be skimming along the edge of the rings.

Is this scientifically possible, however unlikely, and even if not what effects would the close proximity of a moonlike satellite to a planets rings have on the surface of the planet (climate, sunlight reflection/blockage, etc.), the rings (like, would the gravity of the moon pull on the rings and skew them somehow or something?), and the moon itself (like the gravity of the planet causing it to flatten out as it rotates)?

Oh. Also it all has to be stable enough for life to begin and mature unaided on the planet with little to no change in the orbital system.

Edit: It would appear that this scenario is generally accepted as possible (though unlikely, but that's pretty much irrelevant when you look at the likelihood of a planet that supports intelligent life in general), so that satisfies the first part of the question. However I still feel like a shepherd moon that's big enough to act like our own in terms of gravity would still have some unusual effects on rings that are much closer to the moon and smaller than the planet itself. Am I just paranoid, or does that make sense?

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  • $\begingroup$ Do you mean along the top of the rings, so the edge of the moon is touching the rings, or so the rings loop like they're bisecting the moon? (When viewed from the equator) $\endgroup$
    – ckersch
    Commented Dec 5, 2014 at 22:30
  • $\begingroup$ Well yes, the rings look like they're bisecting it, but I'm saying it doesn't have to be touching them, just close enough to them that it would appear that way from the planet's surface. I suppose the angle you're viewing it from affects that factor, but hopefully you get what I'm saying. Really really close to them. $\endgroup$
    – Ricky
    Commented Dec 5, 2014 at 22:32
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    $\begingroup$ Are you talk ring shepherds? en.wikipedia.org/wiki/Moons_of_Saturn#Ring_shepherds $\endgroup$
    – Twelfth
    Commented Dec 5, 2014 at 22:33
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    $\begingroup$ Possibly. I suppose that depends on whether a single shepherd could be large enough to act like Earth's moon. $\endgroup$
    – Ricky
    Commented Dec 5, 2014 at 22:37
  • $\begingroup$ The likelihood of planets that support intelligent life is completely unknown, and subject to much debate. $\endgroup$
    – Innovine
    Commented Mar 18, 2017 at 16:24

2 Answers 2

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The problem here isn't really about whether or not a moon could "skim" a ring, but whether or not rings could form around Earth.

There are three ways planetary rings can form. All three involve material coming inside the Roche limit of the planet. The rings of Saturn have a mass of about $3 \times 10^{19} \text{ kilograms}$, one one-thousandth the mass of Earth's Moon. The formula for the Roche limit shows that the Moon is 41 times as far from the Earth as its Roche limit is, and a less-massive body of the same density would have a smaller Roche limit, so if this object was the same density as the Moon, it would have to have gone really close to Earth. At any rate, the Moon would be too far away from the ring system.

Anyway, here's how rings can form, and why they might not around Earth:

  1. Accretion from the protoplanetary disk. This is simple. Matter from the solar system's protoplanetary disk ventures inside the Earth's Roche limit, gets torn apart, and becomes a ring. I don't think this would happen on Earth because Earth is not massive and never has been, and so would not pull a lot of material towards it. Sure, it could get the required one one-thousandth Moon mass, but would it come close enough to form a ring?
  2. A moon gets hit hard. Collisions happen, and they happened frequently in the early solar system. After all, that's how Earth got its Moon. If a moon got hit hard enough that debris was flung into space and somehow got inside Earth's Roche limit, it could turn into a ring. However, this would necessitate a second moon-like object (or protoplanet) being near Earth at some point in time. It's likely that it would still be here today - after all, only a tiny bit of mass would have been lost. True, an asteroid could have been smashed, but what are the odds Earth would have gotten the debris?
  3. A moon gets ripped apart by tidal forces. This is the most commonly-cited explanation for ring formation. However, it requires a moon to go pretty close to Earth, as we can see from the Earth's Roche lobe with respect to the Moon. Likely? Perhaps not, considering that Earth's gravity was relatively weak in the early solar system, compared to the gas giants.

I doubt rings could form around a planet like Earth.

Let's say rings do form. Could our Moon skim them? Well, remember that they would be near Earth's Roche limit - 1/41 the distance to our Moon! True, the Moon is receding, but it probably never was even close to being that close to Earth. If it was close enough, it would be torn apart by tidal forces. Shepherd moons (as Twelfth helpfully brought up) are exceptions, but they are small, and could be fragments of a larger moon that was torn apart, or broke apart in a collision.

Okay, let's go to the Moon as a shepherd moon. The reason that I assumed that the Moon could not be a shepherd moon is because of the mass discrepancy. Here's what I'm talking about:

  • Mass of the Moon: $7.3477×10^{22}~\rm kg$

Some shepherd moons:

  • Metis: $3.6×10^{16}~\rm kg$
  • Adrastea: $2×10^{15}~\rm kg$
  • Cordelia: $4.4×10^{16}~\rm kg$
  • Ophelia: $5.3×10^{16}~\rm kg$
  • Galatea: ($2.12 ± 0.08) ×10^{18}~\rm kg$
  • Mimas: $(3.7493±0.0031)×10^{19}~\rm kg$
  • Thebe: $4.3×10^{17}~\rm kg$
  • Prometheus: $(1.595±0.015)×10^{17}~\rm kg$
  • Pandora: $(1.371±0.019)×10^{17}~\rm kg$
  • Amalthea: $(2.08±0.15)×10^{18}~\rm kg$
  • Despina: $2.2×10^{18}~\rm kg$

There are almost certainly many more.

The point is, these moons are tiny. They have no chance of drastically perturbing the rings because their absence from a portion of the ring has no negative effect. The Moon, though, is much larger, and will greatly perturb all objects in its vicinity.

So what if (and this is a $\Huge BIG$ if, and I see I've found an interesting feature in $\LaTeX$) the Moon is close to the rings? Then the outcome tends to be that the ringed Earth gets another ring.

The constituents of planetary rings (no, they aren't continuous, although that would be awesome) can greatly vary in size. From Wikipedia:

The composition of ring particles varies; they may be silicate or icy dust. Larger rocks and boulders may also be present, and in 2007 tidal effects from eight 'moonlets' only a few hundred meters across were detected within Saturn's rings.

So we're looking at objects ranging from a few microns in diameter to a few football fields in diameter (and it doesn't matter which side of the Atlantic you're on) - although most of these objects probably aren't big enough to pull themselves into spheres. The Moon, by contrast, is about 3500 kilometers in diameter. It will clearly be the most dominant object in the area.

Look at this unfortunately compressed image of Saturn's rings:

Unfortunately compressed image of Saturn's rings
Image in the public domain.

The zoom on the link is really bad, but this is a good compromise. Long story short, the Moon is big enough to fit through some of Saturn's rings and nearly all of the gaps.

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  • $\begingroup$ So are you saying there's no way a shepherd moon could be big enough to act like our moon given its close proximity? Say, 1/41 the size of our moon? $\endgroup$
    – Ricky
    Commented Dec 6, 2014 at 4:34
  • $\begingroup$ There is no limit (that I'm aware of) to the size of shepherd moons. In most respects they act just like any other moon, the difference is that their position in orbit helps keep the ring together. $\endgroup$
    – Tim B
    Commented Dec 6, 2014 at 17:25
  • $\begingroup$ I see. So in terms of answering my question I take ALL of this as a "yes to the first part". So what about the effects? $\endgroup$
    – Ricky
    Commented Dec 6, 2014 at 18:54
  • $\begingroup$ @Ricky I will expand on that later. I apologize; I'm in a rush at the moment. $\endgroup$
    – HDE 226868
    Commented Dec 6, 2014 at 19:05
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    $\begingroup$ Note that it's possible for a moon to orbit inside its Roche limit if gravity isn't the only thing holding it together - Saturn's moon Pan orbits well within its Roche limit without breaking apart, thanx to its tensile strength. $\endgroup$
    – Vikki
    Commented Jun 14, 2018 at 21:10
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A little late, but no one has asked what seems to be an obvious question:

If you're looking at moon & rings from the planet's surface, why does it matter how close to the rings the moon is?

If the orbits are in the same plane, then (depending on the planet's diameter and how close to the equator you are, shouldn't it look as though the moon is either skimming the ring, or being bisected it? Even if in reality the moon orbits far beyond the ring?

Somewhat like these images Cassini captured, either looking like this or this.

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  • $\begingroup$ If the moon is too far past the rings, wouldn't it appear to people at the poles that the moon is either above or below? $\endgroup$ Commented Mar 18, 2017 at 18:26
  • $\begingroup$ @JarredAllen - Depends on the inclination of the mun's orbit. $\endgroup$
    – Vikki
    Commented Jun 14, 2018 at 21:11

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