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What are the minimum dimensions (length, width, height, etc.) for rings to at least be as significant (as in brightness/visibility) in the night sky as our moon?

Since there are so many variables that come with latitude, and angles in relation to the planet: How big would the rings need to be to be seen from the poles?

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    $\begingroup$ Whether it's a duplicate or not, this is just geometry. Composition can be dust so thick it looks solid from the planet or moon-sized rocks not that close together. Unfortunately, "significant" isn't an objective term. If I hold up a ruler, would a band 8" wide be as "significant" as a moon the size of a pie plate? Regrettably, this all means there's no "right" answer because there are far too many variables (not just one set of numbers). Remember that how wide the band looks will depend 100% on where you're standing on the planet. There's no single solution. $\endgroup$ – JBH Jul 3 '18 at 2:39
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    $\begingroup$ Let me add more perspective. Point the rotation axis at the sun, set the band radius (R2-R1) to 10X the planetary radius starting .25 the radius away from the surface and 1,000 miles thick, and it's that one day of the year when the axis is pointing at the sun. Standing on the pole sunward the rings would look almost infinitely wide as far into space as you can see. Now tip the axis up like Earth, same bands, but you're on the equator. The ring would look like a 3" wide band tapering to nothing to the east and west. Too many variables. Way too many variables. $\endgroup$ – JBH Jul 3 '18 at 2:46
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    $\begingroup$ Do you mean significant in terms of angular size, or in terms of brightness? $\endgroup$ – HDE 226868 Jul 3 '18 at 3:10
  • $\begingroup$ Sorry I wasn't clear guys, I mean significance as in brightness and visibility $\endgroup$ – Thalassan Jul 3 '18 at 3:12
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Before I begin, we need this picture:

enter image description here

That's the Earth and our moon. OK, simple answer:

  • Rather than Earth's tilted axis, we have an axis of rotation that perpendicular to the planet's orbit (0° to the ecliptic plane).

  • Rather than Earth's moon, which does not orbit the equator but has a fairly complex orbit, we're going to assume a ring that's circular and centered on the equator (orbits the planet on the ecliptic plane).

  • No matter what the ring is made of, from the observer on the planet, it looks solid when the sun hits it just right.

Your's standing on the pole, which means you're technically looking down on the ring, but from your perspective (due to the enormous distances involved), it actually appears like you're looking at something low in the sky (aka, "up"). The goal is for the ring to have the same brightness and, I'll assume, the same "vertical impact" as the moon would have if you cut a moon-sized circle out of the picture our observer sees from the pole.

Rings have at least two radii. We'll assume the simplest case of just two. R1 is the "nothing there" radius (also called the "inner radius") between the center of the earth and the inner edge of the ring. We'll make R1, oh, let's say 85% of the distance from the Earth to its moon or about 200,000 miles.

R2 is the outer edge of the ring. The ring's width is R2-R1. We want the part of the ring that appears the thickest (the center of the arch we can see when the sun comes up) to be about the same visual size as Earth's moon. Due to perspective, we need the actual ring width to be larger than the moon's diameter. At a guess, let's say 115% of the lunar orbit or 275,000 miles. That's a ring width of 75,000 miles. (Fair warning, a 30% increase in width to compensate for the perspective shift from a sphere to a flat plane might be quite a bit too much... but it's close enough for government work.)

As morning breaks, you would see an amazing arch, brightest at eastern horizon, darkest at the western horizon, thickest in the middle, and it would (if my guesses are basically correct) have the same standard of visibility as our moon if you just stared at it.

As the day progresses the ring's brightness would appear to shift with the sun from east to west. Because you're at the pole and it's orbiting the equator, the outer edge of the ring would appear much brighter than the inner edge. It would be pretty cool.1

You'll note that I made assumptions about composition and otherwise ignored it

Your composition is important. How much material there is, how thick the ring is, how dense the material is within the ring, what the material is made of... all that (and more) contributes to what happens when sunlight hits the ring. Is there ice? Sparklies. Is it dirt? Kind of a lunar feel, unless there's iron in the dirt, then maybe a redish hue. Is there a lot? High reflection. Is there a little? low reflection and fuzzy borders.

And then you add in axial tilts, orbital tilts, time of the year... Variables....

But, that should be the simple case basics. There are a fair number of people on this site who know more about this than I ever will. If they post with a comment along the lines of, "You know, JBH's answer was OK, but..." then take their word for it.

In the end, for the sake of your story (I'm assuming this is for a story), let me recommend that you don't get bogged down too deeply in the details. You need to be able to reasonably describe what you want the reader to see in their imagination. Exact details are far more than they need for that. Besides, at these scales you can vary a number by what would seem to a mere mortal to be massive amounts (tens of thousands of miles, for example) and still be "basically correct."

Cheers.


1How the rings present in daylight is an assumption on my part. Remember there is o axial tilt in this simple example, which means is should (I believe) do something you wouldn't see on Saturn — the outer edge should be brighter than the inner. This would be due to the outer edge acting to create a shadow on the innter edge. Due to Saturn's axial tilt, the outer edge never shadows the inner edge, it only shadows the planet.

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    $\begingroup$ Photos of saturns ring from cassini do not show any visible difference in brightness between the inner and outer rings. The rings would be of the same brightness except where the Earth's shadow hit it. Given the orbital mechanics of your proposed system, that would happen with the same frequency as a blood moon, and demonstrate the identical lighting effect. The brilliance of rings depend on the albedo of the material they are composed of, and it's dispersal. The Moon's albedo is about 9% (like coal) but ice, like saturns rings, is much brighter. $\endgroup$ – pojo-guy Jul 3 '18 at 12:00
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    $\begingroup$ Photo of saturns rings: saturn.jpl.nasa.gov/system/resources/detail_files/… $\endgroup$ – pojo-guy Jul 3 '18 at 12:16
  • $\begingroup$ @pojo-guy, you're not taking into account Saturn's axial tilt, which presents the rings more evenly toward the sun. My first bullet point assumption removes the tilt for the sake of simplicity. I believe (but obviously can't prove) that would result in the brightness shift between the edges since the leading edge would shadow the trailing edge - but I could be wrong. I'll edit my answer to make this point. $\endgroup$ – JBH Jul 3 '18 at 14:07
  • $\begingroup$ Artist's representations of Earth with rings - cnet.com/news/… $\endgroup$ – VBartilucci Jul 3 '18 at 14:11
  • $\begingroup$ @VBartilucci, THANKS! That is a wonderful depiction that will likely do more for the OP than all my words combined. $\endgroup$ – JBH Jul 3 '18 at 14:16

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