10
$\begingroup$

Recently, I've seen those pictures showing what Earth would be like if it had rings, and that made me wonder, what would the sky map be like? This ring would appear as a fixed object (pretty much as stars) so I'd like to know how could this be represented on a two-dimensional map.

What do you guys think?

enter image description here

Note: The ring is rather rocky than icy (due to the closeness to the Sun) and, since in real life moon is pretty shiny at night and it is made out of rock, I think it would be very visible (during day and night both), as it would reflect a high amount of sunlight.

$\endgroup$
  • $\begingroup$ You are probably thinking of a thick and shiny ring system like Saturn's. If we had a dark and thin ring system like Jupiter, we might not be able to see them clearly from the surface. You could edit to clarify. $\endgroup$ – BobTheAverage Apr 27 '17 at 21:56
  • $\begingroup$ @BobTheAverage Thanks your your feedback, Bob. As far as I know, rings would be made of dust and rock and, since the moon is very bright at night, I can say that this hypothetical ring system would ve very visible. However, I'm not asking about whether we could see them or not, but how would they be represented in a celestial map. $\endgroup$ – MoholyNagy Apr 27 '17 at 22:58
  • $\begingroup$ A note about the moon's shininess: I remember from college that it only has an albedo of 20 -- only 20% of the sun's light reflects back to us. If your particles have a higher albedo than that, and they are "dense" enough, it'll probably be pretty visible during the day! $\endgroup$ – BrettFromLA Apr 27 '17 at 23:46
  • $\begingroup$ @BrettFromLA I've also read that, due to the brightness of the rings, close stars may not be visible because of light pollution. But that's another story. $\endgroup$ – MoholyNagy Apr 27 '17 at 23:52
  • 1
    $\begingroup$ The rings would not be "fixed" in the sky but would vary in their appearance and position against the stars depending in your latitude, and the inclination of the ring plane to the equator. Hence it would not make sense to represent them on a star chart. $\endgroup$ – MikeW Apr 28 '17 at 12:05
9
$\begingroup$

Ring Behavior

Assuming that this ring is visible, it would most certainly need to go into star charts. However, how that ring goes around the earth is important. Is it a simple ring, which hovers around our equator? Are there any complicated behaviors of the ring, such as precession or "wobbling"?

Simple Ring System

If ring forms along the equator, then you get to see star charts like the on you've pictured with a boarder. This is no big deal- you just see the equator marked with the "ring."

If the ring is off-kilter, that is, it's at an angle relative to the equator, this would look different on star charts. It depends on the angle, but it can form a swirl. It'll look like the milky way on your star chart- an "s" shaped curve. The curve goes from super-curvy when the ring is very close, but not quite in line with the equator, to very flat, when it's 90 deg. on the star chart.

This assumes the ring, if it is tilted, rotates with the earth so that the places under the ring are always under the ring. (That is, the individual particles composing the ring do not have a geosynchronous orbit, but the ring structure itself does.)

A Moving Ring

So rings are composed of things swirling around a planet, but what happens if the plane of the ring moves? That is, the ring is experiencing some kind of precession, just like a wobbly top does.

This is very problematic, and would likely qualify it for not appearing on star charts, as the rotating ring could also appear on every point in the sky!

If the ring's precession lets it wobble just a little, star charts may show its maximum and minimum positions and dates corresponding to those. Oh, this also means that measuring the height/position of the ring becomes another easy way to measure time, so calendars in such a world may be based off of that. (And yes, it may or may not line up well with the actual year, so no, you may not get a perfect calendar. Sorry mathematicians!)

If the ring's precession lets it wobble a lot (as in "twice a year, the ring is directly overhead any point on the globe between really far north and really far south, twice a year."), it ought to be omitted from star charts. You may have a small side-chart showing where the ring would be, and what stars are behind it.

$\endgroup$
  • $\begingroup$ That's a very interesting idea. This measurement of years/seasons based on the ring's wobble is really clever! Thank you very much for your answer. $\endgroup$ – MoholyNagy Apr 27 '17 at 23:16
  • $\begingroup$ You mean precession? $\endgroup$ – JDługosz Apr 28 '17 at 8:16
  • 2
    $\begingroup$ The ring is not rigid, but each speck has an independent orbit. The kind of precession you mention is not possible. And the time scale would be thousands of years, not 6 months. $\endgroup$ – JDługosz Apr 28 '17 at 8:20
  • 2
    $\begingroup$ The ring cannot be at an angle. It will be exactly on the equator. Even rocky planets have a bulge due to rotation, and tides will force it into a thin plane. $\endgroup$ – JDługosz Apr 28 '17 at 8:22
  • $\begingroup$ @JDługosz Interestingly enough, "precession" isn't recognized by my browser spell-check. This is what we get when non-physicists don't write the code for spell-checkers. $\endgroup$ – PipperChip Apr 28 '17 at 14:18
7
$\begingroup$

The width would look different depending on your lattitude. So fat bands would not be drawn; rather just a dotted line. Maybe other marks showing the thickness seen from different points.

In fact, the thickness of the ring — which stars are blocked or shown — becomes an important aid to navigation and will have been apprciated since ancient times. So lattitude marks along the ring center will be an important and relevant feature.

The rings spin around, so you won’t have visible asymmetry with ring features that stick to part of the background, as your map seems to show.

$\endgroup$
  • $\begingroup$ Probably not that important to navigation - measuring the angle of the north star can be done fairly precisely and directly corresponds to latitude. Measuring the much smaller difference between edges of a fuzzy ring and then doing the trig to calculate latitude, or looking it up in a book of tables is more complicated and less precise. Sort of reminds me of Halley using tables of moon phase and occulted stars for longitude rather than a chronometer. $\endgroup$ – Pete Kirkham Apr 28 '17 at 9:59
  • $\begingroup$ If specific stars are reached by the ring edge, this can be seen directly without instruments. So maybe the sextent is new high-tech, but the first navigators just read the ring. $\endgroup$ – JDługosz Apr 28 '17 at 19:04
4
$\begingroup$

As I understand it, the star map in your picture is what you'd get if you stood at either pole, looked at the hemisphere of sky around you, and mapped it onto a flat circle. Assuming the ring is equatorial (which I think it has to be, in any normal astronomical situation), from the pole it would appear to circle the horizon; therefore, on your starmap it would appear as a thin band around the outside of each circle.

If the ring was instead in a polar orbit, it would be a straight line or stripe running horizontally through the center of both circles. If somewhere in between, it would be a sine wave following a path similar to (but less wobbly than) the Milky Way in your image.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.