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Thread: Astronomy in Disney's latest cartoon

  1. #31
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    Quote Originally Posted by Hornblower View Post
    My bold. Is that all? From various sources I find upwards of 10,000x, perhaps closer to 100,000 when you include all of the infrared.
    Yeah, I muffed the math. Would the lower temperatures implied for our reddish moon allow the planet to be much closer to the star? IOW, could the planet's atmosphere have some IR blocking molecules of some kind?
    We know time flies, we just can't see its wings.

  2. #32
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    If we use R to designate the ratio of the diameters of the inner and outer circular arcs of the "Turkish crescent" in the cartoon, then we can calculate the necessary apparent diameter of the illuminant as:

    App. Dia. = 2*(90 degrees - asin(R))

    Grant Hutchison

  3. #33
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    Quote Originally Posted by grant hutchison View Post
    If we use R to designate the ratio of the diameters of the inner and outer circular arcs of the "Turkish crescent" in the cartoon, then we can calculate the necessary apparent diameter of the illuminant as:

    App. Dia. = 2*(90 degrees - asin(R))
    Using this, my radial values yields about 20 deg. but only a 5% change in a radial amount measurement error will produce a 7 deg. variation.
    We know time flies, we just can't see its wings.

  4. #34
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    Quote Originally Posted by George View Post
    Using this, my radial values yields about 20 deg. but only a 5% change in a radial amount measurement error will produce a 7 deg. variation.
    So that would imply a twilight zone in a band along the terminator 20 degrees wide - not the sharp terminator of the cartoon.

    Grant Hutchison

  5. #35
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    One way to produce a sharp-edged terminator that intrudes a long way on to the night side is to have a small illuminant, but give our moon very high atmospheric refraction. The superior mirage thus created lifts the illuminant into view on what would geometrically be the dark side, but the visible disc remains small, allowing for a sharp terminator. Rayleigh scattering and extinction might still produce a visible broad twilight zone, however.

    Grant Hutchison

  6. #36
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    Quote Originally Posted by grant hutchison View Post
    So that would imply a twilight zone in a band along the terminator 20 degrees wide - not the sharp terminator of the cartoon.
    True if we were to assume this to be a picture-perfection depiction, but the artist uses sharp lines for everything (e.g. hair), so this may be one item we can discount.
    We know time flies, we just can't see its wings.

  7. #37
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    Quote Originally Posted by grant hutchison View Post
    One way to produce a sharp-edged terminator that intrudes a long way on to the night side is to have a small illuminant, but give our moon very high atmospheric refraction. The superior mirage thus created lifts the illuminant into view on what would geometrically be the dark side, but the visible disc remains small, allowing for a sharp terminator. Rayleigh scattering and extinction might still produce a visible broad twilight zone, however.
    That's clever but I think your latter point the likelihood it would have a broad twilight. If the atmosphere, however, is somehow dense but shallow (ie a very dense moon) then it might work. If so, then we might be able to determine its composition, which would be a significant accomplishment. A high albedo with not atmosphere would seem more likely if it weren't for the extended bright arcs.

    I wonder if the eclipse idea works well if we assume a binary system?
    We know time flies, we just can't see its wings.

  8. #38
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    Quote Originally Posted by George View Post
    True if we were to assume this to be a picture-perfection depiction, but the artist uses sharp lines for everything (e.g. hair), so this may be one item we can discount.
    Well, once you decide one component of this moon is not a perfect depiction, then you've undermined the OP's wish to explain the image in astronomical terms. Because the obvious explanation is, "It's just a cartoon, for pity's sake, not a video record of a real place."
    So we have to be Watsonian, rather than Doylist. Or so it seems to me.

    Anyway, the show-stopper seems (to me) to be that we can't keep this moon in a suitable constant phase, whether it be eclipsed or crescent or full. So either we have to say there's some seasonal reason the events of the story only ever occur at a particular lunar phase, invoke some modification of gravity, or decide it's a tethered balloon or some other unsatisfactory (ie, non-astronomical) object.

    Grant Hutchison
    Last edited by grant hutchison; 2019-Jul-18 at 02:55 PM.

  9. #39
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    Well, it looks like most of this debate is moot at this point because

    The show decided, for the final episode of the season, to make the moon, suddenly, inexplicably, full.

    I kinda don't what to know what a moon's orbit would have to be like to do that...
    "Occam" is the name of the alien race that will enslave us all eventually. And they've got razors for hands. I don't know if that's true but it seems like the simplest answer."

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  10. #40
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    Quote Originally Posted by grant hutchison View Post
    Well, once you decide one component of this moon is not a perfect depiction, then you've undermined the OP's wish to explain the image in astronomical terms. Because the obvious explanation is, "It's just a cartoon, for pity's sake, not a video record of a real place."
    [What happen to Pete for Pete's sake? ] It's inevitable that some artistic license must be allowed, but it's interesting to see if we can come close to a coincidental depiction of something that could be out there.

    Anyway, the show-stopper seems (to me) to be that we can't keep this moon in a suitable constant phase, whether it be eclipsed or crescent or full.
    But why is that? If we can, somehow, present a close orbit for the planet to have a reasonably fast orbital period, then why couldn't its moon have an apparent "phase lock" with the planet, where the orbital period for planet (about star) and moon (about planet) match?

    Using a habitable zone constraint, could something like this work, roughly:

    Mstar = 25% Sun
    Rstar = 10x Sun
    Temp = 1500K
    Mpl = 0.9x Earth
    So...
    a of pl = 0.7 AU [we may need to shrink this with a lower temp. star, admittedly, for tidal lock]
    Per of pl = 9 days
    a of moon = 1500 km (1/2 deg app. size)
    app size of star = 15 deg.

    [Not a main sequence star, but we need lots of time anyway to assure the planet's tidal lock.]
    We know time flies, we just can't see its wings.

  11. 2019-Jul-19, 03:12 PM
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  12. #41
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    Quote Originally Posted by George View Post
    But why is that? If we can, somehow, present a close orbit for the planet to have a reasonably fast orbital period, then why couldn't its moon have an apparent "phase lock" with the planet, where the orbital period for planet (about star) and moon (about planet) match?
    Because in that situation the star's gravity becomes a significant perturber on the moon. The "phase lock" you describe is just the situation in which moon, planet and sun occupy the same fixed rotating reference frame - exactly what Lagrange solved for, which gives the stable L4, L5, the unstable collinear solutions at L1, L2, L3, and no where else you can put a moon and not have it wander off immediately. L4 and L5 give a gibbous phase, and the collinear solutions that would give an eclipse/new/full phase won't persist without station-keeping.

    Grant Hutchison

  13. #42
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    Quote Originally Posted by grant hutchison View Post
    Because in that situation the star's gravity becomes a significant perturber on the moon. The "phase lock" you describe is just the situation in which moon, planet and sun occupy the same fixed rotating reference frame - exactly what Lagrange solved for, which gives the stable L4, L5, the unstable collinear solutions at L1, L2, L3, and no where else you can put a moon and not have it wander off immediately. L4 and L5 give a gibbous phase, and the collinear solutions that would give an eclipse/new/full phase won't persist without station-keeping.
    Ah yes, and an unstable moon, due to, say, capture, would be a strain on credulity, even for a comic illustration. It would have to be an L4 or L5 (always gibbous as you say, which explains why "phase lock" is not a working term.

    Even if a phase-lock stability could be gained with great distance from the star, say 100 AU, then a phase-lock won't help because the moon would be too far from the planet to match the planet's period.
    We know time flies, we just can't see its wings.

  14. #43
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    In all of this discussion of the phase, we have no information on the timing of the scenes where it appears. All we have in this thread are links to two still frames.

  15. #44
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    Quote Originally Posted by Hornblower View Post
    In all of this discussion of the phase, we have no information on the timing of the scenes where it appears. All we have in this thread are links to two still frames.
    The OP claim is that the waning crescent view is constant regardless of time. But, whether the artists vary this or not, the intrigue is whether or not, using our imaginary resources, we can devise a way favoring this circumstance. There are implied variables including the planet's need to be in a habitability zone, a reddish star would be helpful, and a large apparent size for the star would allow the over-arcing of the moon.

    It seems a close orbit is the only logical choice, but the limit on Lagrange points for any stability eliminates the possibility for a locked crescent view of their moon.
    We know time flies, we just can't see its wings.

  16. 2019-Jul-20, 10:38 PM
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  17. #45
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    Superflare?

  18. #46
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    Quote Originally Posted by publiusr View Post
    Superflare?
    What about it?

  19. #47
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    The look of the moon by earthshine from a burning world?

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