3
$\begingroup$

In this scenario, an Earthlike planet is situated in a quaternary solar system clumped into two binaries--a Parent, in which the planet orbits, and a Grandparent, in which the Parent binary orbits.

For now, let's simply call the two Parent stars "Mom" and "Dad". "Mom" is a G-type star 6,000 degrees Kelvin, whereas "Dad" is an F-type star 7,300 degrees Kelvin. Similarly, we will call the two Grandparent stars "Grandma" and "Grandpa". "Grandma" is a B-type star 30,000 degrees Kelvin and "Grandpa" is an O-type star 40,000 degrees Kelvin.

This Earthlike planet orbits the Parent binary at a point where the habitable zones of both binaries overlap. If this were literally Earth, it would be too irradiated to support life at all. But this one has some defense mechanisms. One is an ozone layer one inch thick (which is actually eight times thicker than our own). Two is a magnetic field that measures 12 gauss (approximately 18 times higher than our own). Three, the atmospheric thickness is higher than Earth's. The bets vary between 6.0 and 3.2 Earths. The amount of carbon dioxide in the atmosphere is as high as Earth's was during the Cambrian, and oxygen levels are as high as Earth's was during the Carboniferous.

Now that we've got the concern out of the way, let's discuss a bit about the planet's seasons, in that it has both hemispherical and orbital seasons due to its extreme ellipsis. How extreme? Let's take a look at the list:

  • Axial tilt: 19.7°
  • Rotation: 48 hours
  • Revolution: 1,043 days, or 2,086 Earth days, or almost six Earth years
  • Orbital spring: 134 days, or 268 Earth days, or almost three-quarters of one Earth year
  • Orbital summer: 88 days, or 176 Earth days, or almost one-half of one Earth year
  • Orbital autumn: 134 days, or 268 Earth days, or almost three-quarters of one Earth year
  • Orbital winter: 687 days, or 1,374 Earth days, or nearly four Earth years

Since this planet orbits a binary that orbits another binary, orbital winter is treated more like our summer, in which the Grandparent binary shares the "night" sky with the moon and the Parent binary is so far away that "daylight" is basically 48 hours of twilight. But in the orbital summer, all four stars are in the daytime sky, and the moon has the night sky all to itself.

Speaking of which, the moon has a crust made not of rock, but aluminum, a metal that reflects 95 percent of light, making nights on this orbital season seven to eight times brighter than a night on Earth.

Put all those things together, and it boils down to one simple question: What color would the sky be as a result of all of these?

$\endgroup$
6
  • $\begingroup$ Sky color depends not only on stars but also on the chemical composition of the atmosphere. The easiest way to check this is to calculate "Rayleigh scattering": - sciencedirect.com/topics/medicine-and-dentistry/… - en.wikipedia.org/wiki/Rayleigh_scattering - youtu.be/4HBuHX4-VU8?si=qul2A4dYDh3rjRCD Since practical aspect of science (calculations) are not my forte I am really able to provide you only with theory and a bit of material. Hope this helps. P.S. I am really sorry but the amount of mistakes I make is bordering on having dyscalculia. $\endgroup$
    – Zaćmienie
    Commented Jul 20 at 22:45
  • $\begingroup$ Consider this question. The combination of the stars will present a "white" light, but each star will have its time ascendant. The chart might help you work out what the color would be throughout the year. $\endgroup$
    – JBH
    Commented Jul 21 at 4:52
  • $\begingroup$ @JBH That is for single-star systems with more Earthlike atmospheres. $\endgroup$ Commented Jul 21 at 5:06
  • $\begingroup$ @JohnWDailey True, but it's a reasonable place to start. Light is light and as each star has its moment closer to the planet in question, the effect on the atmosphere can be modeled as that of a single star. Besides, you name it an earthlike planet with a thicker atmosphere, which the chart embraces. I don't consider this a complete duplicate of the linked question - but it's unlikely that you couldn't answer your question based on that chart. $\endgroup$
    – JBH
    Commented Jul 21 at 18:33
  • $\begingroup$ I feel this would be a better question for astronomy Stack Exchange. Also star color is dependent on atmosphere. $\endgroup$
    – Rhymehouse
    Commented Jul 21 at 19:20

1 Answer 1

12
$\begingroup$

Blue, Rayleigh Scattering is a function, primarily, of depth of atmosphere. There are a few gases, notably the various sulfur oxides, that can tint the atmosphere in high enough concentrations but under an Earth normal atmosphere the sky will be blue. All stars emit light that is effectively white at habitable distances, their temperature colour and fine emission spectrum is almost certainly different but rarely by enough to make any difference to the wavelength interactions in their atmosphere.

$\endgroup$
2
  • $\begingroup$ +1. Since this world's atmosphere is thicker I guess the sky will be bluer. Which I guess means daytime will be a bit more like sunset than it is on Earth, with the suns appearing a bit more yellow-tinted and a higher proportion of light coming from scattering, so that shadows look a bit bluer. Sunsets will presumably be even more spectacular than they are on Earth, with the suns becoming deep red even while they're relatively high above the horizon. $\endgroup$
    – N. Virgo
    Commented Jul 22 at 6:32
  • $\begingroup$ @N.Virgo Huh I'd read the Earth-Like tag and missed the atmospheric thickness, from memory the sky will be a lighter blue, tending towards white, due to the increased scattering of longer wavelength light. $\endgroup$
    – Ash
    Commented Jul 22 at 7:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .