What would the climate of this habitable world be like?

Question

Somewhere, far away in the universe, there is a galaxy that has not been made by nature. Whatever kind of hyperadvanced civilization could have created such a titanic project, and for whatever purpose is not relevant in this question. At the center of this artificially constructed galaxy is a black hole 300 trillion times as massive as our sun. Surrounding that hole is an accretion disk one quintillion times as bright as our sun. Surrounding that is a habitable zone three times as wide as the entire Milky Way galaxy. Within this zone is a K5 main-sequence star, 74% as wide, 69% as massive and only 16% as bright as our sun. Such stars burn their hydrogen so slowly that they can last not for ten billion years like our sun, but 34. And they are so hot that they can have a habitable zone that isn’t close enough to tidally lock their planets. Another, equally important, factor about orange dwarves is that they emit very low quantities of ultraviolet radiation.

The planet that orbits that star is its own alien space bat, a planet that blatantly and violently disregards our understanding of astrophysics. At 18,500 miles wide, it is 230% as wide as Earth, resulting in an overall area of 1,075,210,086 square miles. But instead of having a crushing gravity as a super-Earth should, it carries instead 100% of Earth’s gravity. A single day lasts 26 hours, and it rotates 417 times to make up its year. We have measured its magnetic field at no greater than 7.8 gauss, 12 times greater than on Earth. Its atmosphere is just as thick as Earth's. The star rises from the west and sets on the east, yet the planet still orbits it from a prograde, or counterclockwise, direction. Its axial tilt varies from 19.7 to 26.9 degrees. However, it’s not all that this planet has to define its seasons. Regardless of the point in the year, the accretion disk outlining the black hole will always be present in the sky, measuring in at an angular diameter of half a degree. The star, in turn, measures in at 1.88 degrees wide, almost four times as wide as our sun. Both sources of light are present during the day in the “summer” months. Its single moon is so far away and yet so large that it fills in at an angular diameter of 2.61 degrees.

Here is the planet's current map:

And then, here is the planet's current map with mountains labeled and directed for tectonic movement:

The land between the two mountain ranges used to be a basaltic plateau before the collisions squeezed it up into being as high as Tibet.

Currently, carbon dioxide in the atmosphere is 500 parts per million, and oxygen makes up 35% of the atmosphere.

Using all of the information listed above, what would the Köppen classification climate map of this world look like?

Answer 1

The greater the planet's axial tilt angle, the more extreme the seasons are, as each hemisphere receives more solar radiation during its summer, when the hemisphere is tilted toward the K5 main-sequence star, and less during winter, when it is tilted away.

Increasing the planetary radius (in your case 9,250 miles) also leads to a lower planetary albedo and warmer climate, pushing the inner edge of the habitable zone to lower stellar irradiation.

The most immediate effect of changing Earth's axial tilt to 21.5 degrees would be a fast expansion of the north pole ice cap and the freezing to the ocean surrounding Antarctica. In the northern hemisphere there is about a 1000 mile zone starting at just below the polar circle and extending about 1000 miles southward where most of the earth's conifer forests exist. Socratic

Given that earth's axial tilt range is between 22.1 and 24.5 degrees in a cycle of 40,000 years, your planet's axial tilt range of 19.7 to 26.9 degrees is quite massive. The greater the axial tilt, the more extreme the seasons, and less extreme seasons for the lower the axial tilt. So your planet is likely to experience very mild and very extreme weathers, which includes long winters (extremely cold ones) and probably not so long summers.

I would classify the entire top "part" of your map as ET (tundra) or EF (ice cap). Most of the center of the large "island" poking out would have a few BWh (desert) regions, quite a lot of BSk (semi-desert) regions. Your entire continent should consist of mainly DWc or DWd (subarctic) regions due to winters.

But mainly, the temperature should be decently close to Earth's, but with variations in the increase and decrease over seasons.

This is a very vague impression and likely wrong

Answer 2

This is tough to say with the data provided.

You've said the star is somewhere within a habitable zone, whose distance is not specified, defined by a very bright light source (the accretion disk). The planet has a 452 (Earth) day year around a star of 0.69 solar masses: take T = 2 pi a^1.5 u^-0.5 and we have that the planet is just slightly further from the star than Earth, receiving just 16% of the light, so the accretion disk dominates the energy contribution. Assuming it provides 0.84 fraction of Solar light, starting from 1E+18, taking the square root, we get 9.1E+9 AU distance from the accretion disk. That gives me 14500 light years ... hardly "three times the Milky Way", but there are a few ways to take that.

Bigger problem is that the star rises in the west, and the planet's orbit is not retrograde. Normally a star rises in the west because someone uses an unfortunate definition of "north" that goes by the whole star system (and hopes nothing lands closer to edge-on than Uranus). There is a way a star could rise in the west even exluding that ... the planet might orbit faster than it rotates, so that the star creeps up the zodiac faster than it can turn out of the sky. But for that, your planet has to have many sidereal days before the Sun catches up once for the year ... or something. I don't really know how to make it work.

Anyway, given the data, assuming the orbit worked out above, the planet depends entirely on a fixed astronomical feature for light, i.e. the accretion disk. That's different from a planet revolving around a star. The disk could be above one pole of the planet, or the other, in which case it is much like a tidally locked planet, only rotating... However, I'll take your statement that "Regardless of the point in the year, the accretion disk outlining the black hole will always be present in the sky" to mean that it is precisely equatorial ... in which case it does act exactly like a normal star would, except there's one more "day" by that standard (sidereal day, not synodic), and the accretion disk might be eclipsed for part of the year by the dim red sun, if it happens to align with the accretion disk at an equinox. Note that there are no seasons, despite the axial tilt, from the accretion disk, assuming it is at the equator, so only the star provides very weak seasons as it moves north and south in the sky.

Now all that said, I've only assumed the planet is like Earth in temperature overall; the distance isn't given and the atmosphere is different. But if it's like Earth, you have a continent at the top which is like Antarctica, a high plateau at the equator which is like Kilimanjaro, and other terrain that is between but not truly "temperate" due to the weakness of the seasons. I'm reluctant to start talking about which way the winds blow because of the "sun rising in the west" thing!

Answer 3

It will be too hot there for liquid water.

Let us science this up. How can we determine how close this planet is to its star? We know the size of the star and we know how big it appears from the perspective of the planet. We can use those values to calculate the distance. This will be a big help in determining likely climate.

https://lco.global/spacebook/sky/using-angles-describe-positions-and-apparent-sizes-objects/

d = 206,265 D / θ

D = linear size of an object θ = angular size of the object, in arcsec d = distance to the object

Star is 74% the width of our sun. Our sun is 1392000km, *0.74 = 1030080 km width of star. From OP angular size is 1.88 degrees Convert degrees into arc seconds = 1.88 * 3600 = 6768

1030080 * 206265 = 212469451200 212469451200/ 6768 = the planet is 31393240 km away from its star; round to 31 million km = 0.2 AU

Our sun by comparison is 150000000 km away; 150 million km Our sun is 70 million km from Mercury.

To use the habitable zone calculator one needs luminosity and temperature. OP provides luminosity of 16% of our sun. I took temperature from here

https://en.wikipedia.org/wiki/K-type_main-sequence_star ; an average K5 star has 0.17 luminosity so that matches, and wikipedia gives an average of 4440K temperature

Here is a snip from the calculator at http://depts.washington.edu/naivpl/sites/default/files/hz_0.shtml#overlay-context=content/hz-calculator

At 0.2 AU the planet is not inside the habitable zone of its star. It is too close to the star. It is hot there. The answer to the OP as regards the climate: toasty, as in stuff there is toast. Melba toast.

A side question should someone be interested in playing with these calculators is that of the accretion disc. I did not factor that in but it too will illuminate this planet. One quintillion times as bright as the sun is pretty bright. 1800 arc seconds in the sky is not trivial. How far away is the radiation source represented by the disc and what is the calculated habitable zone, treating this object as a star for purposes of calculation. Dailey I am trying to salvage your world and I suspect that maybe it could be a rogue planet, warmed just by the disc.

Anyone interested in crunching those numbers is welcome to addend an edit to this question. I think the temperature of the disc might be hard to estimate.

Answer 4

Ocean currents play a huge role in climate. Air and ocean current are the planets way of trying to equalize the temperature of the whole planet. This is as we all know impossible but nevertheless it is an ongoing process.

Most significant thing that pops out to me in your map, is the southern pole being ice and obstruction free. this will produce and constant and quite large current around the pole region. This in turn will have a, perhaps not dominating effect but certainly over the long term very significant effect on the overall climate. The second is the large ocean, also unobstructed, there should be 2 large currents here separated by and equatorial counter current driven by surface winds west to east. Let's call these two currents the north and south gyre, to give them anilogs to Earths. The south polar current should be swift and constant, I would expect the southern most portion of the continent to be quite cold, and wind blown. Storms would be fequint from warm winds defected from the mountains, also along the southern cost. Little ice buildup would be preset because of the strong current. In Fact I would expect there to be quite a good amount of deformation in the coastal topography of the land in the direction of the current west to east. As the cost recedes north you may see ice build up in the sheltered areas. strong ocean eddies thrown off and rough seas commonly rough seas. The Southern Gyre here Rotates counterclockwise pulling warm water from the equator (Down the east side of the continent, west side of the ocean) and pushing cool water up (Up the west side of the continent)

This is incomplete, not to mention probably inaccurate guestimation, but without any typographical data anything with higher resolution cant be expected.

EDIT: Just reread this part:

The star rises from the west and sets on the east, yet the planet still orbits it from a prograde, or counterclockwise, direction. Blockquote

Im not so sure what to make of this, not sure I understand how this works as the OP has explained. This has less to do with orbit than planetary rotation. But as another post mentions this will reverse the usual prevailing currents.

Answer 5

This may help others

With the counter-ordinary daily rotation, the prevailing winds will reverse. However, I'd expect there to be a cell reversal around $\pm 30^0$ latitude.

The Himalayan-like mountainous zone will completel alter the equatorial southern winds. However, it will enhance the winds at the -30 degree cell.

At the 19.7 degree tilt, here is a description of the artic circle for your world. In these regions, the sun will never set during summer; and it will never rise during winter.

And at the 26.9 degree tilt.

With this information in hand, and using the Earth as a guide :

Scale and Attribution

Since, by fiat, this planet breaks a little known physics to somehow be Earthlike, I've transposed the temperate bands from Earth, applied the only topographic feature provided in the question (the mountain range), and applied the winds relative to that feature.

I thought about applying great deserts, like the Sahara and Australia, but chose against it.