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I'm designing a worldbuilding project and the first major hurdle I'm encountering is the star(s) I want to have. The primary star is an F9V - F8V star with a mass of 1.13 - 1.15 Solar Masses, a radius of 1.1 - 1.2, and a luminosity of 1.5L - ~1.6L

The Secondary star is a K type star with around ~0.725 Solar Masses. With a luminosity of .21L - .18L funnily enough this star is extremely similar to another star I found (That actually exists)
https://en.wikipedia.org/wiki/TW_Piscis_Austrini This system is an S - type orbital system with the main planet (The focus of the worldbuilding project) orbiting around the F type star while the K type star orbits further out. currently I have the K type orbiting at approximately 5.5AU

here is a few parameters I want the system to have.

  1. The Secondary star must be at least ABLE to be visible during the day.(Not Just visible but "decently" bright) I'm aware given the type of orbit, during some parts of the year that will not happen.

  2. It must be able to Light up the night side of the planet (When in position to do so) significantly more than a full moon. Hard limit at around twilight light levels. No Brighter than that.

My Questions Are -

#1 Is this even a possible system? is the K type way to close for the "habitable planet" to retain a stable orbit? I have done some "testing" in Universe Sandbox 2 however I'm concerned to that game's reliability. If this solar system "works" but isn't plausible please present an alternative that works better!

#2 What is the apparent magnitude of the secondary star at this distance? and a Bonus, given an earthlike atmosphere what would it look like from the surface of the planet

#3 And- how much heat will this star give the planet?

And for a bonus if someone wants to find the habitable zone of this F type star, and were the best spot to place the planet will be is, that would be appreciative

if your curious for the worldbuilding planets specs, they are

Mass = ~1.20 Earth Masses
Surface Gravity = 9.92 m/s
Surface Pressure = 1.13 Bar
Radius = 1.09 Earth Radius
Day length = 27 hours
Density = 5.10 g/cm^3

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    $\begingroup$ Hey, Dono -- there are a lot of subquestions inside this question. I'd suggest paring it down to one major one to avoid it being too broad. $\endgroup$
    – HDE 226868
    Feb 17 at 20:28
  • $\begingroup$ I'm very new to this site so please give me an idea on how to do that ;-; I honestly didn't know what title to give this question... $\endgroup$
    – Dono
    Feb 18 at 20:13

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Short Answer:

The K type star should probably be more than 10 times farther away from the F type star and the planet than in the question.

Long answer:

According to my calculations, a star with a luminosity of 1.5 to 1.6 solar luminosity would have what I call an Earth Equivalent Distance or EED at 1.224 to 1.264 or AU. The EED is the distance from the star where the planet would received as much radiation from the star as Earth gets from the Sun.

And you write that the K type star:

currently I have the K type orbiting at approximately 5.5AU

5.5 AU is 4.49 times 1.224 AU and 4.35 times 1.264 AU.

And that seems a bit to close for orbital stability in a non-circumbinary or S-type orbit.

According to the Wikipedia article:

In non-circumbinary planets, if a planet's distance to its primary exceeds about one fifth of the closest approach of the other star, orbital stability is not guaranteed.1 Whether planets might form in binaries at all had long been unclear, given that gravitational forces might interfere with planet formation. Theoretical work by Alan Boss at the Carnegie Institution has shown that gas giants can form around stars in binary systems much as they do around solitary stars.6

https://en.wikipedia.org/wiki/Habitability_of_binary_star_systems#:~:text=Planets%20that%20orbit%20just%20one,planets%20within%20stable%20orbital%20ranges.

Thus it would appear that K type star might be orbiting a little too close to the F type star for a planet orbiting the F type star at the EED to have a stable orbit, and you might want to move the K type farther from the F type star.

Unfortunately I can't access the paper with the orbital stability limit. Here is a link to the claim that gas giant planets can form in an S-type orbit in a binary system.

https://web.archive.org/web/20110515225714/http://carnegieinstitution.org/news_releases/news_0601_10.html

And of course you want a terrestrial type planet, not a gas giant. And that article doesn't say whether there is a minimum separation between the stars necessary for planet formation or what it would be.

Here is a link to a newer article article about planetary formation in binary star systems.

https://www.cam.ac.uk/research/news/astronomers-show-how-planets-form-in-binary-systems-without-getting-crushed

And here is a link to the paper mentioned.

https://www.aanda.org/component/article?access=doi&doi=10.1051/0004-6361/202141139

And no doubt there are other articles about planetary formation in binary star systems.

I wrote above:

Thus it would appear that K type star might be orbiting a little too close to the F type star for a planet orbiting the F type star at the EED to have a stable orbit, and you might want to move the K type farther from the F type star.

One alternative might be to move the planet closer to the F type star than the EED instead of moving the K type star farther out.

In that case the planet would have to be no closer to the F type star than the inner edge of the circumstellar habitable zone or Goldilocks zone of the F type star. How do you calculate the inner and/or outer edges of the habitable zone of a star? A simple rule of thumb method is take the luminosity of the star divided by the luminosity of the Sun, find the square root, and multiply it by the distance of that edge of the Sun's habitable zone.

So where are the inner and outer edges of the Sun's Goldilocks zone?

Here is a link to a list of over a dozen estimates in the last 60 years.

https://en.wikipedia.org/wiki/Habitable_zone#Solar_System_estimates

You will note that one estimate is that the inner edge of the Sun's habitable zone could be as close as 0.38 AU, and another estimate is that the inner edge of the Sun's habitable zone could be as far out as 0.99 AU. An inner limit of 0.38 AU would seem to give lots of room to have your planet closer than the EED, but that probably would require a special atmosphere that wouldn't be breathable for humans. And of course an inner limit of 0.99 AU wouldn't enable the planet to be significantly closer than the EED.

So you should move your K type star farther away from the F star to give the habitable planet a stable orbit, and because it might not be possible for the stars to form planets with they are only 5.5 AU apart.

I guess I would advise to have the two stars twice as far apart, at 11 AU.

But I guess that I will do some calculations with the two stars 55 AU apart. With the planet at the EED between 1.224 and 1.264 AU, the F type star will appear to have a smaller disc than the Sun, and will appear as bright as the Sun in the sky of the planet. Thus the F type star will have an apparent magnitude of about minus 26.832 (lower value magnitudes are brighter than higher value magnitudes).

So if they were at the same distance, the F type star would appear about 7.142 to 8.888 times as bright as the K type star. And at 55 AU, the K type star would be about 43.5 to 44.9 times far away from the planet. That means the apparent brightness of the K type star would be about 1/((43.5 X 43.5)) to 1/(45 X 45), or 1/1,892.25 to 1/2,025, or 0.000528 to 0.000493, as bright as the F type star, divided by the difference the brightness of the two stars when seen at the same distance. So the apparent brightness of the K type star would be about 0.0000739 to 0.0000554 that of the F type star.

And that might seem very dim compared to the F type star.

A magnitude difference of 1corresponds to a brigtness difference of 2.514 times, and a magnitude difference of 5 corresponds to brightness difference of 100 times, and a magnitude difference of 10 corresponds to a brightness difference of 10,000 times.

Since the K type star would be about a bit less than 0.00001 as bright as the Sun as seen from Earth, it would have an apparent magnitude of about minus 16. That is over 11 magnitudes or 250 times as bright as the planet Venus.

The planet Venus can be be seen in broad daylight, and Jupiter and Mars can be seen when the Sun is very low in the sky. The light of the planet Venus on Earth is bright enough to cast shadows.

And the K type star should 11 magnitudes or 250 times as bright in the sky of the planet as Venus in the sky of Earth. It would be about 3 magnitudes, or 15 times, brighter than the full moon, and even brighter than the moon at other phases. It would be visible from the surface of the planet whenever it was above the horizon and not hidden by clouds, in the day and in the night.

At a distance of 55 AU, the K type star would not show a visible disc, but would appear as a very bright point of light.

So the K-type star would not provide much heat to the planet, but would be quite bright and make it easy to see at night.

And at the distance of 5.5 AU asked about in the question, the K type star would be 10 times closer and thus 100 times, or 5 magnitudes, brighter, at about magnitude minus 21. that would make it about 8 magnitudes, or about 650 times, brighter than the full moon. And the K type star would have an angular diameter of about 4 arc minutes and thus appear as a tiny and very bright disc.

You want the brightness of the K type star to be no more than twilight on Earth.

3.4 lux is considered to be the dark limit of twilight.

https://en.wikipedia.org/wiki/Lux#Illuminance

A magnitude of minus 14.2 is considered equal to 1 lux. So the brightest you want the K type star to be should be about minus 15.5 or something. And at 55 AU the K type star should have an apparent magnitude of minus 16, brighter than desired, and at 5.5 AU the K type star would have an apparent magnitude of minus 21, 250 times your limit for brightness.

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  • $\begingroup$ I can never get over this absolute GEM of a website. Thank you SO MUCH for taking the time out of your day to run all of these calculations and link all sources just for some silly little question. So to clarify at 5.5 AU it is WAY to bright for what I'm looking for while at 55 AU it is about what I'm looking for? also the star is a bit close to give stable orbits got it. As long as it looks like a binary system from the surface I'm more than happy. $\endgroup$
    – Dono
    Feb 18 at 20:07

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