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I made a planet orbiting (S-type orbit) a K-type star in a binary system; the other star is F-type.

Star F-type K-type
Mass 1.3 M☉ or 2.6x1030 kg 0.7 M☉ or 1x1030 kg
Radius 1.14 R☉ or 0.8x109 m 0.86 R☉ or 0.6x109 m
Temperature 1.22 T☉ or 7,050 K 0.76 T☉ or 4,390 K
Luminosity 2.86 L☉ or 10x1026 W 0.24 L☉ or 0.92x1026 W
Minimum Maximum Period
55 AU 175.8 AU 888 Earth years
Planet
Mass 1.5 M⊕ or 8,960.4x1021 kg
Density 1.0 M⊕ or 5,514kg/m3
Radius 1.15 M⊕ or 7,294,531.72 m
Gravity 1.15 M⊕ or 11,232 m/s2
Volume 1.5 M⊕ or 1,625.03x1018 m3
Surface Area 668.32x1012 m2 (668,320,024.257 km2)
Orbital Period 156d 19h 13m (157d)
Rotational Period 20h 02m 72.120s (20h)
Perihelion 58,992,000,000 m
Aphelion 61,008,000,000 m
Axial Tilt 20.2°
Orbital Distance 0.4 AU

How does the second star (F-type) influence the climate of the planet orbiting the K-type star at its closest?

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  • $\begingroup$ What is an S-type orbit? In title. $\endgroup$
    – rek
    Mar 26 at 12:41
  • $\begingroup$ @rek I'ts when a planet orbits only one of the stars in a binary star system. $\endgroup$ Mar 26 at 12:48
  • $\begingroup$ How did you make the format? $\endgroup$ Mar 27 at 15:12
  • $\begingroup$ rek did it to me lol it's beautiful! $\endgroup$ Mar 29 at 3:11
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I'm not sure If I am reading every thing you have here correctly, so I will describe my methods before giving my conclusions. Tell me if I got something wrong.

First, lets figure out the amount of solar radiation your planet is getting and from which sources. To find this we need to compare apparent brightness. I used this calculator https://www.omnicalculator.com/physics/luminosity.

For the K-type

We get an apparent brightness of -27.27. This is easy since we know that the planet is 0.4 AU away. Compared to the Sun from Earth, which is about -26.8 that is about 154% as bright, So you get a bit more heat.

For the F-Type

Here I have decided to work a bit harder, and tell you how the energy output should vary throughout the year. We know that theoretically the planet can be as close to the F-Type as the minimums distance between the stars minus the orbital distance and as far as the maximum distance between the stars plus the orbital distance. so that is 54.6 and 176.2 AU, plugging that in we get -19.26 and -16.74 for luminosities respectively. In comparison to the sun that is 1066 times less energy and 10960 times less energy respectively. In comparison to the full moon (-12.74) that is 415 times brighter and 40 times brighter. This means that the energy output you get ranges from very bright full moon to pretty bright full moon. To get the closer amount of energy from the sun you would have to be 32.7 AU from the sun. That is further than Neptune at 30 AU, the furthest planet form the sun.

So, when the stars are furthest apart we get:

K-Type: 154% of the sun

F-type: 0.009% of the sun (that is the correct number, it is 9 thousandths of a percent)

when the stars are closest together we get

K-Type: 154% of the sun

F-type: 0.094% of the sun (again, the correct number)

As you can see, the greatest possible effect from the F-type star is less than that of minor fluctuations in the K-type star. The F-star has about the same effect on the planet as the sun has on Neptune, which stays at a cozy -200 degrees Celsius year round.

If you are looking for a star that could nearly double the heat the the planet receives, a star like Antares B (https://en.wikipedia.org/wiki/Antares) would have roughly the same heat output as the sun at its closest point, but about one tenth of the sun's heat at its highest point. That said, Antares B is an unusually bright star for a M-type. Most non-blue non B/O type stars don't clear 1000 luminosity.

In conclusion

There would be no change to the climate based on the second star.

Extra

The second star may have one effect. While the effect of the second star can't affect plant life it can affect animal life. Since you have an axial tilt, there will be some parts of the year and places when the polar regions of the planet get no sunlight for months called the polar night (https://en.wikipedia.org/wiki/Polar_night). If your planet is in a polar orbit around one of the stars then during about a half of the 888 earth year binary orbit during the polar night one of poles will be constantly under second star's light with no light from the K-type star. The pole that has this light will alternate and depending on how close the stars are during this time they are pointed the right way, this light may be stronger for one pole than the other. Since this is based on the stars this will be regular, not periodic. So one pole may always get more light from the second star in its polar nights.

This will may lead to some creatures becoming semi-nocturnal and semi-hibernating. Instead of hibernating in these winter months since there is no light they may stay awake instead. This may lead to nocturnal animals getting stronger or weaker (more light helps them navigate and operate, but also makes the advantages of night adaptations weaker while helping other creatures that are less night adapted) However, no creature will evolve to be this way permanently, since this only lasts for about 200 years and stops for 600 years. At best this would be a behavior that is learned or evolves to sense the change and then apply the behavior.

While not many people live in these regions, and fewer people visit both poles polar natives will probably have stories that explain why sometimes for hundreds of years there is no darkness, and then most years there are months of darkness. Also, if people do visit both poles before the scientific revolution, like in "Avatar the Last Airbender" expect the polar natives to try and use the relative brightness of their years under the second star to assert their superiority.

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  • $\begingroup$ Charlie, first of all, thank you for your time and effort to help me! So the secondary star is not a big deal to that planet... I was thinking it may cause more than a beautiful view in the night sky, and now i'm ease with this. As you can see in the sheet rek made beautifully to me, the polar circle is at latitude 69.8°, but i made all my land masses below 55°. One more time, thank you very much! $\endgroup$ Mar 29 at 3:45

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