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I am making a setting for a various TTRPGs that me and my friends want to play.

I want this world to have longer seasons than earth. I was thinking about 3-4 earth years each season.

This is difficult, as I know this would not work with our current solar system. I know that as you get further out from the sun, orbits generally get longer. But, the goldilocks zone for our sun does not reach out to the zone that this long orbit would be possible. So, what kind of star would this planet need to be orbiting in order to not end up freezing, but still have long seasons.

As I have it now, it orbits a red giant at about the distance Jupiter is from our sun. But I am unsure of if this is realistic AT ALL. My planet is a little bit bigger than earth (1.2) if that makes a difference.

I am willing to do math to figure all of this stuff out, but I have no idea where to start.

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    $\begingroup$ Welcome to the site! I like this question, but am not going to immediately answer this. However, a small bit of advice: only build the parts of the world that matter. For instance, if this is a fantasy TTRPG, then this can be "hand waved" as one of those things people just don't know. Of course, if planet hopping and/or the actual orbital mechanics matter to play, then it's certainly justifiable to figure this out. I am sure we'll get some good answers soon! $\endgroup$
    – PipperChip
    Commented Sep 26, 2022 at 17:04
  • $\begingroup$ Thanks! That is certainly something I have considered haha. I am planning on running a hard fantasy game in the near future as it is something my players have expressed interest in. Perhaps orbital mechanics are a bit overkill, but I always start on the macro scale and work my way down, so I can't get past it. As of now, my hand wave is the big red giant simply because I don't know how it works :) $\endgroup$
    – Nixk
    Commented Sep 26, 2022 at 17:06
  • $\begingroup$ I second @PipperChip's advice. I've run a couple of games in fantasy settings whose annual seasons and climate were different from Earth's, and when I told the players that "winter has come early this year" (as a thematic echo of the new evil threat), not one of the players asked me whether this occurrence was evidence that the planet's orbit around its star had changed, or whether the star was entering a new stage in its lifecycle. Come to think of it, they didn't even ask if the longer cold season would increase the cost of stabling their horses. They simply bought snowshoes. $\endgroup$
    – Tom
    Commented Sep 26, 2022 at 18:11
  • $\begingroup$ @PipperChip , all true... but a bit of consistency can really bring the room-- uh, I mean world, together $\endgroup$
    – cowlinator
    Commented Sep 26, 2022 at 19:39
  • $\begingroup$ With my lack of hard knowledge for orbital mechanics, I would not presume to offer an Answer. Still and all, I'm wondering if some binary star system might not do the job? The planet does a figure 8 orbit between two normal stars with summer happening when orbiting at the normal distance of either star. A hard winter sets in during the transit from Alpha star to Beta star and back again. $\endgroup$
    – Blaze
    Commented Sep 27, 2022 at 1:04

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Red giants are not as good of candidates for life as you may think

They tend to be much less stable than main sequence stars because thier shells go through a cycle of periodic collapses resulting in nova that would each wipe out all life in the would be Goldilocks zone. Also, stars spend about 10 times as long as main sequence stars as they do as Red Giants.

You might sort of be able to make a main sequence star work better

As the mass of a main sequence star increases (forcing more distant orbits to be a bit faster), its luminosity increases exponentially meaning the goldilocks zone expands out farther from a star in proportion to mass much fast than orbital speeds accelerate.

To get an Earth like 3-4 year season, you need an orbital distance of about 5-7 AU orbiting a main sequence star about 1.9-2.0 times the mass of our sun in order to get this planet in the middle of the Goldilocks zone. Such a star would be about 15-30 times as bright as our sun despite only being about twice its mass.

At this mass you are looking at a type A star just a bit smaller and dimer than Sirius which would appear white or ever so slightly blue compared to our own sun which is in the yellow range, but keep in mind that not only do more massive stars become exponentially brighter, they also become exponentially shorter lived. This star would only have a total life span of a few hundred million years; so, you world would likely not have time for life to evolve. If it is inhabited, all life likely came from another world.

But you can do even better by playing with atmospheric gasses

While a type A star star is not bad, you could probably get away with a heavy type F main sequence star (yellow-white dwarf). At 1.5-1.6 solar masses, you could still get about 1/2 as much sunlight as the Earth receives at an acceptable orbital distance. While Earth is mostly made of iron, oxygen, and silicon, some exoplanets have been observed with far more carbon in place of silicon which could result in more CO2 and less Ozone in the atmosphere increasing the greenhouse effect when all other factors are comparable.

In this case you could have a star that is stable for about 2 billion years... maybe not long enough to see complex life evolve if it follows the same timescale as Earth did, but close enough to say that complex life could maybe evolve there before the star dies.

The higher levels of carbon could also be a good excuse for a world with more mega fauna which is good for a fantasy setting.

As for what the world would be like, it would barely be distinguishable from Earth in most ways. The sun would be a bit more white than yellow, and the sky would be a slightly deeper blue. Diamonds would be as common as quartz and vise versa. UV radiation will be a bit more harsh, but by in large, the differences could be negligible, especially if the local life is adapted to this environment.

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  • $\begingroup$ This is about a dungeon/fantasy scenario, maybe your main sequence Sirius-like star could work fine, when there is no need to let this species evolve on the same planet they live on. They may have immigrated or been planted there. $\endgroup$
    – Goodies
    Commented Sep 26, 2022 at 21:28
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Easiest thing I can imagine would be to give the planet an accelerated axial precession around a stable main seq host star. The wobbling top equivalent of the planet's axis. When the north pole points away from the sun it's winter in the northern hemisphere. On earth the procession is geologically slow, ~26,000 years. And are related to Milankovitch cycles / Ice ages and other cyclic climatological events. If however the procession was fast enough to keep the pole pointed away it's star even as it orbited, you could choose how long you wished your winters to be.

This is a deep rabbit hole if you want to "keep it real" for a fast procession like this, I think it would call for the absence of any moon of substantial size, among other things.

Note: Happy to be fact checked on this. How stable/unstable a wobbly planet like this would be. Perhaps it would have to have been smacked by another body pretty hard in the geologically recent past to evolve this. I'm in no way able to math out an orbital system no matter how big of a chalkboard you give me.

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We have some boundaries to satisfy.

First, there is a relationship between the mass of the Sun and the orbital period of the planet. That is given by Kepler's Third Law, and we can write,

T=SQRT(4PI^2D^3/(G*M))

so, if we want a T of about sixteen times that of Earth, we need to increase (D^3/M) by a factor of 256 with respect to the Solar System. Either we place the planet farther from the sun, or we make the sun smaller.

In either case the planet will be way colder.

But we can do both at once. We can increase the sun mass and increase the distance even more. If we double the sun mass, X^3/2 = 256 yields X^3 = 512, so X=8, which means we need to place the planet eight astronomical units away from the sun.

This means that it will get 1/(8*8) = 64 times less light and heat than the Earth. Is a star twice as big as the Sun also sixty-four times brighter?

For main-sequence stars there is a proportionality law due to Jakob K. E. Halm that states that the increase in the star's luminosity is proportional to the increase in mass, multiplied by 1.4 and raised to the power of approximately 3.5 (for the interval of interest to us).

So a main sequence star twice as big as the Sun will be just 1.4 x 2^3.5 = 15.4 times brighter, which is not enough. But 3.5 is a lot - it means that luminosity increases rapidly with mass.

To make a long story short, let us consider a B9V-type main sequence star.

It will have a mass of 2.8 solar masses, so the planet will need to be at about 8.94 AU to get our 16-year period.

Its luminosity is 80 times that of the Sun, and at a distance of 8.94 AU, it will get 80 times less luminosity - so, it will get exactly the same energy as Earth does.

We still have a problem here: the star surface temperature (which determines its colour) also depends on mass, and in our case it will be around 11000 K to Earth's Sun's 5700 K: this means that the spectrum will be markedly whiter - actually, since the spectrum extends beyond the visible boundary, it will have a much stronger ultraviolet component (yellow = the Sun, scaled; orange = our B9V star).

spectral emission from a B9V star

So, you'll probably want to tweak the atmospheric composition a bit. A denser, smaller planet will have a steeper gravity well, which allows a surface gravity of Earth normal while allowing a deeper atmosphere. You don't need very much, because dangerous ultraviolet is already absorbed by atmospheric gases; you'll end up with a thicker ozone layer.

Another notable difference will be that a star of 2.8 solar masses has a radius just about 40% more than the Sun (mass goes like the cube of the radius, so radius goes up with just the cube root of the mass). But, being nine times farther, it will appear nine times smaller; so, just about 1.4/9 = 16% of the Sun. It will appear as a white-hot, blazing pinpoint of light in the middle of the sky. Since the same apparent radiation comes from a disc about 3% the apparent size of the Sun, its apparent intensity will be thirty times higher; it will be impossible to look at directly with the unprotected eye (painful, and dangerous). For the same reason there should be much more pronounced parhelia for most of the year.

The sky will also be bluer and, farther from the sun, of a much deeper blue than Earth. Sunsets and dawns, by the same token, should be a deeper red.

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Slow it down.

The Sun and orbit distance is the same as Earth. But the orbit speed is slower so one BlipBloop year is 16 Earth years. So Each Season is 4 Earth years long. Now get to work on the calendar.

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    $\begingroup$ That is not really going to work. The square of the orbital period is inversely proportional to the mass of the primary, so to have a "year" of sixteen Earth years, the Sun must be 16^2 = 256 times smaller. That's below the Whitworth limit of 0.08 solar masses; the star would not even ignite. $\endgroup$
    – LSerni
    Commented Sep 27, 2022 at 22:00
  • $\begingroup$ @LSerni Yes this is more complicated than I first thought. $\endgroup$
    – Daron
    Commented Sep 27, 2022 at 22:20
  • $\begingroup$ it should work if we put the planet farther than the Earth. At that point, though, the star needs to be bigger and brighter. I've whipped up a not-too-satisfactory solution with a B-type star. $\endgroup$
    – LSerni
    Commented Sep 27, 2022 at 22:39

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