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Architect of Worlds – Step Nine: Structure of Protoplanetary Disk

Architect of Worlds – Step Nine: Structure of Protoplanetary Disk

Over the next few days, I’ll be posting the current draft of the next section of the Architect of Worlds project. Here, now that we’ve designed and arranged the star(s) of a given system, we can give each star its own family of attendant planets, determining their basic physical and dynamic properties along the way.


Step Nine: Structure of Protoplanetary Disk

In this step, we determine the most important properties of the star’s protoplanetary disk, which will in turn govern the size and placement of the planets that form. These properties include the location of the disk’s effective inner and outer edges, the location of the “snow line,” and the relative mass and density of the disk.

Procedure

Select or compute each of the following parameters, and record the results for later use.

Disk Inner Edge

Astronomers are not clear where the inner edge of a protoplanetary disk will normally be located. In fact, it’s possible that the disk has no inner edge, since material continues to fall onto the star’s surface throughout the period of planetary formation. Planets which migrate strongly inward do seem to stop some distance away from the primary star, but their eventual orbital period may be quite short, on the order of a few days.

Select a radius for the inner edge of the protoplanetary disk. To select a distance at random, roll 2d6:

R=(2d6)\times0.003\times\sqrt[3]{M}

Here, R is the radius of the disk inner edge in AU, and M is the star’s mass in solar masses. Round off to two significant figures.

Snow Line

The “snow line” represents a distance from the star where volatiles, especially water, can freeze and remain solid within the protoplanetary disk. Inside the snow line, any solid matter within the disk will tend to be dry: dust leading up to masses of stone. Outside the snow line, water and other volatiles will be present in the form of ices. Thus, crossing the snow line outward, an observer would see a sharp rise in the amount of solid material available for planetary formation. The most probable location for the formation of the system’s largest planet is at the snow line.

Determine the radius of the snow line as follows. If L0 is the initial luminosity of the star in solar units, then:

R=4.2\times\sqrt{L_0}

Here, R is the radius of the snow line in AU. Round off to two significant figures.

Slow-Accretion Line

On the outer edges of a planetary system in formation, the density of available material is low and the orbital period of that material is long. Protoplanets forming in this region may have difficulty sweeping up all the material that’s theoretically available. Most of that material is likely to remain free when the period of planetary formation ends, to be swept out into interstellar space. We model this effect by establishing a “slow-accretion line,” beyond which the formation of planets is unlikely.

Determine the radius of the slow-accretion line as follows. If M is the mass of the star in solar masses, then:

R=15\times\sqrt[3]{M}

Here, R is the radius of the slow-accretion line in AU. Round off to two significant figures.

Disk Density

Mass available for the formation of planets will be limited, since the bulk of the protostellar nebula will have ended up in the star rather than in the protoplanetary disk. The mass in the disk is dominated by gas, primarily hydrogen and helium. Other constituents of the disk include frozen volatiles (outside the snow line) and dust.

For astronomers, measuring the mass of a protoplanetary disk is rather difficult. Most estimates indicate that a star’s protoplanetary disk will have about 1% of the star’s mass, although this can vary widely. For this design sequence, we will need to determine the disk mass factor. This is a multiplicative factor; for example, a star with mass equal to the Sun and a disk mass factor of 1.0 will have a protoplanetary disk roughly as massive as the Sun’s.

Select a disk mass factor between 0.1 and 1.0, with most protoplanetary disks having a density factor close to 1. To select a disk mass factor at random, roll 3d6 on the Disk Mass Factor Table. Feel free to select a value between two results on the table. Each component in a multiple star system can have its own disk mass factor.

Roll (3d6) Disk Mass Factor
3 0.1
4 0.13
5 0.18
6 0.25
7 0.36
8 0.5
9 0.7
10-11 1.0
12 1.4
13 2.0
14 2.8
15 4.0
16 5.6
17 7.5
18 10.0

Selecting for an Earthlike world: The ideal disk density to produce an Earthlike world depends on many factors. To maximize the probability of an Earthlike world, multiply the selected disk mass factor by the star system’s metallicity; the result should be close to 1.

Planetary Mass Budget

The disk mass factor will determine how much material is available for the formation of planets. We will measure this material as a planetary mass budget, deducting from this budget as planets are placed. The planetary mass budget is an estimate of the metals that will end up in planets, where “metals” is used in the astronomical sense (that is, all elements heavier than helium).

To determine the planetary mass budget, let M be the mass of the star in solar masses, let K be the star’s metallicity, and let D be the disk mass factor determined above. Then:

B=80\times M\times K\times D

Here, B is the planetary mass budget, measured in Earth-masses. Round off to two significant figures.

Special Case: Forbidden Zone

If the star under development is a member of a multiple star system, it is possible that one of its companion stars will approach so closely as to disrupt part of the protoplanetary disk. This will give rise to a forbidden zone, a span of orbital radii in which no stable orbit is possible. As the planetary system is designed, planets will not form within the forbidden zone, and any planet that migrates into the zone will be lost.

To check for the existence of a forbidden zone, compute one-third the minimum distance for the nearest companion star. This will be the radius of the inner edge of any potential forbidden zone.

If a forbidden zone exists, and its inner edge is inside the slow-accretion line, then some of the disk material that might otherwise have formed planets has been stripped away by the companion star’s influence. Adjust the planetary mass budget as follows. If B0 is the planetary mass budget before accounting for the forbidden zone, RF is the radius of the inner edge of the forbidden zone, and RA is the radius of the slow-accretion line, then:

B=B_0\times\sqrt{\frac{R_F}{R_A}}

Here, B is the adjusted planetary mass budget. Round off to two significant figures.

Examples

Arcadia: Alice is working with a single star with a mass of 0.82 solar masses, initial luminosity of 0.28 solar units, and metallicity of 0.63. She decides that the disk inner edge will be at about 0.025 AU. She computes that the snow line will be at about 2.2 AU and the slow-accretion line will be at about 14.0 AU.

Alice makes note of the star’s metallicity of 0.63, and selects a disk mass factor of 2.0. Multiplying the two together yields a result of 1.26, which is reasonably close to 1 and seems likely to yield an Earthlike world at the end of the design process. The planetary mass budget will be:

80\times0.82\times0.63\times2.0\approx83

Since the primary star is a singleton, there will be no forbidden zone, so Alice has all the information she needs to complete this step of the design process.

Beta Nine: Bob is working on the primary star of the Beta Nine system, a red dwarf with a mass of 0.18 solar masses, luminosity of 0.0045 solar units, and metallicity of 2.5. He rolls 2d6 for a result of 8, and determines that the disk inner edge will be at about 0.014 AU. He computes that the snow line will be at about 0.28 AU, and the slow-accretion line will be at about 8.5 AU.

Bob rolls 3d6 on the Disk Mass Factor Table, and gets a roll of 8. This suggests a disk mass factor of 0.5. Bob accepts this result, noticing that the product of 0.5 and the star’s metallicity of 2.5 is 1.25, not far from the value most likely to yield an Earthlike world. The initial planetary mass budget will be about 18 Earth-masses.

Beta Nine is a binary star system, with a brown dwarf companion in a close orbit. The minimum separation of the two stars is 2.0 AU. One-third of this is 0.67 AU, well within the slow-accretion line, but not within the snow line. There is a forbidden zone for this star, with its inner edge at 0.67 AU. The existence of the forbidden zone will (dramatically) reduce the available planetary mass budget:

18\times\sqrt{\frac{0.67}{8.5}}\approx5.1

The Beta Nine primary will have fewer planets, since the brown dwarf companion has stripped away most of the material needed to form them!

Status Report (13 July 2018)

Status Report (13 July 2018)

With the release of “Pilgrimage” I was thinking that my next major project would involve getting the next Aminata Ndoye story (“In the House of War,” a roughly 20,000-word novella) polished up and out the door.

Going back and reviewing the most recent version of that story, though, I think I may need to do some world-building work first. I’ve done a fair amount of research since I first wrote that story, and my ideas about how interstellar civilization is structured have evolved a bit.

So, new plan of action:

  • First item will be to revise and improve the planetary-system design sequence for Architect of Worlds. I’ll be publishing the revised material here over the next week or so.
  • Then I’m going to re-work my current map of the interstellar neighborhood (and the associated database of nearby planetary systems). Along the way I’m going to double-check my computations from about 2014-2015 about the galactic density of habitable planets, sentient life, high-tech civilizations, and so on. It’s possible that my new design sequences will give rise to a somewhat different set of assumptions.
  • I may also do at least a sketch map of the local galactic spiral arm, just to give me a better idea of the “terrain” in khedai space.
  • Once I have all that done, I should be able to revise “In the House of War” for publication, and I might have a clearer picture to support further stories in the setting too.

Looks as if my fantasy novel, The Curse of Steel, will be going on the back-burner for a while. That’s okay. I’ve learned the hard way to let my muse go where it wants to go at the moment. At least I’ll be making progress on Architect of Worlds, and I should be able to get another Human Destiny story out the door at the end.

Building Toswao

Building Toswao

This morning I’ll be focusing on the single Earth-like world in the Karjann star system, Toswao, the focus of my play-through of Bios: Genesis and Bios: Megafauna.

Here, I’m getting into world-design procedures that I haven’t fully documented yet. The part of the Architect of Worlds project that’s giving me the most trouble is procedures for working out the properties of individual worlds. I find that there are a lot of contingent factors, many of which have been completely ignored by the world-design systems that I’ve previously found in print, or written myself. Some of those factors are not well-understood even today, or are so complex that no simple model will really capture them. So it’s a challenge to come up with a design sequence that’s coherent, straightforward to apply, and likely to reflect a wide range of plausible results. Research continues.

Of course, for Toswao, a lot of parameters are already set and it’s just a matter of fleshing out details, while checking to make sure there’s nothing wildly implausible. That’s an easier problem.

Let’s start with what we know. Toswao is a terrestrial planet with a mass of 1.18 Earth-masses. I have  a straightforward model for the density of terrestrial bodies, and with one dice roll I can compute that Toswao has an average density of 1.044 times that of Earth (5.72 g/cc). That immediately gives us a planetary radius of 6635 kilometers (1.04 times that of Earth) and a surface gravity of 1.09 G.

Here’s the first big question that most published world-design sequences would ignore: does Toswao have a strong magnetic field?

It turns out that item is important. A planetary magnetic field is critical for protecting the surface environment from solar and cosmic radiation. It’s also critical for making sure the planet can retain a significant atmosphere. Without a strong magnetic field, the solar wind comes into direct contact with the outer atmosphere, and will tend to strip away the air over fairly short time-scales. This effect turns out to be quite a bit stronger than simple thermal loss, so if you want a habitable planet, you really need to make sure compasses work there.

Toswao is nice and big and dense, so it will certainly have a liquid nickel-iron core whose rotation can create a dynamo. But that leads us to a second question that most world-design sequences probably get wrong: how quickly does Toswao rotate, and where is its rotational axis?

The world-design sequences I’ve seen (and the ones I’ve written in the past) generally assume that terrestrial planets all rotate at similar rates, their rotational axes well-behaved, modified (if at all) only by tidal effects over long periods. Yet even in our own planetary system we can see that this isn’t the case, especially when we look at Venus. Recent models for the formation of terrestrial planets suggest that the process is much more catastrophic than we once assumed. Every terrestrial planet, even Earth, has been shaped by enormous impacts and collisions, so that its final rotation axis and rate are more random than we might expect.

Then, of course, the tidal interactions between a terrestrial planet and its primary star (and any major natural satellite) turn out to be much more challenging to model than we might like. This isn’t because the physics of the situation are poorly understood – they’re not – but because the system is very sensitive to small details. If Earth was a perfect, elastic, and uniform sphere, it would be easy to determine exactly how solar and lunar tides would affect its rotation over eons. Unfortunately, terrestrial planets are quite a bit more complex and varied than that.

In the light of that, I have yet to produce a game-ready model for planetary rotation that I’m happy with. For now, let’s assume that Toswao’s rotation is similar to that of Earth (especially since the planet does have a major natural satellite like our Moon). Toswao is younger than Earth, so I’ll assume that its rotation rate is a bit faster, and that its satellite is a little closer in than the Moon. Without laying out all of my selections and computations in full, here’s some results:

Toswao

  • Mass: 1.18 Earth-masses
  • Density: 1.044 Earth (5.72 g/cc)
  • Radius: 6635 kilometers (1.04 Earth)
  • Surface Gravity: 1.09 G
  • Orbital Radius: 0.99 AU
  • Orbital Eccentricity: 0.08
  • Periastron: 0.91 AU
  • Apastron: 1.07 AU
  • Angular Diameter of Primary Star: 0.55 – 0.64 degrees
  • Orbital Period: 0.9660 standard years (352.84 standard days)
  • Rotation Period: 22.608 standard hours (0.9420 standard days)
  • Day Length: 22.669 standard hours (0.9445 standard days)
  • Apparent Year Length: 373.56 local days
  • Axial Inclination: 24°

Given these values for Toswao’s rotation, we can be confident that it has a nice, strong magnetosphere to protect the air and surface. We can proceed on the assumption that Toswao has a more or less Earth-like atmosphere.

We already know some things about that atmosphere, from the final state of the Bios: Megafauna game, and from our computations when we were determining the planet’s placement in orbit around Karjann. A quick random dice roll gives us an “atmospheric mass” for Toswao of about 1.2, noticeably greater than that of Earth. Along with the known details of composition that I generated earlier, that gives us:

  • Atmospheric Mass: 1.2
  • Surface Atmospheric Pressure: 1.3 atmospheres
  • Atmospheric Composition: Nitrogen 64%, oxygen 34%, argon 1.6%, carbon dioxide 0.2%, other components 0.2%. Nitrogen partial pressure about 0.83 atm. Oxygen partial pressure about 0.44 atm. Carbon dioxide partial pressure about 0.003 atm.
  • Hydrographics: 88% ocean coverage
  • Planetary Albedo: 0.5
  • Greenhouse Effect: 44 K
  • Average Surface Temperature: 292 K (19° C, or 66° F)

That atmosphere looks breathable for unmodified and unprotected humans, but just barely. The partial pressure of oxygen is approaching high enough to be toxic over long exposures, and there’s a lot of CO2 in the air too. We would probably find Toswao’s air rather invigorating in the short term, but causing some damage to our eyes and lungs in the long term. In the meantime, we might find our cognitive function a bit muddled by CO2-triggered changes in blood flow to our brains. Might want to wear a light breather mask just to keep our blood chemistry happy, if we’re going to be spending much time here.

One more set of details. We know from the Bios: Genesis game that Toswao had a “big whack” event like Earth’s, giving rise to a big, Luna-like natural satellite. I double-checked Toswao’s “Hill radius,” the distance at which Karjann’s gravitational influence overwhelms Toswao’s, and found that there’s plenty of room for the planet to retain a moon.

A random roll sets the satellite’s mass, from which I can quickly determine its density, radius, and surface gravity. I made the non-random decision to place this satellite a little closer to Toswao than Luna is to Earth, about 50 Toswao-radii rather than Luna’s distance of 60 Earth-radii. Here are the numbers:

Toswao’s Moon

  • Hill Radius: 2.06 million kilometers
  • Orbital Radius: 320,000 kilometers
  • Orbital Eccentricity: Negligible
  • Mass: 0.0165 Earth-masses
  • Density: 0.64 Earth (3.53 g/cc)
  • Radius: 1880 kilometers
  • Surface Gravity: 0.189 G
  • Orbital Period: 19.180 standard days
  • Apparent Lunar Cycle: 23.776 standard hours (0.9907 standard days)
  • Synodic Month: 20.283 standard days
  • Angular Diameter: 0.69 degrees (from planetary surface)

So there we go. There are a few more physical parameters we could probably generate, but this should give us enough to work with for now.

Toswao is an ocean planet, a little warmer than Earth, with lots of clouds. Visiting humans would find the local gravity heavy, but manageable even over long periods. The planet’s atmosphere is breathable for humans in the short term, although we might find it difficult under long exposures. I haven’t explicitly computed the strength of local tides, but both the primary star and the moon are more massive and closer than their counterparts on Earth, so I would expect stronger tides.

Toswao has Earth-like axial tilt and so exhibits similar seasons, although the situation is complicated by a larger orbital eccentricity. Depending on how the orbital parameters line up with the axial inclination, that might tend to either damp out or to accentuate seasonal variation.

I don’t intend to draw a world map, unless the story emerging in my head turns out to be a lot more extensive than I expect. Still, we can say a few things, based on the end state of the Bios: Megafauna game. I would expect the planet’s small continents to be heavily forested, at least in their natural state. Lots of green in the shallow seas, too, to contribute to that high oxygen concentration. I wouldn’t expect to see a lot of deserts or wastelands.

A useful exercise, not only because it gave me a world to use in my creative work, but also because it gave me a motivating example, bringing out details that I’ll need to address in upcoming sections of Architect of Worlds. In the next couple of posts, I’ll be working out a character template for the dominant sentient species native to this world, and writing up some of their back story.

Building the Karjann System

Building the Karjann System

Okay, for the last few weeks I’ve been logging a play-through of the Phil Eklund games Bios: Genesis and Bios: Megafauna, in a demonstration of how those games can be used to support worldbuilding for science fiction. A quick way to review those posts would be to check out the Worldbuilding by Simulation category and look at all the posts since the beginning of June 2018.

Now it’s time to do some math, and design the star system and main habitable planet compatible with the results of the Bios games. I’ll be using the current draft design sequences from my Architect of Worlds project. In particular, the current draft of the star system design sequence can be found at Architect of Worlds: Designing Star Systems. The design sequence for designing planets hasn’t been published yet, and I need to do a fairly extensive revision pass before that happens, but its current draft should be sufficient for this purpose.

I begin by coming up with a pair of names for the habitable planet (Toswao) and its primary star (Karjann). I have absolutely no constructed language work to back those up, and probably won’t go that far for a single story. Those names simply emerged from the back of my mind under the stimulus of a random-name generator; I think they look and sound pleasant, so there we go.

Primary Star

Looking back on the Bios: Megafauna game, I recall that Toswao has spent most of its history with very warm climate, well above Earth’s present average temperature. That suggests a primary star that’s a touch more massive than Sol, and therefore probably more luminous.

Meanwhile, we also know that Toswao is quite a bit younger than Earth. With adjustments, the Bios: Genesis game covered about 3.75 billion years from planetary formation to the end of the Proterozoic period. The Bios: Megafauna game covered about 240 million years from there to the first appearance of a tool-and-language-using species. Add that up and we get 3.99 billion years, which I’m comfortable rounding off to 4.0 billion. Evolution moved fast here! That doesn’t necessarily indicate anything about Karjann, but in my mind the notion of a somewhat more energetic primary star also fits a faster pace of development. So I decide to non-randomly select a primary star mass of 1.04 solar masses.

With a dice roll, I find that Karjann is a solo star – no need to generate details for any companions. I set the star system’s age at exactly 4.0 billion years, and randomly generate the system’s metallicity, ending up with a value so close to 1.0 that I decide to round that off as well. Working through the design sequence, I end up with the following parameters:

Karjann

  • Mass: 1.04 solar masses
  • Main Sequence Lifespan: 8.6 billion years
  • Current Age: 4.0 billion years
  • Metallicity: 1.0
  • Current Effective Temperature: 5800 K
  • Current Luminosity: 1.23 sols
  • Radius: 0.0051 AU (767,000 km)
  • Spectral Class: G2V

Karjann turns out to be quite similar to Sol, a cheerful yellow star about halfway through its stable lifespan, a touch hotter and noticeably brighter.

Planetary System

Before beginning planetary system design, I need to figure out where the habitable world (Toswao) is going to be placed. Here, I have a few clues.

The final state of the Bios: Megafauna game suggested that the planet’s atmosphere had 34% free oxygen. This is pretty high, equivalent to the highest level ever seen in Earth’s atmosphere, back in the Cretaceous era. Some research tells me that such a high free oxygen level has to be supported by very high levels of carbon dioxide, several times the current value in Earth’s atmosphere. So I pin the current CO2 level as about six times Earth’s pre-industrial level, or about 1800 parts per million.

The final Bios: Megafauna state also suggests a planetary albedo of 0.8, but that isn’t at all plausible. The most reflective water-vapor clouds have about that albedo, so a long-term planetary albedo that high means that the entire planet is covered with the brightest possible cloud canopy. Unlikely over a long period, and how is anything surviving with direct sunlight cut off from the photosynthetic base of the food chain? Still, a planet with more hydrographic surface than Earth is likely to have more cloud cover, and therefore a higher overall albedo. I’ll set the planet’s albedo to 0.5, which is probably still very high, but not utterly implausible.

That albedo also suggests a lot more water vapor in the atmosphere than Earth currently sees. I’ll tentatively assume double the amount.

At low concentrations, greenhouse gases appear to affect the planetary average temperature in a logarithmic fashion: every time you double the amount of a greenhouse gas, the temperature goes up by a fixed amount. This question is hideously complex, and climate scientists don’t have any simple models for it, but for CO2 the effect seems to be about 3 K of increase for every doubling of the concentration in the atmosphere. Assuming that water vapor behaves similarly, the greenhouse effect on Toswao appears to be about 11 K more aggressive than on pre-industrial Earth. That gives us a total greenhouse effect of about 44 K.

In Bios: Megafauna, planetary climate is marked on a scale which varies up and down during the game. Next to the bottom of the scale is a space marked “Ice Age,” which I tentatively interpret as a planetary average temperature of 280 K, equivalent to the middle of the last glacial age. The top space on the scale is marked “Runaway Greenhouse,” which I’ll tentatively take as a planetary average temperature of about 350 K, high enough (assuming standard atmospheric pressure) for the equatorial oceans to start boiling. There are twelve spaces between these two points on the scale, so a rough guess of about 6 K per space makes sense. At the end of my play-through, the climate was in the higher of the two spaces in the “Cool” climate band, two spaces above the “Ice Age” point. That suggests a planetary average temperature of about 292 K, a bit warmer than present-day Earth.

If the actual surface temperature is about 292 K, then a greenhouse effect of 44 K suggests an albedo-adjusted blackbody temperature of about 248 K. The relevant formula is:

T_B=278.8\times\sqrt[4]{\frac{\left(1-A\right)L}{R^2}}

Here, A is the planetary albedo, L is the primary star’s luminosity in sols, and R is the planet’s orbital radius in AU. Plugging in values and solving for R, we get an orbital radius of about 0.99 AU, surprisingly quite close to the value for Earth.

Okay, now that I know where to place Toswao, I can lay out the whole planetary system. In particular, I determine that the primary gas giant (the Jupiter-analogue) engaged in moderate inward migration, but then got caught up in a “Grand Tack” event which pulled it back outward to its present position. This depleted the population of planetesimals in the inner system, leading to smaller planets, more widely spaced. Here’s the basic table of planets:

Radius Planet Type Planet Mass
0.41 AU Terrestrial Planet 1.05
0.68 AU Leftover Oligarch 0.15
0.99 AU Terrestrial Planet (Toswao) 1.18
1.68 AU Terrestrial Planet 0.65
2.95 AU Terrestrial Planet 1.12
4.27 AU Large Gas Giant 350
6.83 AU Large Gas Giant 400

I didn’t meddle too much with the random dice rolls here, aside from ensuring that a Terrestrial Planet would appear at the right orbital radius to become Toswao. I did have two results that I wanted to ensure, given the outcome of the Bios: Genesis game.

First, I needed there to be a Mars-analogue close to Toswao, so that at least some microbial life would make the journey very early in the system’s development. The dice gave me a Leftover Oligarch in the next inward orbit, so I was happy with that. That planet was probably relatively cool and moist in the first hundred million years or so after formation, but while Karjann has heated up over the eons, the small planet has been baked dry and is now more barren than Mars.

Second, I wanted to make sure there was no “late heavy bombardment” (LHB), since that event didn’t take place in the Bios: Genesis game. The best theory we have about the LHB, assuming it happened at all, is that our gas giant planets went through a period of orbital instability that also disrupted the Kuiper Belt. Here, the outermost gas giant is still well within the “slow-accretion line” that represents the nominal start of the system’s Kuiper Belt. Hence the belt has never been disrupted, and is probably much more full and dense than ours. An analogue might be the Tau Ceti system, whose Kuiper Belt appears to be at least ten times as dense as Sol’s.

So here we go, before I head off to work this morning. The Karjann system has a solo star and seven planets, including the Earthlike world Toswao. The inner planets are likely to be a super-hot Venus type, and a very hot, dry Mars-like. Beyond Toswao we have two cold worlds unlike anything in the Sol system, probably with lots of ice, the outermost likely to have a methane-ammonia atmosphere if the temperatures are right. Then two Jovians, which apparently gathered up all the mass that might otherwise have formed ice giants on the fringes of the system. Finally, a dense Kuiper Belt that probably indicates lots of comets. No asteroid belt, although there are likely to be extensive collections of junk in the Trojan points of the two gas giants.

Next time, I’ll focus on Toswao and its major natural satellite, and work out the details of its physical environment.

Architect of Worlds – Step Eight: Stellar Orbital Parameters

Architect of Worlds – Step Eight: Stellar Orbital Parameters

This is the last step in the design sequence for star systems – once the user has finished this step, she should know how many stars are in the system, what their current properties are, and how their orbital paths are arranged.

At this point, I’ve finished the current rewrite of the “Designing Star Systems” section of the book. I don’t plan to make any further mechanical changes to that section, except to correct any errors that might pop up. The instructions and other text might get revised again before the project is complete. A PDF of the current version of this section is now available at the Sharrukin’s Archive site under the Architect of Worlds project heading.


Step Eight: Stellar Orbital Parameters

This step determines the orbital parameters of components of a multiple star system. This step may be skipped if the star system is not multiple (i.e., the primary star is the only star in the system).

Procedure

The procedure for determining the orbital parameters of a multiple star system will vary, depending on the multiplicity of the system.

The important quantities for any stellar orbit are the minimum distance, average distance, and maximum distance between the two components, and the eccentricity of their orbital path. Distances will be measured in astronomical units (AU). Eccentricity is a number between 0 and 1, which acts as a measure of how far an orbital path deviates from a perfect circle. Eccentricity of 0 means that the orbital paths follow a perfect circle, while eccentricities increasing toward 1 indicate elliptical orbital paths that are increasingly long and narrow.

Binary Star Systems

To begin, select an average distance between the two stars of the binary pair.

To determine the average distance at random, roll 3d6 on the Stellar Separation Table.

Stellar Separation Table
Roll (3d6) Separation Base Distance
3 or less Extremely Close 0.015 AU
4-5 Very Close 0.15 AU
6-8 Close 1.5 AU
9-12 Moderate 15 AU
13-15 Wide 150 AU
16 or more Very Wide 1,500 AU

To determine the exact average distance, roll d% and treat the result as a number between 0 and 1. Multiply the Base Distance by 10 raised to the power of the d% result. The result will be the average distance of the pair in AU.

Feel free to adjust the result by up to 2% in either direction. You may wish to round the result off to three significant figures.

Next, select an eccentricity for the binary pair’s orbital path. Most binary stars have orbits with moderate eccentricity, averaging around 0.4 to 0.5, but cases with much larger or smaller eccentricities are known.

To determine an eccentricity at random, roll 3d6 on the Stellar Orbital Eccentricity Table. If the binary pair is at Extremely Close separation, modify the roll by -8. If at Very Close separation, modify the roll by -6. If at Close separation, modify the roll by -4. If at Moderate separation, modify the roll by -2. Feel free to adjust the eccentricity by up to 0.05 in either direction, although eccentricity cannot be less than 0.

Stellar Orbital Eccentricity Table
Roll (3d6) Eccentricity
3 or less 0
4 0.1
5-6 0.2
7-8 0.3
9-11 0.4
12-13 0.5
14-15 0.6
16 0.7
17 0.8
18 0.9

 

Once the average distance and eccentricity have been established, the minimum distance and maximum distance can be computed. Let R be the average distance between the two stars in AU, and let E be the eccentricity of their orbital path. Then:

R_{min}=R\times\left(1-E\right)

R_{max}=R\times\left(1+E\right)

Here, Rmin is the minimum distance between the two stars, and Rmax is the maximum distance.

Trinary Star Systems

Whichever arrangement is selected, design the closely bound pair first as if it were a binary star system (see above). This binary pair is unlikely to have Wide separation, and will almost never have Very Wide separation. Select an average distance for the pair accordingly. If selecting an average distance at random, modify the 3d6 roll by -3. Select an orbital eccentricity normally, and compute the minimum and maximum distance for the binary pair.

Once the binary pair has been designed, determine the orbital path for the pair (considered as a unit) and the single component of the star system. The minimum distance for the pair and single components must be at least three times the maximum distance for the binary pair, otherwise the configuration will not be stable over long periods of time.

If selecting an average distance for the pair and single component at random, use the Stellar Separation Table normally. If the result indicates a separation in the same category as the binary pair (or a lower one), then set the separation to the next higher category. For example, if the binary pair is at Close separation, and the random roll produces Extremely Close, Very Close, or Close separation for the pair and single component, then set the separation for the pair and single component at Moderate and proceed.

Select an orbital eccentricity for the pair and single component normally, then compute the minimum distance and maximum distance. If the minimum distance for the pair and single component is not at least three times the maximum distance for the binary pair, increase the average distance for the pair and single component to fit the restriction.

Quaternary Star Systems

As in a trinary star system, design the closely bound pairs first. Each binary pair is unlikely to have Wide separation, and will almost never have Very Wide or Distant separation. Select an average distance for each pair accordingly. If selecting an average distance at random, modify the 3d6 roll by -3. Select an orbital eccentricity, and compute the minimum distance and maximum distance, for each binary pair normally.

Once the binary pairs have been designed, determine the orbital path for the two pairs around each other. The minimum distance for the two pairs must be at least three times the maximum distance for either binary pair, otherwise the configuration will not be stable.

If selecting an average distance for the two pairs at random, use the Stellar Separation Table normally. If either result indicates a separation in the same category as either binary pair (or a lower one), then set the separation to the next higher category. For example, if the two binary pairs are at Close and Moderate separation, and the random roll produces Moderate or lower separation for the two pairs, then set the separation for the two pairs at Wide and proceed.

Select an orbital eccentricity for the two pairs normally, then compute the minimum distance and maximum distance. If the minimum distance for the two pairs is not at least three times the maximum distance for both binary pairs, increase the average distance for the two pairs to fit the restriction.

Stellar Orbital Periods

Each binary pair in a multiple star system will circle in its own orbital period. The pair and singleton of a trinary system will also orbit around each other with a specific period (probably much longer). Likewise, the two pairs of a quaternary system will orbit around each other with a specific period.

Let R be the average distance between two components of the system in AU, and let M be the total mass in solar masses of all stars in both components. Then:

P=\sqrt{\frac{R^3}{M}}

Here, P is the orbital period for the components, measured in years. Multiply by 365.26 to get the orbital period in days.

Special Case: Close Binary Pairs

Most binary pairs are detached binaries. In such cases, the two stars orbit at a great enough distance that they do not physically interact with each other, and evolve independently. However, if two stars orbit very closely, it’s possible for one of them to fill its Roche lobe, the region in which its own gravitation dominates. A star which is larger than its own Roche lobe will tend to lose mass to its partner, giving rise to a semi-detached binary. More extreme cases give rise to contact binaries, in which both stars have filled their Roche lobes and are freely exchanging mass in a common gaseous envelope.

This situation is only possible for two main-sequence stars that have Extremely Close separation, or in cases where a subgiant or red giant star has a companion at Very Close or Close separation. If a binary pair being considered does not fit these criteria, there is no need to apply the following test.

For each star in the pair, approximate the radius of its Roche lobe as follows. Let D be the minimum distance between the two stars in AU, let M be the mass of the star being checked in solar masses, and let be the mass of the other star in the pair. Then:

R=D\times(0.38+0.2\log_{10}{\frac{M}{M^\prime}})

Here, R is the radius of the star’s Roche lobe at the point of closest approach to its binary companion, measured in AU. Compare this to the radius of the star itself, as computed earlier. If the star is larger than its Roche lobe, then the pair is at least a semi-detached binary. If both stars in the pair are larger than their Roche lobes, then the pair is a contact binary.

The evolution of such close binary pairs is much more complicated than that of a singleton star or a detached binary. Mass will transfer from one star to the other, altering their orbital path and period, profoundly affecting the evolution of both. Predicting how such a pair will evolve goes well beyond the (relatively simple) models applied throughout this book. We suggest treating such binary pairs as simple astronomical curiosities, special cases on the galactic map that are extremely unlikely to give rise to native life or invite outside settlement. Fortunately, these cases are quite rare except among the very young, hot, massive stars found in OB associations.

One specific case that is of interest involves a semi-detached binary in which one star is a white dwarf. Hydrogen plasma will be stripped away from the other star’s outer layers, falling onto the surface of the white dwarf. Once enough hydrogen accumulates, fusion ignition takes place, triggering a massive explosion and ejecting much of the accumulated material into space. For a brief period, the white dwarf may shine with hundreds or even thousands of times the luminosity of the Sun. This is the famous phenomenon known as a nova.

Most novae are believed to be recurrent, flaring up again and again so long as the white dwarf continues to gather matter from its companion. However, for most novae the period of recurrence is very long – hundreds or thousands of years – so nova events from any given white dwarf in a close binary pair will be very rare. Astronomers estimate that a few dozen novae occur each year in our Galaxy as a whole.

Examples

Arcadia: Alice has already decided that the Arcadia star system has only the primary star, so she skips this step entirely.

Beta Nine: Bob knows that the Beta Nine system is a double star. Proceeding entirely at random, he rolls 3d6 on the Stellar Separation Table and gets a result of 7. The two components of the system are at Close separation. He takes a Base Distance of 1.5 AU and rolls d% for a result of 22. The average distance between the two stars in the system is:

1.5\times{10}^{0.22}\approx2.489

Bob rounds this off a bit and accepts an average distance of exactly 2.50 AU. He rolls 3d6 on the Stellar Orbital Eccentricity Table, subtracts 4 from the result since the stars are at Close separation, and gets a final total of 5. The orbital path of the two stars has a moderate eccentricity of 0.2. Bob computes that the minimum distance between the two stars will be 2.0 AU, and the maximum distance will be 3.0 AU.

Bob can now compute the orbital period of the two stars:

P=\sqrt{\frac{{2.50}^3}{(0.18+0.06)}}\approx8.07

The two stars in the Beta Nine system circle one another with a period of a little more than eight years.

The two components of the Beta Nine system form a binary pair, with a minimum separation of 2.0 AU. There is no possibility of the pair forming anything but a detached binary, so Bob does not bother to estimate the size of either component’s Roche lobe.

Modeling Notes

The paper by Duchêne and Kraus cited earlier describes the best available models for the distribution of separation in binary pairs. The period of a binary star appears to show a log-normal distribution with known mode and standard deviation. Generating a log-normal distribution with dice is a challenge without requiring exponentiation at some point, hence the unusual procedure for estimating separation used here.

Architect of Worlds – Step Seven: Stellar Classification

Architect of Worlds – Step Seven: Stellar Classification

Step Seven: Stellar Classification

This step determines the classification of each star in the system being generated, according to the Morgan-Keenan scheme most often used by astronomers. This classification scheme is not strictly necessary for the complete design of a star system, but it can provide useful flavor, and many science fiction readers and players will recognize it.

The classification for any given star is composed of two components, its spectral class and its luminosity class. Spectral class is strongly dependent on the star’s effective temperature. It uses the capital letters O, B, A, F, G, K, and M, in a sequence from the hottest (O-type) to the coolest (M-type) stars. Each letter class is also divided into ten sub-categories, numbered 0 through 9. Our own Sun, for example, is a G2-type star. Luminosity class is (for most stars) marked by a Roman numeral. Almost all stars fall into class III, class IV, or class V. Under this system, a star’s complete classification is given by spectral class, then luminosity class, with no spaces in between. Hence the Sun’s complete classification is G2V.

Brown dwarfs have also been assigned spectral classes, under an extension of the Morgan-Keenan system that uses the capital letters L, T, and Y for progressively cooler objects. These assignments are more tentative, since the study of brown dwarfs is relatively new and very cold examples are hard to observe. The spectral class Y is almost hypothetical at present, with only a few objects appearing to meet the definition.

To determine the spectral class of a star other than a white dwarf, locate the Temperature value on the Spectral Class Table closest to its effective temperature, and read across to the right to find its most likely spectral class.

Spectral Class Table
Temperature Class Temperature Class Temperature Class Temperature Class
9700 A0 5900 G0 3850 M0 1300 T0
9400 A1 5840 G1 3700 M1 1200 T1
9100 A2 5780 G2 3550 M2 1100 T2
8800 A3 5720 G3 3400 M3 1000 T3
8500 A4 5660 G4 3200 M4 950 T4
8200 A5 5600 G5 3000 M5 900 T5
8000 A6 5540 G6 2800 M6 850 T6
7800 A7 5480 G7 2650 M7 800 T7
7600 A8 5420 G8 2500 M8 750 T8
7400 A9 5360 G9 2400 M9 700 T9
7200 F0 5300 K0 2300 L0 600 or less Y0
7060 F1 5130 K1 2200 L1    
6920 F2 4960 K2 2100 L2    
6780 F3 4790 K3 2000 L3    
6640 F4 4620 K4 1900 L4    
6500 F5 4450 K5 1800 L5    
6380 F6 4330 K6 1700 L6    
6260 F7 4210 K7 1600 L7    
6140 F8 4090 K8 1500 L8    
6020 F9 3970 K9 1400 L9    

Main sequence stars have a luminosity class of V. Brown dwarfs also technically fall into luminosity class V, and we will record them as such. Subgiant stars have a luminosity class of IV, and red giant stars (whether on the red giant branch or the horizontal branch) have a luminosity class of III.

White dwarf stars are an exception to this system, with their own classification scheme. We will record all white dwarf stars as having the simple classification D.

Examples

Arcadia: Alice notes that the primary star of the Arcadia system has an effective temperature of 4950 kelvins. The closest value on the Spectral Class Table is 4960 kelvins, associated with a spectral class of K2. Since the star is on the main sequence, its complete classification is K2V.

Beta Nine: Bob notes that the two stars of the Beta Nine system have effective temperatures of 3200 K and 1420 K. It turns out that the two stars are a red dwarf of class M4V and a brown dwarf of class L9V.

Modeling Notes

A star’s spectral class depends on many features of its spectrum, and when it comes to the decimal subclasses, astronomers do not agree on their definitions. The same star is often given slightly different classification depending on the source. Once again, the system described here is a simplification designed for ease of use. The primary source was Mamajek’s compiled set of definitions, cited under Step Six.

Architect of Worlds – Step Six: Stellar Evolution

Architect of Worlds – Step Six: Stellar Evolution

This step is the core of the new design sequence for stars and stellar objects. Most world-design sequences found in tabletop games take no account of that fact that stars evolve and change over time – two stars of the same mass and composition can look very different if they are of different ages. Applying these changes to a randomly generated population of stars will provide increased plausibility, and reflect the greater variety of stars found in the real universe. This step is a trifle complex – it has four separate cases, and requires a bit more math.


Step Six: Stellar Evolution

At this point, we have determined the number of stars in the system, the initial mass of each, and the age and metallicity of the overall system. This step determines how the star system has evolved since its formation, and sets the current effective temperature, luminosity, and physical size of each star.

While on the main sequence, a star will evolve and change rather slowly, due to the “burning” of hydrogen fuel and the slow accumulation of helium “ashes” in the stellar interior. Stars normally grow brighter over time, changing in effective temperature and size as well. Stars which reach the end of their stable main-sequence lifespan go through a subgiant phase, then evolve more as a red giant, eventually losing much of their mass and settling down as a small stellar remnant called a white dwarf.

Procedure

This step should be performed for each star in a multiple star system.

Selecting for an Earthlike world: For there to be an inhabitable world in the star system, at least one star must be on the main sequence or a subgiant. If the mass and age of the primary star were selected to allow for an Earthlike world, that star is likely to fit this criterion as well.

To begin, for any object of 0.08 solar masses or more, refer to the Master Stellar Characteristics Table on the following two pages. Record the Base Effective Temperature (in kelvins) for each star. Record the Initial Luminosity (in solar units) and the Main Sequence Lifespan (in billions of years) as well.

If any star has a mass somewhere between two of the specific entries on the Master Stellar Characteristics Table, use linear interpolation to get plausible values for the star’s base effective temperature, initial luminosity, and main sequence lifespan.

Master Stellar Characteristics Table
Mass Base Effective Temperature Initial Luminosity Main Sequence Lifespan
0.08 2500 0.00047 6400
0.10 2710 0.00087 4200
0.12 2930 0.0016 2800
0.15 3090 0.0029 1900
0.18 3210 0.0044 1300
0.22 3370 0.0070 870
0.26 3480 0.010 630
0.30 3550 0.013 420
0.34 3600 0.017 270
0.38 3640 0.020 170
0.42 3680 0.025 150
0.46 3730 0.031 120
0.50 3780 0.038 110
0.53 3820 0.046 92
0.56 3870 0.054 78
0.59 3940 0.065 68
0.62 4020 0.079 59
0.65 4130 0.095 51
0.68 4270 0.12 43
0.70 4370 0.13 39
0.72 4490 0.15 35
0.74 4600 0.17 32
0.76 4720 0.20 29
0.78 4830 0.22 26
0.80 4940 0.25 24
0.82 5050 0.28 22
0.84 5160 0.31 20
0.86 5270 0.35 18
0.88 5360 0.39 16
0.90 5450 0.44 15
0.92 5530 0.48 14
0.94 5590 0.53 13
0.96 5670 0.59 12
0.98 5700 0.65 11
1.00 5760 0.70 10
1.02 5810 0.78 9.3
1.04 5860 0.85 8.6
1.07 5920 0.97 7.7
1.10 5990 1.10 6.9
1.13 6030 1.30 6.5
1.16 6080 1.50 6.1
1.19 6140 1.70 5.7
1.22 6190 1.90 5.2
1.25 6250 2.10 4.7
1.28 6300 2.40 4.4
1.31 6350 2.70 4.1
1.34 6410 3.00 3.9
1.37 6470 3.30 3.6
1.40 6540 3.70 3.3
1.44 6620 4.10 2.9
1.48 6720 4.70 2.7
1.53 6870 5.50 2.5
1.58 7030 6.30 2.4
1.64 7190 7.30 2.0
1.70 7390 8.60 1.9
1.76 7550 9.90 1.6
1.82 7740 11.00 1.5
1.90 7990 14.00 1.3
2.00 8300 17.00 1.1

Once you have determined the Base Effective Temperature, Initial Luminosity, and Main Sequence Lifespan for each star in the system being designed, examine the following four cases for each star:

  • The first case applies to any brown dwarf with mass less than 0.08 solar masses.
  • The second case applies to any star with mass between 0.08 and 2.00 solar masses, if the system’s age is less than the star’s Main Sequence Lifespan. Such a star is considered a main sequence star.
  • The third case applies to any star with mass between 0.08 and 2.00 solar masses, if the system’s age exceeds the star’s Main Sequence Lifespan by no more than 15%. Such a star will be a subgiant or red giant star.
  • The fourth case applies to any star with mass between 0.08 and 2.00 solar masses, if the system’s age exceeds the star’s Main Sequence Lifespan by more than 15%. Such a star will be a white dwarf.

Apply the guidelines under the appropriate case to determine the current effective temperature, luminosity, and radius for each star.

First Case: Brown Dwarfs

A “star” with less than 0.08 solar masses will be a brown dwarf. Such an object accumulates considerable heat during its process of formation. A very massive brown dwarf may also sustain nuclear reactions in its core (deuterium or lithium burning) for a brief period after its formation, giving rise to additional heat. This heat then escapes to space over billions of years, causing the brown dwarf to radiate infrared radiation, and possibly even a small amount of visible light.

A very young and massive brown dwarf may be hard to distinguish from a small red dwarf star. Older objects will fade through deep red and violet colors, eventually ceasing to radiate visible light at all. A “dark” brown dwarf will eventually resemble a massive gas-giant planet like Jupiter. Ironically, even a very massive brown dwarf will still have about the same physical size as Jupiter, making the resemblance even stronger.

To estimate the current effective temperature of a brown dwarf, let M be the object’s mass in solar masses, and let A be the object’s age in billions of years. Then:

T=18600\times\frac{M^{0.83}}{A^{0.32}}

Here, T is the brown dwarf’s current effective temperature in kelvins. Effective temperature for a brown dwarf can be no higher than 3000 K.

The radius of a brown dwarf will be about 70,000 kilometers, or about 0.00047 AU.

A brown dwarf’s luminosity will be negligible. To estimate its luminosity, let T be its current effective temperature in kelvins. Then:

L=\frac{T^4}{1.1\times{10}^{17}}

Here, L is the brown dwarf’s luminosity in solar units.

Second Case: Main Sequence Stars

Objects with 0.08 solar masses or more will be stars. For each star, first check to see if the star system’s age is less than or equal to the star’s Main Sequence Lifespan, as determined from the Master Stellar Characteristics Table. If so, then the star is still on the main sequence and this case applies.

A main sequence star will have an effective temperature reasonably close to the Base Effective Temperature from the table for a star of that mass. The exact effective temperature will depend on the star’s exact composition, and other factors that are beyond the scope of these guidelines. Feel free to select a current effective temperature within up to 5% of the value on the table for a star of that mass.

Low-mass stars grow hotter extremely slowly, and can be considered to have the same temperature as when they formed no matter how old they are. Select an effective temperature for them without any concern for their age.

Intermediate-mass and high-mass stars change in temperature more noticeably during their main-sequence lifespan. In general, intermediate-mass or high-mass stars will tend to begin life with an effective temperature as much as 3% or 4% below the value on the table, but will reach a peak of about 2% to 3% above that value by about two-thirds of the way through their main sequence lifespan. After that, they will tend to grow cooler again, falling back to the value on the table or even slightly lower by the end of their stable lifespan. Select an effective temperature for such stars accordingly.

Round effective temperature off to three significant figures.

Main sequence stars also grow brighter over time, as temperatures in their core rise and they are forced to radiate more heat. To estimate the current luminosity for a given main-sequence star, use:

L=L_0\times{2.2}^\frac{A}{S}

Here, L is the current luminosity for the star in solar units, L0 is the Initial Luminosity for the star from the Master Stellar Characteristics Table, A is the star system’s age in billions of years, and S is the star’s Main Sequence Lifespan from the table. Feel free to select a final value for the star’s luminosity that is within 5% of the computed value. Round luminosity off to three significant figures.

Note that very low-mass stars have main-sequence lifespans that are far longer than the current age of the universe. Such stars have simply not had enough time to grow significantly brighter since they first formed! You may choose to simply take the Initial Luminosity for such stars without modifying it with the above computation.

Once the effective temperature and luminosity of a star have been determined, its radius can be computed. If T is a star’s effective temperature in kelvins, and L is its luminosity in solar units, then:

R=155,000\times\frac{\sqrt L}{T^2}

The result R is the star’s radius in AU (multiply by 150 million to get the radius in kilometers). Its diameter will be exactly twice this value. Most main-sequence stars will have radii of a small fraction of one AU.

Third Case: Subgiant and Red Giant Stars

If a given star’s age is greater than its Main Sequence Lifespan, but exceeds that value by less than 15%, then it has evolved off the main sequence and is approaching the end of its life. Such a star first evolves through a subgiant phase, during which it loses little of its brightness but grows slowly larger and cooler. At some point its core becomes degenerate, shrinking and increasing dramatically in temperature. This sets off a new form of hydrogen fusion in a shell around the core, releasing considerably more energy and causing the star’s outer layers to balloon dramatically outward. The star becomes much brighter, cooler, and larger, becoming a red giant.

Stars in this stage of their development are often somewhat unstable, and the precise path a star will follow is highly dependent on its mass, composition, and other factors. Rather than attempting to trace the star’s evolution precisely through time, we suggest simply selecting one of the three options described below. To select an option at random, roll d% on the Post-Main Sequence Table.

Post-Main Sequence Table
Roll (d%) Stage
01-60 Subgiant
61-90 Red Giant Branch
91-00 Horizontal Branch

Subgiant stars: During this period, the star remains at about the same luminosity it had at the end of its main-sequence lifespan. Select a luminosity for the star between 2.0 and 2.4 times its Initial Luminosity from the Master Stellar Characteristics Table. The star will also cool to an effective temperature of about 5000 K. Select an effective temperature for the star somewhere between that value and the Base Effective Temperature from the table.

Red giant branch stars: At the end of the subgiant phase, a star is at the “foot” of a structure on the H-R diagram called the red giant branch. From this point, it will grow still cooler, but considerably brighter as well, swelling up to become many times its main-sequence size. The characteristics of stars at the “tip” of the red-giant branch are almost independent of the star’s mass. For stars of moderate metallicity, this implies a luminosity of about 2000 to 2500 solar units, and an effective temperature of about 3000 K.

To select an effective temperature and luminosity for a red giant branch star, select a value between 0 and 1, or roll d% to generate a random value between 0 and 1. If R is the selected value, T is the star’s current effective temperature, and L is its current luminosity, then:

T=5000-(R\times2000)

L={50}^{(1+R)}

Round effective temperature and luminosity to three significant figures, and feel free to select a value for each that is within 5% of the computed value.

Horizontal branch stars: Upon reaching the tip of the red giant branch, a star of moderate mass undergoes a phenomenon called helium flash. Temperatures and pressures at the star’s degenerate core have risen so high that the star can now fuse helium instead of hydrogen. A substantial portion of the star’s mass is “burned” in a few hours, releasing tremendous quantities of energy that (ironically) are almost invisible from a distance. Most of this titanic energy release is used up in lifting the star’s core out of its previously degenerate state, permitting the star to settle into a brief period of relatively stable helium burning. The star’s surface grows hotter, but it shrinks and reduces its luminosity considerably.

The temperature and luminosity of horizontal branch stars again tend to be almost independent of the star’s mass. Select a luminosity between 50 and 100 solar units, and an effective temperature of about 5000 K.

After spending a brief period on the horizontal branch, a star evolves though an asymptotic red giant phase, during which it ejects a substantial amount of its mass into space. This is the primary mechanism by which heavier elements are dispersed back into the interstellar medium, to contribute to the metallicity of later generations of stars. Asymptotic red giant stars are extremely rare, as they normally pass through that stage of their development in less than a million years. They should not be placed at random.

No matter which category the star falls into, its radius can be computed using the same formula as in the case for main sequence stars. Red giant stars are likely to be quite large, with radii of about 1 AU at their greatest extent.

Fourth Case: White Dwarf Stars

At the end of the asymptotic red giant stage, a star’s remaining core is exposed, giving rise to a stellar remnant called a white dwarf. A white dwarf is tiny, only a few thousand kilometers across, and so even if it remains extremely hot it radiates very little energy. The star’s active lifespan is now over – it no longer produces energy through nuclear fusion. Instead, the heat it retains from previous stages of its development will radiate slowly into space over billions of years.

If a given star’s age is greater than its Main Sequence Lifespan, and exceeds that value by 15% or more, then it has become a white dwarf star. As with main sequence stars, the properties of a white dwarf are strongly dependent on its mass and age.

A white dwarf star is only the remnant core of a main sequence star, which will have lost a significant amount of its mass during the transition. Let M0 be the mass of the original main sequence star, as generated in earlier steps. Then:

M=0.43+\frac{M_0}{10.4}

Here, M is the mass of the white dwarf remnant in solar masses. Feel free to select a value within 5% of the one computed. Replace the star’s mass, as generated in previous steps, with this result.

White dwarf stars are formed with very high effective temperatures, and then cool off over time as they radiate heat. To estimate the current effective temperature of a white dwarf star, let A be the age of the white dwarf (that is, the overall age of the system, minus 1.15 times the star’s Main Sequence Lifespan as taken from the Master Stellar Characteristics table). Let M be the mass of the white dwarf in solar masses, as computed above. Then:

T=13500\times\frac{M^{0.25}}{A^{0.35}}

Here, T is the white dwarf’s current effective temperature in kelvins.

The radius of a white dwarf star is almost completely determined by its mass. If M is the mass of the white dwarf, then:

R=\frac{5500}{\sqrt[3]{M}}

Here, R is the approximate radius of the white dwarf star in kilometers. White dwarf stars are extremely small, packing a star’s mass into a sphere no larger than the Earth!

A white dwarf star’s luminosity is usually negligible, although a young (and therefore very hot) white dwarf might have a significant fraction of the Sun’s brightness. To compute a white dwarf’s luminosity, let R be its radius in kilometers and T its effective temperature in kelvins. Then:

L=\frac{R^2T^4}{5.4\times{10}^{26}}

Here, L is the star’s luminosity in solar units.

Examples

Arcadia: Alice records the values from the Master Stellar Characteristics Table for the 0.82 solar-mass primary star in the Arcadia system. It has a Base Effective Temperature of 5050 K, an Initial Luminosity of 0.28 solar units, and a Main Sequence Lifespan of 22 billion years.

The primary is 5.6 billion years old, and so is still rather early in its lifespan as a main sequence star. Alice decides to select an effective temperature for the star about 2% below the value from the table, or 4950 K. To estimate the star’s current luminosity, she uses:

0.28\times{2.2}^\frac{5.6}{22}\approx0.34

She accepts this value for the star’s luminosity, about one-third that of our Sun. To compute the star’s radius, she uses:

155,000\times\frac{\sqrt{0.34}}{{4950}^2}\approx0.0037

This gives the star’s radius in AU, which equates to a radius of about 550,000 kilometers, or about 80% that of our Sun.

Beta Nine: Bob has already determined that the Beta Nine primary star is a low-mass star with 0.18 solar masses. Since low-mass stars evolve very slowly and the whole system is only 2.1 billion years old, Bob decides to take the values for effective temperature and luminosity straight from the Master Stellar Characteristics Table, modifying each of them by less than 5% to allow for a little variety. Bob then computes the radius for the primary star using the formula under the case for main-sequence stars. For the companion, Bob computes the effective temperature and then the luminosity using the formulae under the case for brown dwarfs, and notes the fixed radius. The results are as in the table.

Beta Nine Star System
Component Mass Effective Temperature Luminosity Radius
A 0.18 3200 K 0.0045 0.001 AU
B 0.06 1420 K 0.000037 0.00047 AU

Modeling Notes

The models set out here for various stellar classes are, of course, drastically simplified for the sake of ease of use. For brown dwarfs, useful data were derived from the Burrows and Freeman papers cited below. For white dwarfs, the Catalan paper and Ciardullo’s lecture notes were most useful. Main sequence stars are the easiest to characterize, since they are the most easily observed in large numbers, so there are plenty of detailed models in existence for their properties. The Mamajek data helped to produce the Master Stellar Characteristics Table, as did the EZ-Web application for stellar modeling posted by Townsend.

Burrows, A. et al. (2001). The Theory of Brown Dwarfs and Extrasolar Giant Planets. Reviews of Modern Physics, volume 73, pp. 719-766.

Catalan, S. et al. (2008). The Initial-Final Mass Relationship of White Dwarfs Revisited: Effect on the Luminosity Function and Mass Distribution. Monthly Notes of the Royal Astronomical Society, volume 387, pp. 1692-1706.

Ciardullo, R. White Dwarf Stars. Retrieved from http://personal.psu.edu/rbc3/A414/23_WhiteDwarfs.pdf (2018).

Freeman, R. et al. (2007). Line and Mean Opacities for Ultracool Dwarfs and Extrasolar Planets. The Astrophysical Journal Supplement Series, volume 174, pp. 504-513.

Mamajek, E. A Modern Mean Dwarf Stellar Color and Effective Temperature Sequence. Retrieved from http://www.pas.rochester.edu/~emamajek/EEM_dwarf_UBVIJHK_colors_Teff.txt (2016).

Townsend, R. EZ-Web (Computer software). Retrieved from http://www.astro.wisc.edu/~townsend (2016).

Architect of Worlds – Steps Four and Five: Star System Age and Metallicity

Architect of Worlds – Steps Four and Five: Star System Age and Metallicity

Here’s the next section, a little more math-intensive, but the models are still fairly simple and straightforward. In these steps of the design sequence, the user will generate the age of the star system under development, as well as its metallicity. This last is a measure of the prevalence of heavy elements in the star system’s formation, where “heavy elements” is defined the way astronomers do, as “anything past helium on the periodic table.”


Step Four: Star System Age

This step determines the age of the star system being generated. All the stars in the star system will be the same age, as measured from the moment that the primary star began to fuse hydrogen in its core.

The universe is currently estimated to be 13.8 billion years old. A few stars in our Milky Way Galaxy have been determined to be about the same age, and must have formed very soon after the beginning of the universe. The Galaxy itself is not much younger than that, growing through the accretion of gas and the assimilation of smaller galaxies across billions of years.

The oldest globular clusters appear to have formed about 12.6 billion years ago, and the galactic halo must date to about the same time. Since old halo stars often have orbital paths that carry them through the galactic plane, a few of them are always likely to be found in any given neighborhood of the galactic disk. The disk itself, and the first spiral arms, appear to have formed about 8.8 billion years ago. Most stars in any given neighborhood of the disk will be younger than that.

Procedure

Select an age for the star system being generated, no greater than 13.5 billion years.

To determine an age at random, begin by rolling d% on the Stellar Age Table. Take the unmodified d% roll when generating a region of space like that of our own neighborhood (close to the plane of the Galaxy and within one of the spiral arms, but not in an active star-formation region or inside an open cluster).

Stellar Age Table
Roll (d%) Population Base Age Age Range
01-05 Extreme Population I 0.0 0.5
06-31 Young Population I 0.5 2.5
32-82 Intermediate Population I 3.0 5.0
83-97 Disk Population 8.0 1.5
98-99 Intermediate Population II 9.5 2.5
00 or more Extreme Population II 12.0 1.5

 

Population I stars are relatively young stars which make up most of the galactic disk and the spiral arms. Population II stars are old, metal-poor stars that are normally found in the galactic bulge, the galactic halo, and the globular clusters.

To determine the star system’s exact age, roll d% again, treat the result as a number between 0 and 1, multiply that number by the Age Range, and add the result to the Base Age. The result will be the age in billions of years. You may wish to round the age to two significant figures.

Selecting for an Earthlike world: Instead of determining the age completely at random, assume the star system is in the Intermediate Population I. Stars in this range of ages are most likely to be metal-rich enough to have life-bearing planets, but are also old enough for complex life to have developed there.

Examples

Arcadia: Alice wishes to determine the age of the Arcadia star system. Since she wishes the system to have at least one Earthlike world, she does not roll on the Stellar Age Table, but assumes that the star system will be Intermediate Population I. She takes a Base Age of 2.5 billion years, an Age Range of 5.0 billion years, and rolls d% for a result of 62. The age of the Arcadia system is:

2.5+\left(0.62\ \times5.0\right)=5.6

Alice accepts this value for the age of the Arcadia system. The Arcadia system is apparently about a billion years older than Sol.

Beta Nine: Bob continues to work completely at random while generating the Beta Nine system. He rolls on the Stellar Age Table, getting a result of 20 on the d% roll. The Beta Nine system is in the Young Population I. He rolls another d% for a result of 82. The age of the Beta Nine system is:

0.5+\left(0.82\times2.0\right)=2.14

Bob rounds this off to two significant figures. The Beta Nine system is 2.1 billion years old, a relatively young star system, possibly not old enough to have developed complex life.

Modeling Notes

Most surveys of the solar neighborhood suggest that there are only a few Population II stars in our vicinity. For the model in this book, we assume that this proportion is about 3% of all stars in any given region of the galactic disk. As for the younger stellar populations, astronomers tend to assume that the star-formation rate in the Galaxy has been constant for the past 8-9 billion years, which suggests a “flat” distribution of stellar ages.

Step Five: Star System Metallicity

This step determines the metallicity of the star system being generated.

Most of the matter in the universe is composed of hydrogen and helium, both of which were created in the “Big Bang” at the beginning of time. Heavier chemical elements were almost entirely created by the processes of nuclear fusion in the heart of stars. As it happens, terrestrial planets like Earth, and living beings like us, are largely made up of these heavier elements.

Early in the universe’s history, the supply of such elements was very limited, so very old stars are unlikely to have terrestrial planets capable of supporting life. However, as billions of years passed, stars “baked” the heavier elements and then scattered them back into the interstellar medium. Stars like the Sun formed in interstellar clouds of gas and dust that had been already been enriched in these heavier elements. The presence and relative abundance of these elements is what is measured by metallicity.

We will simplify by assuming that all stars in a star system have the same metallicity. This is not always observed to be the case in multiple star systems, although it is rare for members of the same multiple system to have very different metallicities.

Very old stars can have metallicity as low as zero, composed almost entirely of hydrogen and helium with only tiny traces of heavier elements. A few young stars have been located with metallicity is high as 2.5 or 3.0, with several times as great an abundance of heavy elements as the Sun. The Sun itself seems to be rather metal-rich when compared to other stars of a similar age. In general, metallicity seems to be only weakly correlated with a star’s age – even old stars can turn out to be metal-rich.

Procedure

Select a value for the star system’s metallicity. To determine metallicity at random, apply the following formula, using a roll of 3d6:

M=\frac{3d6}{10}\times(1.2-\frac{A}{13.5})

Here, M is the metallicity value, and A is the age of the star system in billions of years. Modify the result with the following two cases.

  • If the star system is a member of Population II (and is therefore at least 9.5 billion years old), subtract 0.2 from the metallicity, with a minimum metallicity of 0.
  • To account for the occasional unusually metal-rich star, roll 1d6. On a 1, roll 3d6 again, multiply the result by 0.1, and add it to the metallicity value, to a maximum metallicity of 3.0. This step can be applied even to very old stars.

You may wish to round metallicity to two significant figures.

Selecting for an Earthlike world: A star likely to have terrestrial planets like Earth should have a metallicity of at least 0.3. If the star’s age was selected with an Earthlike world in mind in Step Four, it is very likely to have sufficient metallicity.

Examples

Arcadia: Alice wishes to determine the metallicity of the Arcadia star system. Since she has already selected the star system’s age to try to yield at least one Earthlike world, she decides to select the metallicity value at random and see what she gets. She rolls 3d6 for a result of 8, and computes:

\frac{8}{10}\times\left(1.2-\frac{5.6}{13.5}\right)\approx0.63

Rolling 1d6, she gets a value of 3, and leaves the metallicity value where it is. The Arcadia star system is rather metal-poor in comparison to our own, but it should have enough heavy elements to form terrestrial planets.

Beta Nine: Bob continues to work completely at random while generating the Beta Nine system. He rolls 3d6 for a result of 13, and computes:

\frac{13}{10}\times\left(1.2-\frac{2.1}{13.5}\right)\approx1.36

Bob rounds this result off to 1.4. He then rolls 1d6 and gets a 1, indicating that the Beta Nine system formed in an unusually metal-rich region of space. He rolls 3d6 for a result of 11, and adds 1.1 to the result, for a total metallicity of 2.5. He decides that the Beta Nine system might be a good location for a mining or industrial colony.

Modeling Notes

For this book, we assume that the average metallicity of newly formed stars has been rising at a constant rate since the formation of the Galaxy. Most studies have shown that metallicity can vary widely even for stars of similar age, indicating that the distribution of heavy elements in the Galaxy is often very uneven. Data from the following paper was used to produce a rough estimate for the age-metallicity relation for stars in the solar neighborhood:

Edvardsson, B. et al. (1993). The Chemical Evolution of the Galactic Disk – Part One – Analysis and Results. Astronomy and Astrophysics, volume 275, pp. 101-152.

Architect of Worlds – Steps Two and Three: Multiple Stars

Architect of Worlds – Steps Two and Three: Multiple Stars

The next section of the star-system design sequence follows. Here, we determine whether the star system is a single or multiple star, and in the case of a multiple star we determine how many stellar components are present.


Step Two: Stellar Multiplicity

This step determines how many stars exist in the star system being generated.

Our own sun is a single star, traveling through the galaxy with no other stars as gravitationally bound companions. Many stars do have such companions. Double stars, gravitationally bound pairs, are very common. Multiple stars, groups of three or more stars traveling together, are much less common but do occur. Most multiple stars are trinary stars, sets of three. Sets of four or more are possible – in fact, star systems with up to seven stellar components are known – but they are quite rare.

Multiple stars are almost always found arranged in a hierarchy of pairs. That is, the stars in a system can usually be divided into pairs of closely bound partners. Each pair circles around its own center of mass, and the pairs themselves follow (usually much wider) orbital paths around the center of mass of the entire system. Any odd star is usually bound with one of the pairs. This kind of arrangement is very stable over long periods of time.

Very young multiple star systems can form trapezia, in which three or more stars follow chaotic, closely spaced paths around the system’s center of mass. This arrangement is highly unstable, and is unlikely to last very long after the stars’ original formation. Some members of a trapezium will normally be ejected, to travel as singletons. The remaining stars soon settle down into a more stable hierarchy-of-pairs arrangement.

Procedure

To begin, select whether the system being generated is a multiple star. To determine this at random, roll 3d6 and refer to the Multiplicity Threshold Table.

Multiplicity Threshold Table
Primary star’s mass (in solar masses) is . . . Then the star is multiple on a 3d6 roll of . . .
Less than 0.08 14 or higher
At least 0.08, less than 0.70 13 or higher
At least 0.70, less than 1.00 12 or higher
At least 1.00, less than 1.30 11 or higher
At least 1.30 10 or higher

If the star system is multiple, roll d% on the Stellar Multiplicity Table. Star systems with five or more components are possible, but so rare that they should not be selected at random.

Stellar Multiplicity Table
Roll (d%) Number of Stars
01-75 2
76-95 3
96-00 4

Selecting for an Earthlike world: Earthlike planets can appear in single or multiple star systems, although the arrangement of components in a multiple star system (determined in the next step) will affect the presence of such worlds.

Examples

Arcadia: Alice determines the multiplicity of the Arcadia star system. She knows that multiple stars might still have Earthlike worlds, but decides not to take any chances. She determines that Arcadia’s primary star will be a singleton, and does not roll on the Stellar Multiplicity Table.

Beta Nine: Bob continues to work at random while generating the Beta Nine system. He rolls 3d6 to determine whether the Beta Nine system is multiple, and gets a result of 15. Even with the primary’s star’s low mass of 0.18 solar masses, this suggests that the system will, in fact, be a multiple star system. Bob rolls d% and refers to the Stellar Multiplicity Table. His result of 46 indicates that the Beta Nine system will be a double star system.

Modeling Notes

There has been considerable recent work on the frequency of multiple star systems, leading to the (rather surprising) discovery that most stars, especially low-mass stars, are not multiple. Two of the most useful sources for this are:

Lada, C. (2006). Stellar Multiplicity and the Interstellar Mass Function: Most Stars are Single. The Astrophysical Journal Letters, volume 640, pp. 63-66.

Duchêne, G. and A. Kraus (2013). Stellar multiplicity. Annual Review of Astronomy and Astrophysics, volume 51, pp. 269-310.

Step Three: Arrange Stellar Components

This step determines how the components of a multiple star system are arranged into a hierarchy of pairs, and the initial mass of each companion star in the system. This step may be skipped if the star system is not multiple (i.e., the primary star is the only star in the system).

Astronomers normally tag the various stellar components in a multiple star system with capital letters in the Latin alphabet: A, B, C, and so on. So, for example, the famous trinary star Alpha Centauri has three components: the bright yellow-white star Alpha Centauri A, its relatively close orange companion Alpha Centauri B, and a distant red dwarf companion Alpha Centauri C (also called Proxima Centauri, since it is noticeably closer to Sol than the A-B pair).

Unfortunately, astronomers are not always consistent about which component is given which alphabetic tag. In this book, we will always tag the primary star, the most star with the highest initial mass in the system, as the A-component. The other components will be tagged in order of their distance from the primary star.

 

Procedure

The procedure for arranging stars in a system varies, depending on the multiplicity of the system.

Stars other than the primary in a multiple star system are sometimes called companion stars. These stars can have any mass, from tiny brown dwarfs up to stars almost as massive as the primary, although there is a clear tendency toward the latter.

Binary Star Systems

There is only one possible arrangement for the two stars of a binary system. There are two components, A and B, and the primary star or A-component is in a gravitationally bound pair with the B-component.

Select the mass for the companion star. To generate its mass at random, roll d% on the Companion Star Mass Table to determine a mass ratio for the companion.

Companion Star Mass Table
Roll (d%) Mass Ratio
04 or less 0.05
05-08 0.10
09-12 0.15
13-16 0.20
17-20 0.25
21-24 0.30
25-28 0.35
31-32 0.40
35-36 0.45
37-40 0.50
41-45 0.55
46-50 0.60
51-55 0.65
56-60 0.70
61-65 0.75
66-71 0.80
72-78 0.85
79-87 0.90
88 or more 0.95

In each case, feel free to select a mass ratio that is just above the result of the table, increasing the ratio by less than 0.05. For example, if the result on the table indicates a mass ratio of 0.60, it would be appropriate to select an actual ratio greater than 0.60 but less than 0.65. The mass ratio cannot be lower than 0.05 or higher than 1.00.

In a binary star system, the companion star’s mass will be equal to the mass of the primary star, multiplied by the companion’s mass ratio. Round the companion’s mass off to the nearest hundredth of a solar mass unit. You may wish to round the companion’s mass off further, to match one of the entries in the Stellar Mass Table (see Step One). In no case will the mass of a companion star be less than 0.015 solar masses; round any such result up to that number.

Trinary Star Systems

There are two possible configurations for the three stars (components A, B, and C) of a trinary system.

One possibility is that the primary star (the A-component) has no close companion, but the B and C components move some distance away as a gravitationally bound pair of close companions (A and B-C).

The other is that the primary star and the B-component move as a bound pair of close companions, with the C-component moving alone at a greater distance (A-B and C).

Both arrangements appear to be about equally common. When designing a trinary star system, select either one. To select one at random, flip a coin.

In a trinary star system which is composed of a single A-component and a close B-C pair, the mass of the B component is computed using the Companion Star Mass Table, based on the mass of the primary star. The mass of the C-component is computed based on the mass of the B-component. When rolling on the Companion Star Mass Table, add 30 to the roll for the C component.

In a trinary star system which is composed of an A-B close pair and a C distant companion, the mass of each of the B and C components is computed using the Companion Star Mass Table, based on the mass of the primary star. When rolling on the table, add 30 to the roll for the B component.

Quaternary Star Systems

There are many possible arrangements for the four stars (components A, B, C, and D) of a quaternary system. However, by far the most common arrangement, and the most stable over long periods of time, is one in which two binary pairs (A-B and C-D) orbit one another at a wide separation.

In a quaternary star system, the mass of each of the B and C components is computed using the Companion Star Mass Table, based on the mass of the primary star. The mass of the D-component is computed based on the mass of the C-component. When rolling on the Companion Star Mass Table, add 30 to the roll for both the B component and the D component.

Examples

Arcadia: Alice skips this step for the Arcadia star system, since she already knows that the primary star is a singleton.

Beta Nine: Bob continues to work at random while generating the Beta Nine system. Since he has already established that the system is binary, he knows that there will be an A component (the primary star) and a B component (its companion). To determine the mass of the companion star, he rolls on the Companion Star Mass Table, and gets a result of 27 on the d%, for a mass ratio of 0.35. The mass of the companion star will be:

0.18\times0.35\approx0.06

In this case, rounding the companion star’s mass off to the nearest hundredth of a solar mass unit means that it will exactly match one of the entries for brown dwarfs on the Stellar Mass Table. Bob decides to accept this result as is. The companion “star” in the Beta Nine system will be a brown dwarf.

Modeling Notes

Studies have found that the ratio of mass between the components of a binary star appears to be evenly distributed, although there seems to be a statistically significant peak for ratios of 0.95 or higher in the data.

Mass ratios seem to be somewhat dependent on the orbital period. In particular, binary pairs which orbit one another at a short distance seem more likely to be close matches in mass. For simplicity’s sake, the model set out in this book largely ignores this effect, although in trinary and higher-multiplicity star systems we do assume that the close pairs are more likely to be matched. The paper by Duchêne and Kraus (cited under Step Two) discusses these statistical phenomena in some detail.

Architect of Worlds – Step One: Primary Star Mass

Architect of Worlds – Step One: Primary Star Mass

Here’s the first section of the world-design system laid out in the Architect of Worlds project.

As a preview: the design sequence begins by walking the user through the parameters of a star system, one or more stars in a gravitationally-bound group that move together through the Galaxy. We begin by determining the mass of the primary star, which we define here as either the only star in the group, or the one that begins its life with the greatest mass. In later steps we will determine whether there are any additional stars in the system, the mass of any companion stars, the age and metallicity of the overall system, the current status of each star, and finally the orbital parameters of the system.

In later sections of the design sequence, the user will be able to place planetary systems around a given star, and design the physical parameters of individual worlds.

Readers may be a little confused as to why we’re beginning by generating the primary star’s mass. Most design sequences like this one (including at least one previous version of Architect of Worlds) start by determining how many stars are in a given system, and then move on to generate the details of each one. It turns out that a star system’s multiplicity is strongly dependent on the primary star’s mass; more massive stars are significantly more likely to appear in pairs or larger groups. That dependence is complex enough to require we take things in this order if we want plausible results.

One more thing I’d like to point out (the final book will be explicit about this): what we’re generating here is the initial mass of a given star. It’s entirely possible that the object will end up with different mass than what we have here, specifically if we find that it has aged past its red-giant phase and is now a stellar remnant. That detail will be addressed in a later step of the sequence.


Step One: Primary Star Mass

This step determines the initial mass of the primary star in the star system being generated. We will measure the mass of stars in solar masses.

The lowest-mass objects to be generated here are brown dwarfs, substellar objects massive enough to have planetary systems of their own, but not massive enough to sustain hydrogen fusion. Brown dwarfs are not stars, but they are sometimes referred to as such, and for the purposes of setting design they can be treated that way. Brown dwarfs have masses between about 4,000 and 25,000 times that of Earth, or between about 0.15 and 0.08 solar masses.

At 0.08 solar masses and above, objects can sustain hydrogen fusion and are considered stars. Most stars, by far, form with between 0.08 and 2.0 solar masses.

Stars can be extremely massive, up to a theoretical maximum mass of about 150 solar masses, but such gigantic stars are quite rare. Very massive stars also tend to burn through their hydrogen fuel and die very quickly, which means that they rarely get the chance to move far from the open clusters or OB associations where they were formed. Most local neighborhoods of the galaxy will have no such massive stars.

Procedure

Select a mass for the primary star of the star system being generated. To determine a mass at random, begin by rolling d% on the Primary Star Category Table.

Primary Star Category Table
Roll (d%) Category
01-03 Brown Dwarf
04-82 Low-Mass Star
83-95 Intermediate-Mass Star
96-00 High-Mass Star

Depending on the category the primary star falls into, roll d% on the pertinent columns of the Stellar Mass Table on the next page. The result will be in solar mass units.

Feel free to select a mass for the star that is somewhere between two specific entries on the table. For example, if the result on the table indicates an intermediate-mass star of 0.92 solar masses, it would be appropriate to select an actual value greater than 0.92 but less than 0.94 solar masses. Such a selection will require you to do interpolation of several table entries in later steps.

Selecting for an Earthlike world: Instead of determining the primary star’s mass completely at random, assume it is an intermediate-mass star, and go directly to those columns on the table to determine its mass. Stars in this range are bright enough that they can have Earthlike worlds at a distance sufficient to avoid tide-locking, but are also long-lived enough that complex life is likely to have time to evolve.

Stellar Mass Table
Brown Dwarfs Low-Mass Stars Intermediate-Mass Stars High-Mass Stars
Roll (d%) Mass Roll (d%) Mass Roll (d%) Mass Roll (d%) Mass
01-10 0.015 01-13 0.08 01-07 0.70 01-06 1.28
11-29 0.02 14-23 0.10 08-13 0.72 07-12 1.31
30-45 0.03 24-34 0.12 14-19 0.74 13-18 1.34
46-60 0.04 35-43 0.15 20-24 0.76 19-23 1.37
61-74 0.05 44-52 0.18 25-29 0.78 24-30 1.40
75-87 0.06 53-59 0.22 30-34 0.80 31-36 1.44
88-00 0.07 60-65 0.26 35-39 0.82 37-43 1.48
    66-70 0.30 40-43 0.84 44-50 1.53
    71-74 0.34 44-47 0.86 51-58 1.58
    75-77 0.38 48-51 0.88 59-65 1.64
    78-80 0.42 52-55 0.90 66-71 1.70
    81-83 0.46 56-59 0.92 72-77 1.76
    84-86 0.50 60-62 0.94 78-84 1.82
    87-89 0.53 63-65 0.96 85-93 1.90
    90-92 0.56 66-68 0.98 94-00 2.00
    93-95 0.59 69-71 1.00    
    96-97 0.62 72-74 1.02    
    98-99 0.65 75-78 1.04    
    00 0.68 79-82 1.07    
        83-85 1.10    
        85-89 1.13    
        90-92 1.16    
        93-95 1.19    
        96-97 1.22    
        98-00 1.25    

Examples

Alice is aiming for a star system in which an Earthlike planet will appear, so she ignores the Primary Star Category Table and assumes the primary star will of intermediate mass. She rolls d% for a result of 36 and consults the appropriate columns on the Stellar Mass Table. The primary star’s mass is 0.82 solar masses.

Bob has no preconceived ideas about the nature of the Beta Nine system, and indeed he is designing a setting in which even small red dwarf or brown dwarf stars might be significant. He therefore rolls on the Primary Star Category Table and gets a result of 10 on the d%. The Beta Nine primary is a low-mass star. He rolls on the Stellar Mass Table, consulting the columns for low-mass stars, and gets a result of 48 on the d%. The primary star’s mass is 0.18 solar masses.

Modeling Notes

Astronomers have developed several different empirical rules for the distribution of stellar mass, each of which follows one or more power laws. In other words, the frequency of stars of a given mass seems to be proportional to that mass raised to a given power. The specific distribution we observe is called the initial mass function, and it appears to be consistent no matter where in the Galaxy we take a census of stars.

The Primary Star Category Table and Stellar Mass Table here are derived from an estimate for the initial mass function developed by the astronomer Pavel Kroupa. Citation:

Kroupa, P. (2001). On the variation of the initial mass function. Monthly Notices of the Royal Astronomical Society, volume 322, pp. 231–246.