This is the last chunk of the “design planetary systems” section of the Architect of Worlds draft. Before I post it, a small status report.

One of the things I’ve been doing over the past week is applying the current draft model to some real-world data – the HIPPARCOS catalog of the nearest stars to Sol. The next major project on my Gantt chart, after all, is to draw a map of nearby space and take a census of plausible habitable worlds in our neighborhood. This effort is helping me to test out the model, and it’s giving me data to suggest how to outline the next section (on designing specific planets).

So far, the results have actually been rather encouraging. I’ve been able to apply the model and the design sequence even to nearby stars for which we have strong candidate exoplanets. In those cases, I’ve been able to demonstrate not only that the model permits the known exoplanets, but that it also plausibly extrapolates what other planets may exist in the system, as yet undetected! That’s a good check on whether the model and design sequence are actually going to hold up against the real universe.

On the other hand, I have come across a few bits of data indicating that I need to make some revisions. In particular, the current design sequence doesn’t deal with red dwarf stars very well. I’m getting crowded arrays of Mercury- or Mars-sized rocky worlds, even in cases where what I should be seeing is icy “failed cores” out past the snow line. It’s a small logic problem, mostly in Step Eleven, which makes sense since that’s the most complex step in the sequence thus far. I think I already see how to fix it – but it will take a little bit of testing and rewriting.

I know a few of you have been experimenting with the draft sections that I’ve posted. I think what I will do, when I’ve made the necessary changes to the draft, is to revise those specific posts in place. When I’ve done that, I’ll also make a new status-report post here, with links to the altered posts. Of course, once I’m satisfied with this draft section for the time being, I’ll post a PDF to the Sharrukin’s Archive site.

Okay, with that out of the way, let’s have a look at the draft Step Fourteen.

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Step Fourteen: Place Natural Satellites

In this step, we will determine the number and placement of major satellites. We define a major satellite as a natural satellite which is large enough to have formed a sphere under its own gravitation. A major satellite will be at least 200 kilometers in radius if it is mostly made of ice, or at least 300 kilometers in radius if it is mostly stone.

In general, the rocky planets close to a star are unlikely to form major satellites during the process of planetary accretion. A terrestrial planet which suffers a specific kind of massive impact event may form one major satellite, as the planetary material scattered into orbit coalesces. Gas giant planets are likely to have several major satellites. For example, planets with major satellites in our own system include Earth (1), Jupiter (4), Saturn (7), Uranus (5), and Neptune (1).

Many planets will also have moonlets, much smaller satellites that are irregular in shape. Some moonlets may be remnants of the process of planetary formation. Others are likely to be captured asteroids or comets. Even very small objects may have moonlets of their own. In our system, the planet Mars has two moonlets, and many asteroids and Kuiper Belt objects have been found to have moonlets of their own. A gas giant planet will often have dozens of moonlets in a wide variety of orbits.

Procedure

To determine the number and arrangement of natural satellites for a given planet, begin by computing the planet’s Hill radius. This is the distance from the planet within which its gravitation dominates over that of the primary star. The Hill radius defines the region of space where a satellite is likely to form or be captured, and where it can maintain a stable orbit around the planet for long periods of time.

If R_min is the minimum distance from the planet to the primary star (in AUs), M_P is the mass of the planet (in Earth-masses), and M_S is the mass of the star (in solar masses), then:

Here, H is the Hill radius in kilometers. Round off to three significant figures.

First Case: Satellites Forming via Natural Accretion

To estimate how many major satellites will form with the planet, as part of the original accretion process, evaluate the following formula:

Here, H is the Hill radius in kilometers, and R is the average distance from the planet to its primary star in AU. Round N down to the nearest integer. If the result is greater than 0, then the planet will have one or more major satellites that formed with the planet itself. No planet is likely to have more than about 8 major satellites.

If N is greater than 0, feel free to adjust N upward or downward by up to 2, so long as the result is still greater than 0 and no greater than 8. To make this adjustment at random, roll 1d6: subtract 2 from N (minimum 1) on a result of 1, subtract 1 from N (minimum 1) on a result of 2, add 1 to N (maximum 8) on a result of 5, and add 2 to N (maximum 8) on a result of 6.

The innermost major satellite will have an orbital radius equal to about 1d+2 times the radius of the planet (feel free to adjust this by up to half the planet’s radius). Major satellites after the first can be placed using the same procedure as in Step Eleven, assuming tight orbital placement. The eccentricity of major satellite orbits will be very small (less than 0.01).

The total mass of all major satellites formed during planetary accretion will be about one ten-thousandth of the mass of the planet. The mass of each major satellite can be generated at random with a 3d6 roll:

Here, M_S is the mass of a major satellite in Earth-masses, M_P is the mass of the planet in Earth-masses, and N is the number of major satellites. Round the satellite’s mass off to two significant figures.

To determine the density of any of these major satellites, roll 3d6:

Here, D is the satellite’s density, and K is a constant that depends on whether the planet is inside or outside the system’s snow line. Inside the snow line, use K of 0.5. Outside the snow line, use K of 0.25. The radius and surface gravity of these major satellites can be determined as in Step Thirteen.

If a planet has one or more major satellites formed during planetary accretion, it will probably also have many moonlets. We will not generate these in detail, but they may be of interest in general.

Close to the planet will be a family of inner moonlets. These can range in number from a handful up to dozens. Their orbital radii usually range from about 1.8 times the planet’s radius, out to just inside the radius of the innermost major satellite. In some cases, inner moonlets can be interspersed between the orbits of the first few major satellites as well.

If there are inner moonlets, the planet is also likely to have a ring system. A few inner moonlets usually mean a thin system of rings, not easily visible from any distance. More inner moonlets imply a thicker and more visible set of rings. Ring systems usually hug close to the planet, reaching out to about twice the planet’s radius.

To generate the ring system at random, roll 3d6. On a result of 6-9, the planet will have a thin, wispy ring system comparable to that of Jupiter or Neptune, barely visible even at close range. On a result of 10-13, the ring system will be moderate, comparable to that of Uranus, visible from a distance through a telescope. On a result of 14 or higher, the planet will have many inner moonlets, supporting a dense ring system comparable to Saturn’s: easily visible from anywhere in the star system through a telescope, and spectacular at close range.

Beyond the major satellites will be one or more families of outer moonlets. As with the inner moonlets, these can range in number from a handful up to dozens. They are usually captured planetoids or other debris, following orbits that are eccentric, strongly inclined to the planet’s equator, or even retrograde. Their orbital radii usually begin at over 100 times the planet’s radius, and continue outward to about one-fifth to one-third of the planet’s Hill radius.

Second Case: Satellites Forming via Major Impact

Leftover Oligarchs and Terrestrial Planets, late in their formation process, are subject to repeated massive impact events. The planetary material scattered into orbit by such impacts is likely to form a large natural satellite. However, that satellite is in turn likely to spiral back into collision with the parent planet, or to migrate outward and escape the planet’s Hill radius entirely. In our own planetary system, only Earth still has a major satellite because of this process. While Mercury, Venus, and Mars all appear to have suffered similar massive impacts, none of them have retained any resulting major satellite.

To determine whether a Leftover Oligarch or Terrestrial Planet can have a large natural satellite, divide the planet’s Hill radius by its own radius. If the result is 300 or greater, then the planet can retain a large natural satellite over the long term. In this case, roll 1d: on a 5 or 6 the planet will have one large natural satellite.

If it exists, the major satellite will form quite close to the planet, but will quickly move outward due to tidal interactions. Its current orbital radius will usually be between 40 and 100 times the planet’s radius. To generate an orbital radius at random, roll 3d6+7 and multiply by 4 times the planet’s radius. The eccentricity of the major satellite’s orbit will be small (no more than about 0.05).

The major satellite formed by a massive impact event will have a mass about one hundredth of the mass of the planet. The mass of the major satellite can be generated at random with a 3d6 roll:

Here, M_S is the mass of the major satellite, and M_P is the mass of the planet, both in Earth-masses. Round the satellite’s mass off to two significant figures.

To determine the density of the major satellite, roll 3d6:

The radius and surface gravity of the major satellite can be determined as in Step Thirteen.

Third Case: Terrestrial Planet Moonlets

If a Leftover Oligarch or Terrestrial Planet has no major satellite, it may acquire one or more moonlets through a variety of means. For example, in our own planetary system, Mars has two moonlets.

A Leftover Oligarch or Terrestrial Planet may have moonlets if it can have a major satellite (that is, its Hill radius is at least 300 times the planet’s own radius) but it has no such satellite. In this case, roll 1d: on a 4-6 the planet will have at least one moonlet. Roll 1d-3 (minimum 1) for the number of moonlets.

The innermost moonlet will have an orbital radius equal to about 1d+2 times the radius of the planet (feel free to adjust this by up to half the planet’s radius). Moonlets after the first can be placed using the same procedure as in Step Eleven, assuming wide orbital placement. The eccentricity of these moonlets will be very small (less than 0.02).

Examples

Arcadia: Alice adds more columns to her table and computes the Hill radius for each planet. She then selects or randomly generates the number of major moons for each.

Radius	Planet Type	Planet Mass	Radius	Eccentricity	Hill Radius	Satellites
0.09 AU	Terrestrial Planet	0.88	6280	0.03	194000	None
0.17 AU	Terrestrial Planet	1.20	6680	0.10	377000	None
0.30 AU	Terrestrial Planet	0.95	6220	0.18	561000	None
0.57 AU	Terrestrial Planet	1.08	6450	0.05	1290000	None
0.88 AU	Terrestrial Planet	0.65	5670	0.02	1730000	1 moonlet
1.58 AU	Leftover Oligarch	0.10	3380	0.38	1050000	2 moonlets
2.61 AU	Planetoid Belt	N/A	N/A	0.00	N/A	N/A
4.40 AU	Large Gas Giant	480	83000	0.00	79900000	7 major satellites
5.76 AU	Medium Gas Giant	120	70000	0.00	65900000	4 major satellites
9.50 AU	Small Gas Giant	22	30000	0.08	56800000	2 major satellites

As it happens, none of the first four inner planets can have major satellites or moonlets. In particular, the habitable-planet candidate has too small a Hill radius to permit it. Alice rolls at random for the fifth and sixth planets, and gets the results she tabulated above. She decides not to bother generating these moonlets in detail, unless her story visits one of these two planets.

With very wide Hill radii and plenty of distance between them and the primary star, the three gas giants all have extensive systems of satellites. Alice uses the formula and a random die roll to determine how many major satellites each planet has. She doesn’t bother to set up their systems of moonlets, although she does roll at random to see whether either planet has a prominent ring system. She rolls a 10, an 11, and another 10, so all three gas giants have ring systems that are visible at a distance through telescopes, but none of them have a remarkable Saturn-like set of rings.

Beta Nine: Bob computes the Hill radii for his two planets. The inner planet has a Hill radius of about 863,000 kilometers, which is only about 160 times the planet’s radius. The outer planet has a Hill radius of about 1.42 million kilometers, which is better but still comes to only about 260 times the planet’s radius. Bob concludes that neither planet has any satellites.

Step Thirteen: Determine Planetary Density, Radius, and Surface Gravity

The physical size of a planet depends on not only its mass, but its physical composition. In this step, we will determine the density of each planet, which will immediately give us its radius and surface gravity.

Procedure

The density of a planet is a measure of its mass per unit volume. We will express planetary density in comparison with Earth. Thus, a planet with a density of 1.0 is exactly as dense as Earth. To convert to the usual units, multiply by 5.52 to get grams per cubic centimeter.

Leftover Oligarchs and Terrestrial Planets which form inside the snow line are made primarily out of silicate compounds (i.e., rocks). The heat of the accretion process tends to cause heavy metals, especially nickel and iron, to separate out and settle into a planetary core. The density of the fully formed planet will largely depend on the amount of nickel-iron available, and so on the size of this metallic core. To estimate the density of a Leftover Oligarch or Terrestrial Planet, roll 3d6 and apply the following:

Here, D is the planet’s density, and M is its mass in Earth-masses. Round the density off to two significant figures.

In some cases, a rocky Leftover Oligarch may have significantly higher density due to a massive impact event late in the process of planetary formation. The impact scatters most of the lower-density rocky material out into space, leaving behind a planetary body dominated by the nickel-iron core. In our own planetary system, Mercury appears to have undergone such a process. To determine whether a rocky Leftover Oligarch has unusually high density, roll 1d: on a 5 or 6 the planet will be dominated by its metallic core. Add 0.4 to the density computed above.

Failed Cores, and Leftover Oligarchs and Terrestrial Planets that form outside the snow line, incorporate a great deal of water and other ices. The density of the fully formed planet will be significantly lower than that of a rocky planet of the same mass. Determine the density of these planets by rolling 3d6:

Again, D is the planet’s density, and M is its mass in Earth-masses. Round the density off to two significant figures.

Gas Giants are constructed almost entirely out of hydrogen and helium gas. Although a gas giant will have a solid core of stone and ice, the factor dominating its density is the degree to which its gaseous envelope is compressed under gravity. To determine the density of a gas giant, let M be its mass. Then use the appropriate formula below:

Here, D is the density of the gas giant planet. Round the density off to two significant figures.

Radius

The radius of a planet (normally measured in kilometers) is dependent solely on its mass and density. If M is the planet’s mass (in Earth-masses), and D is the planet’s density, then:

Here, R is the planet’s radius in kilometers. Round off to three significant figures.

Surface Gravity

The surface gravity of a planet, measured in comparison to standard gravity at Earth’s surface, is again dependent solely on its mass and density. If M is the planet’s mass (in Earth-masses), and D is the planet’s density, then:

Here, G is the planet’s surface gravity. Round off to the nearest hundredth of a gravity. Note that under our model, a Gas Giant of 200 Earth-masses or less will always have a surface gravity of exactly 1.

Examples

Arcadia: The computations here are very straightforward. Alice adds more columns to her table and generates planetary densities, adjusting these to taste and then computing planetary radius and surface gravity for each planet.

Radius	Planet Type	Planet Mass	Density	Radius	Gravity
0.09 AU	Terrestrial Planet	0.88	0.92	6280	0.91
0.17 AU	Terrestrial Planet	1.20	1.04	6680	1.09
0.30 AU	Terrestrial Planet	0.95	1.02	6220	1.00
0.57 AU	Terrestrial Planet	1.08	1.04	6450	1.05
0.88 AU	Terrestrial Planet	0.65	0.92	5670	0.82
1.58 AU	Leftover Oligarch	0.10	0.67	3380	0.36
2.61 AU	Planetoid Belt	N/A	N/A	N/A	N/A
4.40 AU	Large Gas Giant	480	0.22	83000	2.85
5.76 AU	Medium Gas Giant	120	0.091	70000	1.00
9.50 AU	Small Gas Giant	22	0.21	30000	1.00

Beta Nine: Bob also has no difficulty with the necessary computations.

Radius	Planet Type	Planet Mass	Density	Radius	Gravity
0.27 AU	Terrestrial Planet	0.63	0.98	5500	0.85
0.45 AU	Terrestrial Planet	0.59	0.89	5550	0.78

Step Twelve: Determine Eccentricity of Planetary Orbits

The procedures in Step Eleven will generate a stack of planetary orbits which are likely to be stable, if all of them are perfectly circular. However, few planets follow such carefully arranged orbital paths. In this step, we will assign eccentricity values to the planetary orbits generated in Step Eleven, in such a way that the whole ensemble remains stable.

Starting with the innermost planet and working outward, select an eccentricity for the planet’s orbit. Planetoid Belts will have orbital eccentricity of 0. Planetary orbits tend to have low eccentricity, averaging around 0.1 to 0.2, but cases with eccentricity up to about 0.6 are known.

After each planet’s eccentricity has been determined, the eccentricity of the next planet’s orbit will be bounded. Let R₀ and E₀ be the orbital radius and eccentricity of the planet that has just been checked, and let R₁ and E₁ be the orbital radius and eccentricity of the next planet. Then:

If this inequality holds, then the two orbits will not cross at any point. Select eccentricity values with this requirement in mind, except for the last planetary orbit to be placed. If the last two planetary orbits are in resonance, the two orbits can cross if the outermost planet’s orbit is inclined at a significant angle. This will make no significant difference when determining the properties of that planet, but it may be of interest as a distinctive feature.

To determine an eccentricity at random, roll 3d6 on the Planetary Orbital Eccentricity Table. If the orbital spacing is tight, modify the roll by -4. If the orbital spacing is moderate, modify the roll by -2. Feel free to adjust the eccentricity by up to 0.05 in either direction. Check randomly generated eccentricity values using the inequality above, and increase or reduce eccentricity as needed.

Planetary Orbital Eccentricity Table
Roll (3d6)	Eccentricity
6 or less	0
7-9	0.1
10-12	0.2
13-14	0.3
15	0.4
16	0.5
17	0.6
18	0.7

Once the average distance and eccentricity have been established, the planet’s minimum distance and maximum distance from the primary star can be computed. As before, let R be the planet’s orbital radius in AU, and let E be the eccentricity of its orbital path. Then:

Here, R_min is the minimum distance, and R_max is the maximum distance.

If a planet’s minimum distance implies an approach to its primary star more closely than the inner edge of the protoplanetary disk, this is acceptable. If its maximum distance implies that the planet moves out into a forbidden zone at some point on its orbit, this is not a stable situation; reduce the planet’s eccentricity to ensure that this does not occur.

Selecting for an Earthlike world: Human-habitable worlds are not likely to have high orbital eccentricity, although a moderate value (no greater than 0.2) is probably not incompatible with Earthlike conditions.

Examples

Arcadia: Alice adds another column to her table and generates orbital eccentricities at random, adjusting each result to taste.

Radius	Planet Type	Planet Mass	Eccentricity
0.09 AU	Terrestrial Planet	0.88	0.03
0.17 AU	Terrestrial Planet	1.20	0.10
0.30 AU	Terrestrial Planet	0.95	0.18
0.57 AU	Terrestrial Planet	1.08	0.05
0.88 AU	Terrestrial Planet	0.65	0.02
1.58 AU	Leftover Oligarch	0.10	0.38
2.61 AU	Planetoid Belt	N/A	0.00
4.40 AU	Large Gas Giant	480	0.00
5.76 AU	Medium Gas Giant	120	0.00
9.50 AU	Small Gas Giant	22	0.08

Random generation yielded an eccentricity of 0.18 for the third planet, which concerned Alice, since the fourth planet was her Earthlike-world candidate. She therefore evaluated the inequality to determine how the fourth planet’s eccentricity was bounded:

Apparently, even a very large value for the fourth planet’s eccentricity would still be within bounds. Without making a random roll, she selected a small value of 0.05 (comparable to that of Earth) and proceeded.

Alice’s roll for the sixth planet (the Leftover Oligarch) was a modified 15, suggesting an eccentricity of close to 0.4. She reduced this to 0.38, checked the minimum and maximum distances for this orbit, and determined that the planet that moves from 0.98 AU out to 2.18 AU during its “year.” This did not appear to cross the orbit of the fifth planet, nor did it venture into the heart of the planetoid belt just outward. Alice decided to accept this result as a distinctive feature of the planetary system.

Beta Nine: Bob adds another column to his table. Random rolls give him an eccentricity of 0.0 each time, which he tweaks upward for variety. Both planets in the Beta Nine primary’s system have nearly circular orbits.

Radius	Planet Type	Planet Mass	Eccentricity
0.27 AU	Terrestrial Planet	0.63	0.03
0.45 AU	Terrestrial Planet	0.59	0.02

Step Eleven: Place Planets

Beginning close to the primary star and working outward, use the following procedure to determine the orbital radius, type, and mass for each planet in the system.

Procedure

At any given point in this process, the spacing of planetary orbits will be tight, moderate, or wide. Tight orbital spacing tends to occur when the protoplanetary disk was very dense, encouraging many planets to form and migrate inward together, until they fall into a stable but closely packed arrangement. Moderate orbital spacing normally occurs when the disk is less dense, or in the outer reaches of the disk. Wide orbital spacing occurs in very thin disks, or in regions of the disk that have been disturbed by the migration of a dominant gas giant.

Each planetary system will be governed by two orbital spacing regimes, one from the inner edge of the protoplanetary disk to the final location of the dominant gas giant, and another once the dominant gas giant has been placed. Select an orbital spacing regime when beginning the process of placing planets, then choose again immediately after the dominant gas giant has been placed.

To select an orbital spacing regime at random, roll 3d6 and apply the following modifiers:

• -3 if the disk mass factor is 6.0 or greater
• -2 if the disk mass factor is at least 3.0 but less than 6.0
• -1 if the disk mass factor is at least 1.5 but less than 3.0
• +1 if the disk mass factor is greater than 0.3 but no greater than 0.6
• +2 if the disk mass factor is greater than 0.15 but no greater than 0.3
• +3 if the disk mass factor is 0.15 or less
• +1 if there is a dominant gas giant which underwent weak inward migration
• +2 if there is a dominant gas giant which underwent moderate inward migration
• +3 if there is a dominant gas giant which underwent strong inward migration
• +3 if outward from a dominant gas giant that did not undergo a Grand Tack

The current orbital spacing regime will be tight on a final modified roll of 7 or less, moderate on a roll of 8-13, and wide on a roll of 14 or more.

Before beginning, make a note as to how many gas giant planets must appear in this planetary system. If a dominant gas giant was generated in Step Ten, then there must be at least one gas giant; if a Grand Tack event has taken place, there must be at least two.

Selecting for an Earthlike world: To maximize the probability of an Earthlike world, select orbital placements and planetary types so that a Terrestrial Planet will fall close to the critical radius , as discussed under Step Ten.

Sub-Step Eleven-A: Determine Orbital Radius

If no planets have already been placed, determine the first orbital radius as follows:

• If an epistellar gas giant was generated in Step Ten, then it is automatically the first planet to be placed, and its orbital radius has already been established.
• If there is no epistellar gas giant, and the spacing is currently tight, then the first orbital radius is equal to the disk inner edge radius.
• If there is no epistellar gas giant, and the spacing is currently moderate, then if M is the star’s mass, the first orbital radius will be:

• If there is no epistellar gas giant, and the spacing is currently wide, then if M is the star’s mass, the first orbital radius will be:

After placement of the first planet, each orbital radius will be based on the previous one. Roll 3d6, and subtract 2 if the previous orbital radius was resonant. The next orbit will be resonant if:

• The spacing is currently tight, and the 3d6 roll was 14 or less.
• The spacing is currently moderate, and the 3d6 roll was 10 or less.
• The spacing is currently wide, and the 3d6 roll was 6 or less.

Roll 3d6 on either the Stable Resonant or the Stable Non-Resonant Orbit Spacing Table, depending on whether the next orbital radius is resonant or not. In either case, multiply the previous orbital radius by the ratio from the table to determine the new orbital radius. Round each orbital radius off to the nearest hundredth of an AU.

Stable Resonant Orbit Spacing Table
Roll (3d6)	Ratio	Resonance
3-7	1.211	4:3
8-9	1.251	7:5
10-12	1.310	3:2
13	1.368	8:5
14	1.406	5:3
15	1.452	7:4
16-18	1.587	2:1 (see Note)

Note: Immediately after one 2:1 resonance appears, the next orbit will automatically be resonant as well, and will also exhibit a 2:1 resonance. After that, determine the spacing of further orbits normally. Single 2:1 resonances are normally unstable, but a stack of two or more such resonances can be very stable; this is a special case called a Laplace resonance.

Stable Non-Resonant Orbit Spacing Table
Roll (3d6)	Ratio
3	1.34
4	1.38
5	1.42
6	1.50
7	1.55
8	1.60
9-10	1.65
11-12	1.70
13	1.75
14	1.80
15	1.85
16	1.90
17	1.95
18	2.00

When rolling on the Stable Non-Resonant Orbit Spacing Table, feel free to select a ratio between two of the values on the table, but be careful not to match any of the precise ratio values from the Stable Resonant Orbit Spacing Table.

In any case, if there exists a dominant gas giant that has not yet been placed, and the new orbital radius is at least 0.7 times the orbital radius of the dominant gas giant as established in Step Ten, then skip to the dominant gas giant instead. Select or randomly generate a new orbital spacing scheme after this point.

If the new orbital radius is greater than the radius of the inner edge of a forbidden zone, then stop placing planets and move on to Step Twelve. Any remaining planetary mass budget is lost.

Sub-Step Eleven-B: Determine Planet Type

For each planet, roll on the Planet Type Table. Refer to the Inner Planetary System column for all planets before the dominant gas giant (if any), or the appropriate Outer Planetary System column for the dominant gas giant and all subsequent planets.

If this planet is the dominant gas giant, or a Grand Tack event took place and this planet is the first one after the dominant gas giant, then roll 2d6+8 on the table. Otherwise, roll 3d6.

If the maximum possible number of gas giants for this planetary system have already been placed, then any subsequent planets will be Terrestrial Planets (inside the snow line) or Failed Cores (outside the snow line).

Planet Type Table
Roll (3d6)	Inner Planetary System	Outer Planetary System (Inside Snow Line)	Outer Planetary System (Outside Snow Line)
3-7	Leftover Oligarch	Terrestrial Planet	Failed Core
8-11	Terrestrial Planet	Small Gas Giant	Small Gas Giant
12-14		Medium Gas Giant	Medium Gas Giant
15 or higher		Large Gas Giant	Large Gas Giant

Sub-Step Eleven-C: Determine Planet Mass

The mass M_P of a Leftover Oligarch can be generated randomly as:

The mass M_P of a Terrestrial Planet can be generated randomly as:

Here, M is the mass of the star in solar masses, K is the star’s metallicity, and D is the disk mass factor. Adjust this result by all the following which apply:

• If there is at least one gas giant in the system, the dominant gas giant underwent at least weak migration during Step Ten, and the current orbital radius is less than 0.7 times the orbital radius of the dominant gas giant after inward migration, then multiply the Terrestrial Planet’s mass by 0.75 for weak migration, 0.5 for moderate migration, and 0.25 for strong migration.
• If there is at least one gas giant in the system, the dominant gas giant underwent at least weak migration during Step Ten, and the current orbital radius is at least 0.7 times the orbital radius of the dominant gas giant after inward migration, but less than the current orbital radius of the dominant gas giant, then multiply the Terrestrial Planet’s mass by 0.1. Note that this case should only occur if a Grand Tack event took place in Step Ten.

The minimum mass for a Terrestrial Planet is 0.18 Earth-masses. If the estimated mass of a Terrestrial Planet is less than this:

• If there is at least one gas giant in the system, and the current orbital radius is at least 0.5 times the current orbital radius of the dominant gas giant, then the current orbit will automatically be filled by a Planetoid Belt rather than an actual planet.
• If there is a forbidden zone in the system, and the current orbital radius is at least 0.85 times the radius of the inner edge of the forbidden zone, then the current orbit will automatically be filled by a Planetoid Belt rather than an actual planet.
• Otherwise, treat the planet as a Leftover Oligarch instead, and re-roll its mass as above.

The mass M_P of a Failed Core can be generated randomly as:

The mass M_P of a Small Gas Giant can be generated randomly as:

The mass M_P of a Medium Gas Giant can be generated randomly as:

The mass M_P of a Large Gas Giant can be generated randomly as:

For the last three, M is the mass of the star in solar masses and D is the disk mass factor. If this is the dominant gas giant, then R is the planet’s original orbital radius before any migration or Grand Tack. Otherwise, R is the current orbital radius, or the slow-accretion radius, whichever is less.

In all cases, feel free to adjust the result upwards or downwards by up to one-half of the amount associated with one point on the dice. Round off the planet’s mass to the nearest hundredth of an Earth-mass for Leftover Oligarchs and Terrestrial Planets, and to two significant figures for all other types.

Sub-Step Eleven-D: Adjust Planetary Mass Budget

Mass Cost Table
Planet Type	Mass Cost
Planetoid Belt	0
Leftover Oligarch Terrestrial Planet Failed Core
Small Gas Giant
Medium Gas Giant
Large Gas Giant

Once the new planet’s type and mass have been determined, determine that planet’s mass cost using the appropriate formula from the Mass Cost Table. In these formulae, M_P is the mass of the planet. Round the mass cost for a given planet off to two significant figures.

Deduct the planet’s mass cost from the current planetary mass budget. “Spending” more than remains in the budget is allowed. However, if the planetary mass budget has now been exhausted, and the minimum number of gas giants has been placed, then stop placing planets and move on to Step Twelve. Otherwise, return to Sub-Step Eleven-A and continue to place planets.

Examples

Arcadia: Alice applies the looped procedure described above to place the planets of the Arcadia system. She has few preferences as to the placement of planets, other than a probably habitable world near an orbital radius of 0.58 AU. She decides to use moderate orbital spacing throughout. She has already determined that she has a planetary mass budget of 83.

Alice uses a combination of random rolls and minor adjustments to suit her taste, and builds a table of planets that looks something like the following:

Radius	Planet Type	Planet Mass	Mass Cost	Remaining Mass Budget
0.09 AU	Terrestrial Planet	0.88	0.88	82.12
0.17 AU	Terrestrial Planet	1.20	1.20	80.92
0.30 AU	Terrestrial Planet	0.95	0.95	79.97
0.57 AU	Terrestrial Planet	1.08	1.08	78.89
0.88 AU	Terrestrial Planet	0.65	0.65	78.24
1.58 AU	Leftover Oligarch	0.10	0.10	78.14
2.61 AU	Planetoid Belt	N/A	0.00	78.14
4.40 AU	Large Gas Giant	480	48.9	30.14
5.76 AU	Medium Gas Giant	120	24.0	6.14
9.50 AU	Small Gas Giant	22	19.8	-13.66

After the third planet, Alice noticed that she was approaching the critical orbital radius for the Earthlike world she wanted, so rather than roll a new orbital radius at random she simply selected a ratio of 1.90 and recorded the result. She also selected a Terrestrial Planet for that orbit, rather than rolling at random and possibly getting a Leftover Oligarch.

Recall that the dominant gas giant in the Arcadia system migrated inward to 1.7 AU before undergoing a Grand Tack, which means that the materials to build terrestrial planets are depleted from about 1.19 AU outward. For the sixth orbit, at 1.58 AU, Alice rolled a Terrestrial Planet whose mass turned out to be below the minimum of 0.18 Earth-masses. Since this orbit was not close enough to the dominant gas giant at 4.4 AU, she substituted a Leftover Oligarch instead. The same thing occurred for the orbit at 2.61 AU, but this orbital radius was greater than half that of the dominant gas giant, so that orbit acquired a Planetoid Belt instead.

The next orbital radius after the Planetoid Belt was very close to that of the dominant gas giant, so Alice skipped to that radius instead. Since the dominant gas giant went through a Grand Tack, the next planet outward was guaranteed to be another gas giant. The third gas giant exhausted the planetary mass budget, so Alice stopped generating planets at that point. Notice that even if the first gas giant had exhausted the mass budget, Alice would have been required to place the second, since there was a Grand Tack event. Also, notice that the first two gas giant planets are in a 3:2 resonance.

Alice concludes that the Arcadia planetary system somewhat resembles ours, with rocky planets close in, gas giant planets further out, and a planetoid belt in between. On the other hand, the system’s denser protoplanetary disk meant that the planets were more tightly packed, yielding more substantial rocky planets and fewer gas giants.

Beta Nine: Bob generates the Beta Nine planetary system entirely at random, curious to see what results he will get. He already knows that the system has a planetary mass budget of 5.1, and that a forbidden zone exists at 0.67 AU.

There is no gas giant in the system, and the disk mass is 0.5. Bob makes a modified 3d6 roll of 16, and determines that the planets will have wide orbital spacing. Random rolls generate the following:

Radius	Planet Type	Planet Mass	Mass Cost	Remaining Mass Budget
0.27 AU	Terrestrial Planet	0.63	0.53	4.57
0.45 AU	Terrestrial Planet	0.59	0.59	3.98

The next orbital radius turns out to be at 0.74 AU, which is beyond the edge of the forbidden zone, so no more planets will be placed, even though part of the planetary mass budget remains available.

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.

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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:

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:

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:

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:

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, R_F is the radius of the inner edge of the forbidden zone, and R_A is the radius of the slow-accretion line, then:

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:

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:

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

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.

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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:

Here, R_min is the minimum distance between the two stars, and R_max 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:

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 M´ be the mass of the other star in the pair. Then:

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:

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:

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.

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.

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.

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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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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

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).