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Architect of Worlds: Reality Ensues

Architect of Worlds: Reality Ensues

One element of work on the Architect of Worlds project is that real-world science in this area has been advancing by leaps and bounds for years. Not once but several times, I’ve settled on a method for modeling some feature of planetary formation, only to see some new result published that seems to demand a different model. On one occasion, while I was testing the current model by generating planetary systems for nearby stars, I literally saw the first publication of new exoplanets in a star system I had just done randomly a few days before. This can be frustrating, although I have to admit it’s also rather exciting!

Now it seems to have happened again. The current model I use for planetary system formation implements a combination of the Nice model and the Grand Tack hypothesis, which together describe how Jupiter may have formed and migrated through the early solar system before settling down in its current location. The assumption is that a planetary system’s primary gas giant will normally form near the “snow line” and then migrate inward (and possibly outward) to its final position. This is all implemented in Step 10 of the current sequence.

Now there seems to be evidence that these models aren’t telling the whole story. In a new paper, “The consequences of planetary migration on the minor bodies of the early Solar System,” computer simulation seems to suggest that the core of Jupiter must have formed much further out than the Nice model suggests. It may have formed as far out as four or five times the snow-line radius, and then migrated inward very quickly.

I need to read the Pirani paper in more detail, and see if there’s been any discussion as to how to reconcile those results with the Nice model. To be honest, the Nice model isn’t entirely consensus in the community, so a new model that fills in some of its problematic details might work.

Still, if the Pirani paper seems supportable, it may be necessary to do a significant rewrite of the second chunk of the design sequence. It’s possible that the result may actually be simpler than what I’ve got now. I’ve thought of a way to cut out some of the current sub-steps and computation, making the process a little smoother, that might work. We’ll see what develops.

Architect of Worlds – Step Fourteen: Place Natural Satellites

Architect of Worlds – Step Fourteen: Place Natural Satellites

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.


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 Rmin is the minimum distance from the planet to the primary star (in AUs), MP is the mass of the planet (in Earth-masses), and MS is the mass of the star (in solar masses), then:

H=2170000\times R_{min}\times\sqrt[3]{\frac{M_P}{M_S}}

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:

N=\frac{H^2}{(5\times{10}^{14})\times\sqrt R}

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:

M_S=\frac{(3d6)\times M_P}{100000\times N}

Here, MS is the mass of a major satellite in Earth-masses, MP 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:

D=K+\frac{(3d6)}{100}

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:

M_S=\frac{(3d6)\times M_P}{1000}

Here, MS is the mass of the major satellite, and MP 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:

D=0.5+\frac{(3d6)}{100}

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.

Architect of Worlds – Step Thirteen: Determine Planetary Density, Radius, and Surface Gravity

Architect of Worlds – Step Thirteen: Determine Planetary Density, Radius, and Surface Gravity

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:

D=(0.90+\frac{\left(3d6\right)}{100})\times\sqrt[5]{M}

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:

D=(0.50+\frac{\left(3d6\right)}{100})\times\sqrt[5]{M}

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:

D=\frac{1}{\sqrt M}\ \left(M\le200\right)

D=\frac{M^{1.27}}{11800}\ (M>200)

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:

R=6370\times\sqrt[3]{\frac{M}{D}}

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:

G=\sqrt[3]{MD^2}

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
Architect of Worlds – Step Twelve: Determine Eccentricity of Planetary Orbits

Architect of Worlds – Step Twelve: Determine Eccentricity of Planetary Orbits

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:

\left(1+E_0\right)\frac{R_0}{R_1}-1<E_1<\left(-1+E_0\right)\frac{R_0}{R_1}+1

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:

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

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

Here, Rmin is the minimum distance, and Rmax 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:

\left(1+0.18\right)\frac{0.30}{0.57}-1<E_1<\left(-1+0.18\right)\frac{0.30}{0.57}+1

-0.38<E_1<0.57

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
Architect of Worlds – Step Eleven: Place Planets

Architect of Worlds – Step Eleven: Place Planets

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 R=\sqrt L , 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:

R=(2d6)\times0.01\times\sqrt[3]{M}

  • 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:

R=(2d6)\times0.04\times\sqrt[3]{M}

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 MP of a Leftover Oligarch can be generated randomly as:

M_P=(3d6)\times0.01

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

M_P=(3d6)\times0.2\times M\times K\times D

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 MP of a Failed Core can be generated randomly as:

M_P=(3d6)\times0.25

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

M_P=4+\left(3d6\right)\times0.25\times M\times D\times\sqrt R

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

M_P=4+\left(3d6\right)\times3\times M\times D\times\sqrt R

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

M_P=4+\left(3d6\right)\times15\times M\times D\times\sqrt R

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

M_P
Small Gas Giant 0.9\times M_P
Medium Gas Giant 0.2\times M_P
Large Gas Giant 0.1\times M_P

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, MP 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.

Architect of Worlds – Step Ten: Place Dominant Gas Giant

Architect of Worlds – Step Ten: Place Dominant Gas Giant

Step Ten: Place Dominant Gas Giant

The evolution of the protoplanetary disk, and the formation of planets, will be dominated by the presence of the first gas giant planet to form. This planet may or not be the most massive, but it is usually the gas giant planet that forms closest to the primary star, and its movement through the disk will tend to affect the formation of other planets. In this step, we determine where the dominant gas giant planet (if any) forms, and how it migrates across the protoplanetary disk. This will, in turn, tell us how many gas giants may form.

Procedure

Begin by checking whether the dominant giant forms in a “hot” or “cold” region of the protoplanetary disk, or whether a dominant gas giant will form at all. Then determine how many gas giants can form in the planetary system, and how the dominant gas giant migrates to its final position.

First Case: Hot Dominant Gas Giant

If the protoplanetary disk has very high density of dust, or if the primary star is very bright and so has a distant snow line, the dominant gas giant may form in the warm, dry region inside the snow line. To check for this possibility, compute the following. If M is the mass of the star in solar masses, K is the system’s metallicity, and D is the disk mass factor:

R=\frac{16}{{(M\times K\times D)}^2}

R is measured in AU. If R is less than the disk inner radius, set R to be the disk inner radius instead. The result is the radius at which the dominant gas giant will form, based solely on the accretion of stony planetesimals. This will occur only if:

  • R is less than the snow line radius;
  • R is less than the slow-accretion radius; and
  • R is less than the radius of any forbidden zone.

If all three conditions hold, make a note that the dominant gas giant forms at this radius. Otherwise, check the second case.

Second Case: Cold Dominant Gas Giant

In most cases, the dominant gas giant will form outside the snow line, where ice as well as dust is available for the formation of planetesimals. To check for this possibility, compute the following. If M is the mass of the star in solar masses, K is the system’s metallicity, and D is the disk mass factor:

R=\frac{1}{{(M\times K\times D)}^2}

For a quick check, if the radius for the first case above was computed, simply divide it by 16 to get the radius for this case. If R is less than the radius of the snow line, set R to be the radius of the snow line instead. The result is the radius at which the dominant gas giant will form, based on the accretion of icy planetesimals. This will occur only if:

  • R is less than the slow-accretion radius; and
  • R is less than the radius of any forbidden zone.

If both conditions hold, make a note that the dominant gas giant forms at this radius. Otherwise, no gas giant will form in this planetary system; skip ahead to Step Eleven.

Number of Possible Gas Giants

Depending on the size and mass of the protoplanetary disk, more gas giants may form at larger orbital radii. To determine the further evolution of the planetary system, we need to estimate how many gas giants are possible.

To make this estimate, compute the following. Let R be the radius at which the dominant gas giant forms, as determined in the two cases above. Let Rmax be the slow-accretion radius or the radius of any forbidden zone, whichever is less. Then:

N=1+(6\times\log_{10}{\frac{R_{max}}{R}})

Round N down to the nearest integer. The result is the estimated number of possible gas giants in the planetary system. Note that N must be at least 1, otherwise no gas giant can form in the system and we should already have skipped ahead to Step Eleven.

Disk Migration

Once the dominant gas giant begins to form, it is likely to migrate through the protoplanetary disk. At first, it will migrate inward due to interactions with the gas of the disk. As it migrates inward, its gravity will disrupt the orbits of any planetesimals it approaches or passes, affecting the later evolution of inner planets. To estimate the extent of the dominant gas giant’s inward migration, roll 3d6 on the Planetary Migration Table. Modify this roll by -3 if the disk mass factor is 4 or greater, or by +3 if the disk mass factor is less than 1.

Planetary Migration Table
Roll (3d6) Status After Inward Migration
3-6 Epistellar Gas Giant – Dominant gas giant migrates inward to the disk inner edge radius.
7-9 Strong migration – Dominant gas giant migrates inward to about 0.25 times its initial orbital radius, or to the disk inner edge radius, whichever is greater.
10-12 Moderate migration – Dominant gas giant migrates inward to about 0.5 times its initial orbital radius, or to the disk inner edge radius, whichever is greater.
13-15 Weak migration – Dominant gas giant migrates inward to 0.75 times its initial orbital radius, or to the disk inner edge radius, whichever is greater.
16-18 No migration – Dominant gas giant fails to migrate inward at all.

The table indicates how to estimate the orbital radius to which the dominant gas giant migrates during this phase of its evolution. For the strong migration, moderate migration, or weak migration cases, feel free to adjust the multiplying factor by up to 0.1 in either direction. Make a note of the resulting orbital radius.

The Grand Tack

At some point in its formation, the dominant gas giant may fall into a strong resonance interaction with one or more gas giants forming further away from the primary star. This is likely to halt inward migration through the protoplanetary disk, and may lead to outward migration back away from the star.

A Grand Tack will take place only if there are at least two possible gas giants in the planetary system, as estimated above. If this is the case, roll 3d6. A Grand Tack takes place if the result is 13 or higher.

If a Grand Tack takes place, estimate the final orbital radius of the dominant gas giant by rolling 3d6 and applying the following:

R=(1+\frac{3d6}{10})\times R_M

Here, RM is the planet’s orbital radius after any inward migration is applied, and R is its orbital radius after the Grand Tack is finished. If R is greater than half the radius of any forbidden zone, then set R to that value. Otherwise, feel free to adjust the final orbital radius by up to 5% in either direction.

Selecting for an Earthlike world: The critical orbital radius for an Earthlike world depends on the current (rather than initial) luminosity of its primary star. To estimate this orbital radius, if L is the star’s current luminosity, then:

R=\sqrt L

Here, R is the most likely orbital radius for an Earthlike world. A terrestrial planet with enough mass to support an Earthlike environment is only likely to form in one of three cases:

  • The dominant gas giant migrated inward completely past this radius, and no Grand Tack event took place to pull it back outward;
  • The dominant gas giant migrated inward but entered a Grand Tack event before reaching about 1.5 times this radius; or
  • The dominant gas giant did not migrate inward at all.

Of these three cases, the second (moderate inward migration followed by a Grand Tack) is the most likely to give rise to an Earthlike world.

Examples

Arcadia: Alice suspects that the primary star of the Arcadia system is too dim to promote the formation of a hot dominant gas giant, but she checks anyway. The star’s mass is 0.82 solar masses, its metallicity is 0.44, and the disk mass factor she selected earlier is 2. She computes:

\frac{16}{{(0.82\times0.44\times2.0)}^2}\approx30.7

This is, as Alice expected, well past the slow-accretion line. Moving on to the case of a cold dominant gas giant, she divides the above result by 16 and gets a radius of about 1.9 AU. This is inside the snow line, so Alice resets the radius to be equal to the snow line at 2.2 AU; this is where the dominant gas giant will form.

Alice now needs to estimate how many gas giants could form in the Arcadia planetary system. She computes:

1+\left(6\times\log_{10}{\frac{14}{2.2}}\right)\approx5.8

Rounding down to the nearest integer, she finds that the Arcadia system could have as many as five gas giants in it. Depending on subsequent results, it may have fewer than this number, but it cannot have more.

Rather than generate the dominant gas giant’s evolution at random, Alice wants to maximize the probability of an Earthlike planet forming at the critical orbital radius, which she computes from the primary star’s current luminosity of 0.34 solar units:

\sqrt{0.34}\approx0.58

She therefore decides that the Arcadia system’s primary gas giant exhibited weak inward migration, moving from its initial orbital radius of 2.2 AU to about 1.7 AU, inside the snow line but nowhere near the eventual Earthlike world’s position. Then she decides that the planet underwent a Grand Tack event, plausible since there are at least two possible gas giants in the system. She decides that the dominant gas giant migrated back outward to an orbital radius of 4.4 AU. She makes note of all three radii for future reference.

Beta Nine: Bob knows that a red dwarf star will almost certainly not develop a hot gas giant, so he moves directly to the second case, computing the radius at which a cold gas giant will form:

\frac{1}{{(0.18\times2.5\times0.5)}^2}\approx19.8

This is far beyond the inner edge of the forbidden zone created by the star’s brown-dwarf companion. The primary star of the Beta Nine system will not have any gas giant planets at all. Bob moves on to the next step in the design sequence.

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 (16 May 2018)

Status Report (16 May 2018)

A short note, since it’s been a few days since I last posted anything here. I’ve been up to my eyebrows at the day job, teaching a course on risk management and cybersecurity. After a full day on the platform I’m rarely in good condition to get a lot of creative work done in the evening. Still, my brain has been percolating along on the Curse of Steel project.

I’m currently beginning work on some maps, to give the story some structure. The overall plot of the novel is very much in the “heroic quest” vein, with Kráva and a few companions going on a long journey across unexplored and dangerous countryside to reach an objective. So I need to at least sketch out the geography.

This, as usually happens with me, turns out to be more complicated than it might appear at first glance. Knowing too much about world-building often means you can’t be satisfied with the simple or naïve approach to any problem.

In this case, my brain got stuck on the question of how to draw regional and world maps on a sphere. I keep thinking back to the classic Baynes-Tolkien poster map of Middle-earth, which has been the inspiration for a hundred thousand fantasy-world maps since then. It’s a beautiful map, but the big unspoken problem with it is that it’s flat. The map legend indicates both constant directions and a constant distance scale, and that just cannot be done with any flat projection of a spherical surface. That’s a subtle flaw in the world-building for Middle-earth, especially if (as I suspect) Tolkien did his meticulous measurements of distance and travel times on a similarly flat map.

So, since this piece at least of my world-building is decidedly in the same mold, I want to draw a similar map – but I want to envision my world as a sphere and do my regional map-making on that basis. Which means I need to expand my cartographic tool set.

I usually do map-building with Photoshop, but it’s a challenge to draw on a sphere with that tool, and there’s no way to easily do the standard map projections. However, I’ve recently come across one of the superb world-building YouTube videos produced by Artifexian, in which he discusses a work-flow he’s developed to do just this kind of thing. Here’s a link to the specific video I’m talking about.

So I’ve gotten started on this piece of the project by downloading a couple of freeware tools (GPlates and G.Projector), and will be sketching out global and regional maps over the next few days. I’ll post some of the interim results here.

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.