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

July 25, 2018 Sharrukin Comments 0 Comment

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

July 22, 2018 Sharrukin Comments 2 comments

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, 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:

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

$-0.38<E_1<0.57$

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

July 21, 2018 Sharrukin Comments 6 comments

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 M_P of a Leftover Oligarch can be generated randomly as:

$M_P=(3d6)\times0.01$

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

$M_P=(3d6)\times0.25$

The mass M_P 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 M_P 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 M_P 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, 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.

Architect of Worlds – Step Ten: Place Dominant Gas Giant

July 20, 2018 Sharrukin Comments 0 Comment

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 R_max 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, R_M 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

July 18, 2018 Sharrukin Comments 0 Comment

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, 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:

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

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

Examples

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

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

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

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

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

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

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

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

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

Status Report (13 July 2018)

July 14, 2018 Sharrukin Comments 0 Comment

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

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

So, new plan of action:

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

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

Building Toswao

June 28, 2018 Sharrukin Comments 0 Comment

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

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

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

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

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

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

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

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

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

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

Toswao

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

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

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

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

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

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

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

Toswao’s Moon

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

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

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

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

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

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

Building the Karjann System

June 27, 2018 Sharrukin Comments 0 Comment

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

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

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

Primary Star

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

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

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

Karjann

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

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

Planetary System

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Architect of Worlds – Step Eight: Stellar Orbital Parameters

April 22, 2018 Sharrukin Comments 3 comments

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, 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:

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

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 M´ 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

April 19, 2018 Sharrukin Comments 0 Comment

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.