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Architect of Worlds – Step Twenty-Three: Determine Atmospheric Mass and Pressure

Architect of Worlds – Step Twenty-Three: Determine Atmospheric Mass and Pressure

In this step, we determine the atmospheric mass of the world under development. The atmospheric mass is measured relative to that of Earth – a world with surface gravity of exactly 1, and atmospheric pressure of exactly 1 “atmosphere” at the surface, will have an atmospheric mass of 1.

Some worlds will have a Trace atmosphere – enough to provide climate and weather effects on the world’s surface, but not enough to support any form of complex life. Still other worlds will have no atmosphere at all (or at least no atmosphere that can be detected without sensitive instruments). In both of these cases, the atmospheric mass will effectively be zero.

Atmospheric mass depends on a large number of factors: the world’s blackbody temperature and M-number (determined in Step Nineteen), its prevalence of water (determined in Step Twenty), whether it has undergone a runaway greenhouse event (also determined in Step Twenty), the degree of ongoing vulcanism (determined in Step Twenty-One), and the presence and strength of a magnetic field (determined in Step Twenty-Two).

Once a world’s atmospheric mass has been fixed, the pressure of the atmosphere at the surface (sea level or some other convenient “datum”) can also be determined.

Procedure

Begin by building a list of the likely major components of the world’s atmosphere. Refer to the following table, which lists a number of volatile compounds which might make up a large and stable portion of an atmosphere.

Atmospheric Components Table
Possible Major ComponentMaximum M-NumberMinimum Blackbody Temperature
Molecular Hydrogen (H2)220 K
Helium (He)45 K
Molecular Nitrogen (N2)2880 K
Carbon Dioxide (CO2)44195 K

For each item on the Atmospheric Components Table, check to see whether the world under development has an M-number that is no higher than the one given in the table, and a blackbody temperature that is no lower than the one given in the table. If the M-number is too high, that potential component of the atmosphere will undergo thermal escape. If the blackbody temperature is too low, that component will tend to “freeze out” and form liquid or solid layers on the surface. Either way, that volatile will not be available to make up a substantial atmosphere.

Make a list of the atmospheric components that meet both conditions, and then refer to the following three cases.

First Case

This case holds if one or more of molecular hydrogen, helium, or molecular nitrogen meet both conditions from the table.

In this case, roll 3d6 and modify the result as follows:

  • +6 if the world has Massive prevalence of water
  • +6 if the world has undergone a runaway greenhouse event
  • +6 if the world has a Molten Lithosphere
  • +4 if the world has a Soft Lithosphere
  • +2 if the world has an Early Plate Lithosphere
  • -2 if the world has an Ancient Plate Lithosphere
  • -4 if the world has a Solid Lithosphere
  • -2 if the world has a Moderate Magnetic Field
  • -4 if the world has a Weak Magnetic Field
  • -6 if the world has no Magnetic Field

If the modified dice roll is 0 or less, then the world will have a Trace atmosphere, with an atmospheric mass of zero. Otherwise, multiply the modified dice roll by:

  • 10 if the world has undergone a runaway greenhouse event
  • 1 if the world has blackbody temperature less than 125 K and Massive prevalence of water
  • 0.1 otherwise

The final result is the world’s atmospheric mass. Feel free to adjust this result by up to half of the multiplier.

Second Case

This case holds if the first case does not, but carbon dioxide meets both conditions from the table.

In this case, the world will automatically have a Trace atmosphere, with an atmospheric mass of zero.

Third Case

This case holds if neither the first case nor the second case is in effect (that is, none of the volatiles listed on the table meet both conditions).

In this case, the world will automatically have no significant atmosphere, and an atmospheric mass of zero.

Surface Atmospheric Pressure

To determine the atmospheric pressure at a world’s surface, multiply the atmospheric mass by its surface gravity.

Examples

Arcadia IV has blackbody temperature of 281 K, an M-number of 5, Extensive water with no runaway greenhouse, a Mature Plate Lithosphere, and a Strong Magnetic Field. Major components of the atmosphere will include both molecular nitrogen and carbon dioxide, so the planet falls squarely into the first case. Alice rolls an unmodified 3d6 and gets a result of 9, so Arcadia IV has an atmospheric mass of 0.9. Since the planet has surface gravity of 1.05, the atmospheric pressure at sea level is abut 0.95, very comparable to that of Earth.

Arcadia V has a blackbody temperature of 226 K, an M-number of 6, Moderate water with no runaway greenhouse, a Mature Plate Lithosphere, and no magnetic field. Major components of the atmosphere will include molecular nitrogen and carbon dioxide, so this planet also falls into the first case. Alice rolls 3d6-6 (modified due to the lack of a magnetic field) for a result of 7, so Arcadia V has an atmospheric mass of 0.7. The planet has surface gravity of 0.82, so atmospheric pressure at the surface is about 0.57.

Architect of Worlds – Step Twenty-Two: Determine Magnetic Field

Architect of Worlds – Step Twenty-Two: Determine Magnetic Field

In this step, we will estimate whether the world under development has a significant magnetic field.

The possible cases for a world’s magnetic field will be sorted into four categories: None, Weak, Moderate, and Strong, defined as follows.

  • None: The world has no detectable magnetic field and is completely unprotected from the stellar wind. Examples: Venus, Earth’s moon, Mars, most of the gas giant planets’ major satellites.
  • Weak: The world has a detectable magnetic field (about 1% as strong as Earth’s), but it offers no significant protection from the stellar wind. Examples: Mercury or Ganymede.
  • Moderate: The world’s magnetic field is strong enough to offer limited protection against the stellar wind (about 10% as strong as Earth’s). Examples: None in our planetary system.
  • Strong: The world’s magnetic field is at least comparable to that of Earth, sufficient to provide adequate protection against the stellar wind. Examples: Earth, the gas giant planets.

A world’s magnetic field seems to depend on several items:

  • The world needs to have a hot, liquid outer core of significant mass, composed largely of iron
  • There must be convection taking place in that iron outer core, causing rising and falling currents
  • The world must rotate on its axis

If all three of these conditions hold, the iron outer core forms a dynamo which creates a significant magnetic field. This, in turn, helps protect the world’s atmosphere from being stripped away by stellar wind, and also protects the surface of the world from some harmful radiation. In our own planetary system, only Earth and the gas giant planets have strong magnetic fields.

Note that the third condition – that the world must rotate on its axis – is almost universal. Even a tide-locked world still rotates on its axis, and physical modeling seems to indicate that even slow rotation is enough to support a working dynamo. The existence of strong convective heat transfer through a world’s outer core seems to be the critical factor.

Procedure

To determine the strength of a world’s magnetic field at random, roll 3d6 modified as follows:

  • +4 if the world has a Soft Lithosphere
  • +8 if the world has Early Plate Lithosphere or Ancient Plate Lithosphere and also has Mobile Plate Tectonics
  • +12 if the world has Mature Plate Lithosphere and also has Mobile Plate Tectonics

Refer to the Magnetic Field Table entry for the modified roll.

Magnetic Field Table
Modified Roll (3d6)Magnetic Field
14 or lessNone
15-17Weak
18-19Moderate
20 or greaterStrong

Examples

Arcadia IV has a Mature Plate Lithosphere and Mobile Plate Tectonics, so Alice rolls 3d6+12 and gets a result of 25. Arcadia IV has a Strong Magnetic Field.

Arcadia V has a Mature Plate Lithosphere but has Fixed Plate Tectonics. Alice rolls an unmodified 3d6 and gets a result of 13. Arcadia V has no significant magnetic field.

Architect of Worlds – Step Twenty-One: Geophysical Parameters

Architect of Worlds – Step Twenty-One: Geophysical Parameters

Before we get started with this step in the design sequence, be aware that the modeling here is even more pragmatic and “rule-of-thumb” than usual. I think the following material will work properly, but it’s going to need some rigorous testing and tweaking before I’m satisfied with it.


In this step, we will determine some of the geological history of the world under development. In particular, we will estimate the world’s internal heat budget, characterize the presence and degree of active plate tectonics, and estimate the level of vulcanism.

A world’s internal heat will normally derive from three different sources:

  • The primordial heat of the world’s formation
  • Radiogenic heat derived from the decay of radioactive isotopes
  • Tidal heat generated by friction due to any tidal forces acting on the world

The structure and behavior of the world’s lithosphere will be strongly determined by the amount of heat remaining in the world’s deep interior. The hotter the world’s mantle and core, the more likely it is that heat will escape to and through the world’s surface, softening or melting surface rocks and possibly giving rise to volcanic eruptions.

Procedure

Primordial and Radiogenic Heat Budget

Begin by estimating the primordial and radiogenic heat budget of the world under development.

Evaluate the following quantity for all worlds:

H_P\ =\ (66.4\times{(log}_{10}{(K\times R\times(M+1))))-(8\times A)-182.5}

Here, K is the density of the world compared to Earth, R is the world’s radius in kilometers, M is the metallicity of the star system (as determined in Step Five), and A is the age of the star system in billions of years. HP is a rough measure of the total amount of primordial heat and radiogenic heat a world possesses, on a logarithmic scale. On this scale, Earth had an HP value of about 90 immediately after its formation (and has an HP value of about 54 today).

Tidal Heat Budget

Not all worlds will have a significant budget of internal heat due to tidal friction. Or each world under development, check to see whether the world falls into any of the following two cases. If so, compute the quantity HT according to the formula given.

First Case: Major Satellites of Gas Giants

A major satellite of a gas giant planet only (not a Leftover Oligarch, Terrestrial Planet, or Failed Core) will experience significant tidal heating if and only if:

  • There is at least one other major moon in the next outward orbit from the gas giant, as established in Step Fourteen, the first case, and
  • That “next outward” major moon is in a stable resonance with the moon being developed (that is, the ratio of their two orbital radii was derived from the Stable Resonant Orbit Spacing Table in Step Eleven).

In this case, the resonance between the two orbital periods will tend to maintain a small degree of eccentricity in the first moon’s orbit. This in turn will cause tidal forces imposed by the gas giant to increase and decrease slightly during the moon’s orbital period, causing the moon’s body to “flex” and create friction. In our own planetary system, two of the satellites of Jupiter fall into this case (Io and Europa).

If a moon falls into this case, evaluate the following:

H_T\ =(66.4\times{log}_{10}{(\frac{M\times D}{R^3}))+818}

Here, M is the mass of the gas giant in Earth-masses, D is the moon’s radius in kilometers, and R is the moon’s orbital radius in kilometers. HT is a rough estimate of the moon’s tidal heat budget, on the same logarithmic scale as HP.

Second Case: Spin-Resonant Planets Without Major Satellites

A Leftover Oligarch, Terrestrial Planet, or Failed Core which has no major satellite may experience significant tidal heating due to its primary star, if and only if the planet is in a spin-orbit resonance with its primary star, as determined in Step Sixteen, and at least one of the two following cases is correct:

  • The spin-orbit resonance is not 1:1 (that is, the planet is not tide-locked to its primary star), or
  • Both of the following are true:
    • There is at least one other planet in the next outward orbit from the primary star, as established in Step Eleven, and
    • That “next outward” planet is in a stable resonance with the planet being developed (that is, the ratio of their two orbital radii was derived from the Stable Resonant Orbit Spacing Table).

In either case, tidal forces imposed by the primary star will cause the planet’s body to flex slightly during its orbital period, giving rise to internal friction and heat. In practice, the effect is likely to be minimal unless the planet orbits very close to its primary star.

If a planet falls into this case, evaluate the following:

H_T\ =(66.4\times{log}_{10}{(\frac{M\times D}{R^3}))-444}

Here, M is the mass of the primary star in solar masses, D is the planet’s radius in kilometers, and R is the planet’s orbital radius in AU. HT is a rough estimate of the planet’s tidal heat budget, on the same logarithmic scale as HP.

Once you have computed HP and (possibly) HT, make a note of the greater of the two – that is, the heat budget associated with the source that is currently providing more internal heat for the world – for use in the rest of this step.

Status of Lithosphere

The lithosphere of a world is the top layer of its rocky structure. A world’s lithosphere usually begins as a global sea of magma, but it will soon cool, forming a solid crust that provides a (more or less) stable surface. Over time, as the world cools, the crust will tend to become thicker and more rigid, eventually forming a single immobile plate that covers the entire sphere.

Note that on a world with Massive prevalence of water, the lithosphere is effectively inaccessible, submerged beneath deep ice sheets or liquid-water oceans. In this case, the actual surface of the world will be atop the water layers (the hydrosphere). Determine the status of the lithosphere in any case since it will still affect several other properties of the world.

The possible cases will be sorted into six categories: Molten, Soft, Early Plate, Mature Plate, Ancient Plate, and Solid. These categories are defined as follows.

  • Molten Lithosphere: Large portions of the world’s lithosphere are still covered by magma oceans. A thin solid crust may form in specific regions. Active volcanoes are extremely common and may appear anywhere on the lithosphere. Examples: Earth in the Hadean Eon.
  • Soft Lithosphere: A solid lithosphere has formed, and no magma oceans remain. However, the lithosphere is not strong enough to resist the upwelling of magma from the world’s mantle, so active volcanoes remain very common and continue to appear anywhere on the lithosphere. Examples: Earth in the early Archean Eon.
  • Early Plate Lithosphere: The lithosphere is becoming strong enough to resist the upwelling of magma from the mantle. The crust is organizing into solid plates. Volcanoes remain common, but (depending on the presence of active plate tectonics) may be limited to certain locations. Examples: Earth in the later Archean Eon.
  • Mature Plate Lithosphere: The organization of the crust into solid plates is complete, with most or all of the crust now integrated into the system. Some of the crustal plates are now thicker and more durable. Volcanoes are less common. Examples: Earth today.
  • Ancient Plate Lithosphere: The lithosphere is becoming thick and rigid, and the system of crustal plates is becoming stagnant. Vulcanism is increasingly rare. Examples: Earth billions of years from now, Mars today.
  • Solid Lithosphere: The lithosphere is solid and completely stagnant. Vulcanism is vanishingly rare or extinct. Examples: Earth’s moon.

To determine the current status of a world’s lithosphere, roll 3d6 and add HP or HT, whichever is greater. Then refer to the Lithosphere Table.

Lithosphere Table
Modified Roll (3d6)Lithosphere Status
96 or higherMolten Lithosphere
88-95Soft Lithosphere
79-87Early Plate Lithosphere
45-79Mature Plate Lithosphere
31-44Ancient Plate Lithosphere
30 or lessSolid Lithosphere

Plate Tectonics

Even if a world’s crust is organized into solid plates, those plates may or may not be able to move and interact in an active system of plate tectonics. In our own planetary system, several worlds show some sign of plate tectonics. However, only on Earth is the entire crust arranged into a clear set of plates that move across the mantle and actively recycle crust material. The decisive factor seems to be Earth’s extensive prevalence of water, which permeates the crustal rocks and reduces friction among the tectonic plates.

Determine the status of the world’s plate tectonics only if its lithosphere is in an Early Plate, Mature Plate, or Ancient Plate status as determined above.

The possible cases will be sorted into two categories: Mobile Plate Tectonics and Fixed Plate Tectonics, defined as follows.

  • Mobile Plate Tectonics: The crust’s tectonic plates are able to move freely past or against one another. As tectonic plates collide, some of them experience subduction, moving down into the mantle and recycling the crustal material. Orogeny, or the formation of mountain ranges, takes place in such areas as well. Volcanic activity is likely to take place at plate boundaries. Volcanoes may also appear in plate interiors, at the top of magma plumes rising from the deep mantle.  Such shield volcanoes will tend to form arcs or chains, as the tectonic plate moves across the top of the plume.
  • Fixed Plate Tectonics: The crust’s tectonic plates are unable to move freely. Little or no subduction takes place to recycle crustal material. Orogeny is rare. As with Mobile Plate Tectonics, volcanoes are likely to appear at plate boundaries. Shield volcanoes are also possible, but since the tectonic plates are nearly immobile, such volcanoes can grow very large over time.

In general, a world is likely to have Mobile Plate Tectonics if it is younger (and therefore still has a hot mantle and core) and has plenty of surface water to reduce friction among the plates. To determine the status of a world’s plate tectonics at random, roll 3d6 and modify the result as follows:

  • +6 if the world has Extensive or Massive prevalence of water
  • -6 if the world has Minimal or Trace prevalence of water
  • +2 if the world has an Early Plate Lithosphere
  • -2 if the world has an Ancient Plate Lithosphere

A world will have Mobile Plate Tectonics on a modified roll of 11 or greater, and Fixed Plate Tectonics otherwise.

Special Case: Episodic Resurfacing

If a world has an Early Plate or Mature Plate Lithosphere, and has Fixed Plate Tectonics, then vulcanism will follow an unusual pattern of episodic resurfacing.

In this case, the lithosphere is too strong to permit magma to reach the surface under normal conditions. Since any tectonic plates are fixed in place, subduction and orogeny are very rare. Active volcanoes are also uncommon. However, heat built up in the mantle periodically breaks through, causing massive volcanic outbursts that “resurface” large portions of the lithosphere before the situation restabilizes.

For an Early Plate Lithosphere, these resurfacing events will take place millions of years apart. For a Mature Plate Lithosphere, resurfacing becomes much less frequent, tens or even hundreds of millions of years apart. In our own planetary system, Venus is an example of this case.

Examples

Both Arcadia IV and Arcadia V are planets without major satellites, and both of them are at a significant distance from the primary star, so Alice assumes that tidal heating will be insignificant for both of them. She evaluates HP for both. The star system is 5.6 billion years old and has metallicity of 0.63.

Arcadia IV has density of 1.04 and radius of 6450 kilometers, and so has HP of 41 (rounded to the nearest integer). With a 3d6 roll of 9, the total is 50. Arcadia IV has a Mature Plate Lithosphere.

Arcadia V has density of 0.92 and radius of 5670 kilometers, and so has HP of 34 (rounded to the nearest integer). With a 3d6 roll of 13, the total is 47. Arcadia V also has a Mature Plate Lithosphere. Notice that both of these planets have total scores fairly low in the range for a Mature Plate result, indicating that they have rather “old” geology and may be transitioning to an Ancient Plate configuration.

Arcadia IV has Extensive water, so Alice rolls 3d6+6 for a result of 15. Arcadia IV has Mobile Plate Tectonics, resembling Earth in this respect.

Arcadia V only has Moderate water, so Alice makes an unmodified roll of 3d6 for a result of 7. Arcadia V has Fixed Plate Tectonics and exhibits episodic resurfacing on a timescale of tens or hundreds of millions of years.

Architect of Worlds – Step Twenty: Determine Prevalence of Water

Architect of Worlds – Step Twenty: Determine Prevalence of Water

Water is one of the most common substances in the universe. Its special properties will lead it to have a profound effect on the surface conditions of any world, from its initial geological development, to its eventual climate, and finally to the evolution of life. Some worlds may never have much water, others will tend to lose whatever water they begin with, and still others will retain massive amounts of water throughout their lives.

In this step, we will estimate how much water can be found on a given world. The possible cases will be sorted into five categories: Trace, Minimal, Moderate, Extensive, and Massive. These categories are defined as follows.

  • Trace: No liquid water or water ice remains on the vast majority of the surface. If there is a substantial atmosphere, it may carry traces of water vapor. Small pockets of water ice may remain on the surface, in permanently shadowed craters or valleys, or on the night face of a world tide-locked to its primary star. Small deposits of water may be locked in hydrated minerals deep below the surface. Examples: Mercury, Venus, Earth’s moon, or Io.
  • Minimal: Liquid water is vanishingly rare on the surface, but large deposits of water ice may exist in the form of polar caps, in sheltered craters or valleys, or on the night face of a tide-locked world. Substantial aquifers or ice deposits may exist close beneath the surface. Hydrated minerals can be found in the world’s interior. Examples: Mars.
  • Moderate: A substantial portion of the world’s surface, but not a majority, is covered by some combination of liquid-water seas and water ice, depending on local temperature. The liquid-water oceans or ice deposits are up to a few kilometers in depth. Far away from the oceans or ice deposits, water becomes vanishingly rare. Hydrated minerals are common in the world’s interior. Examples: Mars a few billion years ago.
  • Extensive: Most of the world’s surface is covered by some combination of liquid-water oceans and water ice, up to several kilometers in depth. Water is common in most areas of the surface, even away from the oceans or ice deposits. Hydrated minerals are plentiful far into the world’s interior. Examples: Earth, Venus a few billion years ago.
  • Massive: The entire surface is covered by some combination of liquid-water oceans and water ice, up to hundreds of kilometers deep. Deeper layers of this world-ocean may be composed of higher-level crystalline forms of water (Ice II and up). Hydrated minerals are plentiful far into the world’s interior. Examples: Europa, Ganymede, Callisto, Titan, some “super-Earth” exoplanets.

The amount of water available on a given world will depend upon its M-number (Step Nineteen), its blackbody temperature (Step Nineteen), its location with respect to the protoplanetary disk (Step Nine), and (in some cases) the arrangement of any gas giant planets elsewhere in the planetary system (Steps Ten and Eleven).

Procedure

Begin by noting which of the following three cases the world being developed falls under, based on its M-number.

First Case: M-number is 2 or less

In this case, the world’s prevalence of water is automatically Massive.

Second Case: M-number is between 3 and 28

In this case, determine whether the world is outside or inside the protoplanetary nebula’s snow line, as determined in Step Nine. If the world’s orbital radius (or that of its planet, in the case of a major satellite) is exactly on the snow line, assume that it is outside.

If the world in this case is outside the snow line, then its prevalence of water is automatically Massive.

If the world in this case is inside the snow line, then roll 3d6, modified as follows:

  • Subtract the world’s M-number.
  • Add +6 if there exists a dominant gas giant in the planetary system, it experienced a Grand Tack event, and it is currently outside the protoplanetary nebula’s snow line.
  • Add +3 if any gas giants in the planetary system are currently outside the protoplanetary nebula’s slow-accretion line.

Take the modified 3d6 roll and refer to the Initial Water Prevalence table:

Initial Water Prevalence Table
Modified Roll (3d6)Prevalence
-5 or lessTrace
-4 to 3Minimal
4 to 11Moderate
12 to 19Extensive
20 or higherMassive

If the result on the table is Moderate or higher, and the world’s blackbody temperature is 300 K or greater, then the presence of water vapor in the world’s atmosphere has given rise to a runaway greenhouse event. Make a note of this event for later steps in the design sequence and reduce the prevalence of water to Trace.

Otherwise (the world’s blackbody temperature is less than 300 K) the prevalence of water is as indicated on the table.

Third Case: M-number is 29 or greater

In this case, determine whether any of the three following cases is true:

  • The world’s blackbody temperature is 125 K or greater.
  • The world is the major satellite of a Large gas giant, and its orbital radius is no more than 8 times the radius of the gas giant.
  • The world is the major satellite of a Very Large gas giant, and its orbital radius is no more than 12 times the radius of the gas giant.

If any of these three cases are true, then the world’s prevalence of water is Trace. Otherwise, its prevalence of water is Massive.

Examples

Both Arcadia IV and Arcadia V fall into the second case. The Arcadia system has a dominant gas giant, which underwent a Grand Tack and ended up outside the snow line. The outermost gas giant (at 9.50 AU) is not outside the system’s slow-accretion line (at 14.0 AU). For both planets, therefore, she will roll 3d6, minus the planet’s M-number, plus 6. Her rolls are 13 for Arcadia IV and 10 for Arcadia V, so Arcadia IV has Extensive water while Arcadia V has only Moderate water.

Architect of Worlds – Step Nineteen: Determine Blackbody Temperature

Architect of Worlds – Step Nineteen: Determine Blackbody Temperature

The blackbody temperature of a world is the average surface temperature it would have if it were an ideal blackbody, a perfect absorber and radiator of heat. Real planets are not ideal blackbodies, so their surface temperatures will vary from this ideal, but the blackbody temperature is a useful tool for determining a variety of other surface conditions.

In particular, the blackbody temperature is useful in determining what atmospheric gases the world can retain over billion-year timescales. Simple thermal escape (also called Jeans escape) isn’t the only mechanism by which a world can lose atmospheric gases, but it is a strong influence on the stable mass and composition of the atmosphere.

In this step, we will compute the blackbody temperature and the M-number for the world under development. The M-number is equal to a minimum molecular weight that can be retained over long timescales.

Procedure

To determine the blackbody temperature for a world, evaluate the following:

T=278\times\frac{\sqrt[4]{L}}{\sqrt R}

Here, L is the current luminosity of the primary star in solar units, R is the orbital radius of a planet (or the planet that a satellite orbits) in AU, and T is the blackbody temperature in kelvins. Note that the blackbody temperature will be the same for a planet and all of its satellites.

To estimate the M-number for a world, evaluate the following:

M=\ 676300\ \times\frac{T}{K\times R^2}

Here, T is the blackbody temperature, K is the world’s density compared to Earth, R is the world’s radius in kilometers, and M is the M-number. Round the result up to the nearest integer.

Example

Alice computes the blackbody temperature and the M-number for Arcadia IV and Arcadia V:

WorldOrbital RadiusMassDensityRadiusBlackbody TemperatureM-Number
Arcadia IV0.57 AU1.081.046450 km281 K5
Arcadia V0.88 AU0.650.925670 km226 K6

Comparing both planets to Earth (with a blackbody temperature of 278 K and an M-number of 5), Alice finds that both of these worlds are somewhat Earthlike. Arcadia IV is just a little warmer than Earth, while Arcadia V is significantly colder.

Both worlds seem likely to have atmospheres broadly similar to that of Earth. An M-number of 5 or 6 indicates that a planet can easily retain gases such as water vapor (molecular weight 18), nitrogen (molecular weight 28), oxygen (molecular weight 32), and carbon dioxide (molecular weight 44) against simple thermal escape. It’s possible that other factors will impact the atmospheres of these worlds, but for now, Alice is satisfied that she still has two somewhat hospitable environments to use in her stories.

Dependencies in a Model

Dependencies in a Model

Still working on the next sections of Architect of Worlds.

What’s interesting is that we’re coming to a number of items that fall into a web of dependencies. Some items affect the most likely outcome of others, and it’s not a nicely linear process. Just to give you a sample, here’s some storyboarding I’ve been doing using the Miro application online:

Where this mostly comes into play is with the order that the steps need to come in the design sequence. Fortunately, I have yet to come across any dependency loops. As long as the sequence moves more or less left to right, everything should work properly . . .

Architect of Worlds – Step Eighteen: Local Calendar

Architect of Worlds – Step Eighteen: Local Calendar

In this step, we will determine elements of the local calendar on the world being developed: the length of the local day, the length of any “month” determined by a major satellite, and so on.

Length of Local Day for a Planet

To determine the length of a planet’s day – the planet’s rotation period with respect to its primary star rather than with respect to the distant stars – compute the following:

T=\frac{P\times R}{P-R}

Here, P is the planet’s orbital period as determined in Step Fifteen, while R is the planet’s rotational period as determined in Step Sixteen, both in hours. T is the apparent length of the planet’s day, also in hours.

Note that this equation is undefined in cases when the orbital period and rotational period are equal (that is, the planet is in a spin-orbital resonance of 1:1 and is “tide locked”). In this case, the length of the local day is effectively infinite – the sun never moves in the sky!

At the other extreme, if the orbital period is much longer than the rotational period, then the day length and the rotational period will be very close together.

To determine the length of the local year in local days, simply divide the planet’s orbital period by the length of the local day as computed above.

Length of Apparent Orbital Period for a Satellite

To determine the length of a satellite’s orbital period, from any position on the planet’s surface, use the same equation:

T=\frac{P\times R}{P-R}

Here, P is the satellite’s orbital period as determined in Step Fifteen, while R is the planet’s rotational period as determined in Step Sixteen, both in hours. T is the apparent length of the satellite’s orbital period, also in hours.

Again, this equation is undefined in cases when the satellite’s orbital period and the planet’s rotational period are equal (that is, the planet is tide-locked to its satellite, or the satellite happens to orbit at a geosynchronous distance). In this case, the length of the satellite’s apparent orbital period is effectively infinite – the satellite never moves in the sky.

At the other extreme, if the satellite’s orbital period is much longer than the planet’s rotational period, then the apparent orbital period and the rotational period will be very close together. Earth’s moon is a familiar example – its apparent motion in the sky is dominated by Earth’s rotation.

It’s possible for a satellite’s orbital period to be shorter than the planet’s rotational period. For example, a moonlet that orbits very close is likely to fall into this case. The apparent orbital period will therefore be negative, indicating that the satellite appears to move backward over time. The satellite will rise in the west and set in the east.

Length of Synodic Month for a Satellite

To determine the length of a satellite’s synodic month, use the same equation once more:

T=\frac{P\times R}{P-R}

Here, P is the planet’s orbital period as determined in Step Fifteen, while R is the satellite’s orbital period as determined in Step Fifteen, both in hours. T is the length of the satellite’s synodic month.

It’s very unlikely for a satellite to have the same orbital period around its planet as the planet does around its primary star, so the undefined or negative cases almost certainly will not occur. T will indicate the period between (e.g.) one “full moon” and the next, as observed from the planet’s surface.

Examples

Arcadia IV has no satellite, so the only item of interest will be the length of its local day. Alice computes:

T=\frac{4170\times22.5}{4170-22.5}\approx22.62

The local day on Arcadia IV is only slightly longer than its rotation period. Alice can also determine the length of the local year in local days, by dividing the orbital period by this day length. Arcadia IV has a local year of about 184.35 local days.

Arcadia V has a satellite, so that satellite’s apparent orbital period and synodic month might be of interest. For the apparent orbital period, Alice computes:

T=\frac{16.4\times34}{16.4-34}\approx-31.68

Arcadia V’s moonlet appears to move retrograde or “backwards” in the sky, rising in the west and setting in the east, with an apparent period of about 31.7 hours.

Meanwhile, for the synodic month, Alice computes:

T=\frac{7990\times16.4}{7990-16.4}\approx16.43

The satellite’s synodic month – the period between “full moon” phases – is much shorter than its apparent orbital period. From the surface of Arcadia V, the moonlet will appear to move slowly through the sky, its phase visibly changing as it moves, passing through almost a complete cycle of phases before setting once more in the east. Very strange, for human observers accustomed to the more sedate behavior of Earth’s moon!

Where to Find “Architect of Worlds”

Where to Find “Architect of Worlds”

Just a quick post today, while I continue to work on the next few steps of the Architect of Worlds design sequence. I’m noticing some renewed interest in this project, which I suppose shouldn’t surprise me given that I’m finally getting back to work on it.

It looks as if people are coming to the blog and doing a tag-search for old Architect of Worlds posts. That’s fine, but you should be aware that the earlier steps as originally posted to the blog may not be the most current version of the system. Not to mention, the blog posts aren’t always formatted so as to be easy to read or use.

For the time being, I maintain PDFs of the current “official” version of the draft on the Architect of Worlds page in the sidebar. If you’re interested in what’s been developed so far, you might want to look there rather than try to page through the old blog posts.

So long as everyone respects my copyrights, you’re welcome to download copies for your personal use. That will probably change as the book gets closer to actual publication, but that won’t be for some time yet. Of course, if you work with the system and get some interesting results, I’d be pleased to hear about that.

Architect of Worlds – Step Seventeen: Determine Obliquity

Architect of Worlds – Step Seventeen: Determine Obliquity

The obliquity of an object is the angle between its rotational axis and its orbital axis, or equivalently the angle between its equatorial plane and its orbital plane. It’s often colloquially called the axial tilt of a moon or planet. Obliquity can have significant effects on the surface conditions of a world, affecting daily and seasonal variations in temperature.

Procedure

Begin by noting the situation the world being developed is in: is it a major satellite of a planet, a planet with its own major satellite, or a planet without any major satellite? Notice that these three cases exactly parallel those in Step Sixteen.

First Case: Major Satellites of Planets

Major satellites of planets, as placed in Step Fourteen, will tend to have little or no obliquity with respect to the planet’s orbital plane. To determine the obliquity of such a satellite at random, roll 3d6-8 (minimum 0) and take the result as the obliquity in degrees.

Note that the major satellites of gas giants, distant from their primary star, may be an exception to this general rule. For example, in our own planetary system, the planet Uranus is tilted at almost 90 degrees to its orbital plane. Its satellites all orbit close to the equatorial plane of Uranus, so their orbits are also at a large angle, and their obliquity is very high. Cases like this are very unlikely for the smaller planets close to a primary star – tidal interactions will tend to quickly “flatten” the orbital planes of any major satellites there.

Second Case: Planets with Major Satellites

A Leftover Oligarch, Terrestrial Planet, or Failed Core which has a major satellite is likely to have its obliquity stabilized by the presence of that satellite.

To select a value of the planet’s obliquity at random, roll 3d6. Add the same modifier that was computed during Step Sixteen for the Rotation Period Table, based on the degree of tidal deceleration applied by the major satellite. Refer to the Obliquity Table.

Obliquity Table
Modified RollObliquity
4 or lessExtreme (see Extreme Obliquity Table)
548 degrees
646 degrees
744 degrees
842 degrees
940 degrees
1038 degrees
1136 degrees
1234 degrees
1332 degrees
1430 degrees
1528 degrees
1626 degrees
1724 degrees
1822 degrees
1920 degrees
2018 degrees
2116 degrees
2214 degrees
2312 degrees
2410 degrees
25 or higherMinimal (3d6-8 degrees, minimum 0)

Feel free to adjust a result from this procedure to any value between the next lower and next higher rows on the table.

If the result is Extreme, the obliquity is likely to be anywhere from about 50 degrees up to almost 90 degrees. To select a value at random, roll 1d6 on the Extreme Obliquity Table.

Extreme Obliquity Table
Roll (1d6)Obliquity
1-250 degrees
360 degrees
470 degrees
580 degrees
698-3d6 degrees, maximum 90

Again, feel free to adjust a result from this procedure to any value between the next lower and next higher rows on the table.

Third Case: Planets Without Major Satellites

A Leftover Oligarch, Terrestrial Planet, or Failed Core which has no major satellite will be most affected by its primary star.

However, without the stabilizing presence of a major satellite, the planet’s obliquity is likely to change more drastically over time. Minor perturbations from other planets in the system may lead to chaotic “excursions” of a planet’s rotation axis. For example, although at present the obliquity of Mars is about 25 degrees (comparable to that of Earth), some models predict that Mars undergoes major excursions from about 0 degrees to as high as 60 degrees over millions of years.

To select a value for obliquity at random, begin by rolling 3d6 on the Unstable Obliquity Table.

Unstable Obliquity Table
Roll (3d6)Modifier
7 or lessRoll 1d6 – High Instability
8-13No modifier
14 or higherRoll 5d6 – High Instability

Make a note of any result indicating High Instability for later steps in the design sequence. The planet is likely to be undergoing drastic climate changes on a timescale of millions of years.

Now make a roll on the Obliquity Table, but if High Instability was indicated, roll 1d6 or 5d6 on this table, rather than the usual 3d6. Finally, add the same modifier that was computed during Step Sixteen for the Rotation Period Table, based on the degree of tidal deceleration applied by the primary star. Refer to the Obliquity Table, and possibly the Extreme Obliquity Table, as required.

Examples

Both Arcadia IV and Arcadia V are planets without major satellites, so they both fall under the third case in this section, as they did in Step Sixteen.

For Arcadia IV, Alice begins by rolling a 4 on the Unstable Obliquity Table, indicating that she will need to roll 1d6 rather than 3d6 on the Obliquity Table. That roll will therefore be 1d6+1, and Alice gets a final result of 3. Arcadia IV apparently has extreme obliquity in the current era. Rather than roll at random, Alice selects a value for the planet’s obliquity of about 58 degrees.

Alice makes a note of the “high instability” of the planet’s obliquity. Its steep axial tilt may be a relatively recent occurrence, taking place over the last few million years. Arcadia IV, the Earth-like candidate in her planetary system, will have very pronounced seasonal variations, and may be undergoing an era of severe climate change. Any native life has probably been significantly affected, and human colonists would need to adapt!

Meanwhile, for Arcadia V, Alice rolls a 12 on the Unstable Obliquity Table, indicating that the planet’s rotational axis is currently relatively stable. She rolls an unmodified 3d6 on the Obliquity Table, getting a result of 15. She selects a value for this planet’s obliquity of about 28.5 degrees.

Architect of Worlds – Step Sixteen: Determine Rotation Period

Architect of Worlds – Step Sixteen: Determine Rotation Period

A quick note before I drop the next section of the draft: I caught myself making several errors in the mathematics while developing this step. I think I’ve weeded all of those out, but if anyone is experimenting with this material as it appears, let me know if you come across any odd results.


Step Sixteen: Determine Rotation Period

The next three steps in the sequence all have to do with planetary rotation. Every object in the cosmos appears to rotate around at least one axis, and in fact some objects appear to “tumble” by rotating around more than one.

Planets and their major satellites usually have simple rotation, spinning in the same direction as their orbital motion, around a single axis that is more or less perpendicular to the plane of their orbital motion. There are, of course, a variety of exceptions to this general rule.

In this step, we will determine the rotation period of a given world. In this case, we will be dealing with what’s called the sidereal period of rotation – the time it takes for a world to rotate once with respect to the distant stars.

Worlds appear to form with wildly varying rotation periods, the legacy of the chaotic processes of planetary formation. However, many worlds will have been affected by tidal deceleration applied by the gravitational influence of nearby objects. Tidal deceleration may cause a world to be captured into a special status called a spin-orbital resonance, in which the world’s orbital period and its rotational period form a small-integer ratio.

Procedure

Begin by noting the situation the world being developed is in: is it a major satellite of a planet, a planet with its own major satellite, or a planet affected primarily by its primary star?

First Case: Major Satellites of Planets

Major satellites of planets, as placed in Step Fourteen, will almost invariably be in a spin-orbit resonance state. Most models of the formation of such satellites suggest that they are captured into such a state almost immediately after their formation.

Since a major satellite’s orbit normally has very small eccentricity, the spin-orbit resonance will be 1:1. The satellite’s rotation period will be exactly equal to its orbital period.

Second Case: Planets with Major Satellites

A Leftover Oligarch, Terrestrial Planet, or Failed Core which has a major satellite may be captured into a spin resonance with the satellite’s orbit. This is actually somewhat unlikely; for example, Earth is not likely to become tide-locked to its own moon within the lifetime of the sun. However, a satellite’s tidal effects on the primary planet will tend to slow its rotation rate.

To estimate the probability that a planet has become tide-locked to its satellite, and to estimate its rotation rate if this is not the case, begin by evaluating the following:

T={10}^{25}\times\frac{M_S^2\times R^3}{A\times M_P\times D^6}

Here, A is the age of the star system in billions of years. MS and MP are the mass of the satellite and the planet, respectively, in Earth-masses. R is the radius of the satellite, and D is the radius of the satellite’s orbit, both in kilometers.

If T is equal to or greater than 2, the planet is almost certainly tide-locked to its satellite. Its rotation period will be exactly equal to the orbital period of the satellite.

Otherwise, to generate a rotation period for the planet at random, multiply T by 12, round the result to the nearest integer, add the result to a roll of 3d6, and refer to the Rotation Period Table.

Rotation Period Table
Modified Roll (3d6)Rotation Rate
34 hours
45 hours
56 hours
68 hours
710 hours
812 hours
916 hours
1020 hours
1124 hours
1232 hours
1340 hours
1448 hours
1564 hours
1680 hours
1796 hours
18128 hours
19160 hours
20192 hours
21256 hours
22320 hours
23384 hours
24 or higherResonance Established

Feel free to adjust a result from this procedure to any value between the next lower and next higher rows on the table.

The planet will be tide-locked to its satellite on a result of 24 or higher, or in any case where the randomly generated rotation rate is actually longer than the satellite’s orbital period. In these cases, again, its rotation period will be exactly equal to the orbital period of the satellite.

Third Case: Planets Without Major Satellites

A Leftover Oligarch, Terrestrial Planet, or Failed Core which has no major satellite may be captured into a spin-orbit resonance with respect to its primary star. Even if this does not occur, solar tides will tend to slow the planet’s rotation rate.

To estimate the probability that such a planet has been captured into a spin-orbit resonance, and to estimate its rotation rate if this is not the case, begin by evaluating the following:

T={(9.6\ \times10}^{-14})\times\frac{M_S^2\times R^3}{{A\times M}_P\times D^6}

Here, A is the age of the star system in billions of years, MS is the mass of the primary star in solar masses, MP is the mass of the planet in Earth-masses, R is the radius of the planet in kilometers, and D is the planet’s orbital radius in AU.

Again, if T is equal to or greater than 2, the planet has almost certainly been captured in a spin-orbit resonance. Otherwise, to generate a rotation period for the planet at random, multiply T by 12, round the result to the nearest integer, add the result to a roll of 3d6, and refer to the Rotation Period Table. The planet will be in a spin-orbit resonance on a result of 24 or higher, or in any case where the randomly generated rotation rate is actually longer than the planet’s orbital period.

Planets captured into a spin-orbit resonance are not necessarily tide-locked to their primary star (or, in other words, the resonance is not necessarily 1:1). Tidal locking tends to match a planet’s rotation rate to its rate of revolution during its periastron passage. If the planet’s orbit is eccentric, this match may be approximated more closely by a different resonance. To determine the most likely resonance, refer to the Planetary Spin-Orbit Resonance Table:

Planetary Spin-Orbit Resonance Table
Planetary Orbit EccentricityMost Probable ResonanceRotation Period
Less than 0.121:1Equal to orbital period
Between 0.12 and 0.253:2Exactly 2/3 of orbital period
Between 0.25 and 0.352:1Exactly 1/2 of orbital period
Between 0.35 and 0.455:2Exactly 2/5 of orbital period
Greater than 0.453:1Exactly 1/3 of orbital period

On this table, the “most probable resonance” is the status that the planet is most likely to be captured into over a long period of time. It’s possible for a planet to be captured into a higher resonance (that is, a resonance from a lower line on the table) but this situation is unlikely to be stable over billions of years.

Examples

Both Arcadia IV and Arcadia V are planets without major satellites, so they both fall under the third case in this section. The most significant force modifying their rotation period will be tidal deceleration caused by the primary star.

The age of the Arcadia star system is about 5.6 billion years. Arcadia IV has mass of 1.08 Earth-masses and a radius of 6450 kilometers. Alice computes T for the planet and ends up with a value of about 0.083. Arcadia IV is probably not in a spin-orbit resonance, but tidal deceleration has had a noticeable effect on the planet’s rotation. Alice rolls 3d6+1 for a result of 11 and selects a value slightly lower than the one from that line of the Rotation Period Table. She decides that Arcadia IV rotates in about 22.5 hours.

Meanwhile, Arcadia V has mass of 0.65 Earth-masses and a radius of 5670 kilometers. Alice computes T again and finds a value of about 0.007. (Notice that the amount of tidal deceleration is very strongly dependent on the distance from the primary.) Alice rolls an unmodified 3d6 for a value of 12, this time selecting a value slightly higher than the one from the table. She decides that Arcadia V rotates in about 34.0 hours.

Citations

Gladman, Brett et al. (1996). “Synchronous Locking of Tidally Evolving Satellites.” Icarus, volume 122, pp. 166–192.

Makarov, Valeri V. (2011). “Conditions of Passage and Entrapment of Terrestrial Planets in Spin-orbit Resonances.” The Astrophysical Journal, volume 752 (1), article no. 73.

Peale, S. J. (1977). “Rotation Histories of the Natural Satellites.” Published in Planetary Satellites (J. A. Burns, ed.), pp. 87–112, University of Arizona Press.