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

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

Step Twelve: Determine Eccentricity of Planetary Orbits

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

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

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

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

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

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

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

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

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

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

Here, Rmin is the minimum distance, and Rmax is the maximum distance.

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

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

Examples

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

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

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

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

-0.38<E_1<0.57

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

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

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

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

2 thoughts on “Architect of Worlds – Step Twelve: Determine Eccentricity of Planetary Orbits

  1. I’m enjoying following this and generating a star system as you post new articles. The “Planetary Orbital Eccentricity Table” has overlapping values of “8 or less” and “7-9” – I’m assuming “6 or less” is correct for the first row?

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