Mneme Placement Logic: Formula Reference Guide

This document isolates and explains the core mathematical formulas used to place planets and determine orbital stability within the Mneme Generator.

1. The Anchor Formulas (System Mapping)

Before placing any planets, the system's "zones" must be defined. These are entirely dependent on the Star's Luminosity ($L_{Star}$).

Formula 1.1: The Snow Line ($D_{snow}$)

Purpose: Determines the exact distance (in AU) where water freezes in a vacuum. This is the "seed point" where Gas Giants are most likely to form.

The Formula: $$D_{snow} = 2.7 \times \sqrt{L_{Star}}$$

Variables:

$L_{Star}$: The Star's Luminosity relative to Sol (Sol = 1.0).

Formula 1.2: The Habitable Zone Core ($HZ_{center}$)

Purpose: Finds the "Goldilocks" center where Earth-like temperatures are most likely.

The Formula: $$HZ_{center} = \sqrt{L_{Star}}$$

Edges:

Inner Edge: $0.95 \times HZ_{center}$

Outer Edge: $1.37 \times HZ_{center}$

2. The Dominance Formulas (Hill Stability)

These formulas determine if two planets can coexist at their specific distances without colliding or ejecting one another.

Formula 2.1: The Hill Sphere Ratio ($\mu$)

Purpose: Calculates the raw "Gravitational Weight Class" of a planet relative to its star. This ratio remains constant regardless of where the planet is placed.

The Formula: $$\mu = \sqrt[3]{\frac{m_1 + m_2}{3 \times M_{Star}}}$$

Variables:

$m_1, m_2$: Mass of the two planets interacting (in Earth Masses).

$M_{Star}$: Mass of the central star (in Earth Masses. 1 Solar Mass = 333,000 Earth Masses).

Formula 2.2: The Mutual Hill Radius ($R_H$)

Purpose: Translates the Hill Sphere Ratio ($\mu$) into an actual physical distance (in AU) based on where the planets are currently sitting.

The Formula: $$R_H = a_{avg} \times \mu$$ (Where $a_{avg}$ is the average distance of the two planets: $\frac{a_1 + a_2}{2}$)

Concept: A planet's gravitational "bubble" physically widens the further it gets from the star.

Formula 2.3: The Required Safety Space ($W_{req}$)

Purpose: Determines the absolute minimum gap (in AU) needed between two orbits to prevent chaos over a billion years.

The Formula: $$W_{req} = K \times R_H$$

The "K" Constant (The Rule of K):

$K = 5$ (The Mneme Standard / "Rule of Five"). This creates a densely packed, but stable system.

$K = 8$ (Sol Standard). Creates a relaxed system like our own Jupiter and Saturn.

3. The Action Formulas (The Shove)

When a heavy planet forces its way into an occupied zone, the lighter/outer planet must be moved outward to maintain stability.

Formula 3.1: The Shove Adjustment ($a_{new\_outer}$)

Purpose: Calculates the exact new orbital distance for a planet that has been "pushed" by a heavier inner neighbor.

The Formula: $$a_{new\_outer} = a_{inner} + W_{req} + Variance$$

Variables:

$a_{inner}$: The position of the heavy planet that claimed the spot.

$W_{req}$: The Required Safety Space calculated in Formula 2.3 (using $K=5$).

$Variance$: A small random addition (e.g., $+0.01$ to $+0.1$ AU) so the pushed planet isn't sitting exactly on the mathematical edge of destruction.

4. Multi-Star Stability Formulas (The Rule of Five & Barycenters)

When a system has two or more stars, their massive gravities dictate the orbital structure. Stars orbit a shared center of mass called a Barycenter.

Formula 4.1: The Barycenter Calculation (Center of Mass)

Purpose: Determines the exact point in space two stars orbit around, and how far each star sits from that central point.

The Formulas:

Distance of Star 1 to Barycenter ($r_1$): $$r_1 = D_{total} \times \frac{M_2}{M_1 + M_2}$$

Distance of Star 2 to Barycenter ($r_2$): $$r_2 = D_{total} \times \frac{M_1}{M_1 + M_2}$$

Variables:

$D_{total}$: The total distance between the two stars.

$M_1, M_2$: The mass of Star 1 and Star 2.

Note: If $M_1$ is vastly heavier than $M_2$, the Barycenter will be deep inside $M_1$ (like the Earth and the Sun). If they are equal mass, the Barycenter is exactly in the middle.

Formula 4.2: The Binary Forbidden Zone ($FZ$)

Purpose: Defines the orbital space rendered uninhabitable/unstable by the Companion Star.

The Formulas:

Inner Limit: $FZ_{in} = \frac{D_{total}}{5}$

Outer Limit: $FZ_{out} = D_{total} \times 5$

Rule: No planet can exist between $FZ_{in}$ and $FZ_{out}$ relative to the Barycenter.

Formula 4.3: S-Type Placement (Far Companion / Circumstellar)

Purpose: Use this formula if you want planets to orbit only one specific star. The Companion must be pushed far enough away to leave the inner system safe.

The Formula: $$D_{min} = a_{max\_desired} \times 5$$

Outcome: To keep a planet safe orbiting Star A at $4.0$ AU, Star B MUST be placed at $\ge 20.0$ AU.

Formula 4.4: P-Type Placement (Close Companion / Circumbinary)

Purpose: Use this formula if you want planets to orbit both stars at once (like Tatooine). The stars must be placed extremely close together to act as a single gravitational point.

The Formula: $$D_{max} = \frac{a_{min\_desired}}{5}$$

Outcome: To keep a planet safe orbiting the Barycenter at $0.8$ AU, Star A and Star B MUST be separated by no more than $0.16$ AU.

Formula 4.5: Hierarchical Multi-Star (The Alpha Centauri Model)

Purpose: To safely map a stable 3-star (Trinary) system.

The Logic:

Calculate the Inner Binary (e.g., Alpha Centauri A & B). They orbit their Barycenter ($BC_{AB}$) at distance $D_{AB}$.

Treat the Inner Binary as a Single Star Point with mass $M_{AB} = M_A + M_B$.

Place the Tertiary Star (e.g., Proxima Centauri / Star C).

Stability Check: Star C must be placed far outside the Forbidden Zone of the Inner Binary. $$D_C \ge (D_{AB} \times 5) \times \text{Safety Factor}$$

Real World Example: Alpha Centauri A & B are ~11-35 AU apart. Proxima Centauri is ~13,000 AU away, making it incredibly stable as a hierarchical trinary.

Example Walkthrough: "The Shove" in Action

The Setup: A star ($M_{Star} = 333,000$). We have a Gas Giant ($m_1 = 300$) trying to settle at 3.0 AU ($a_1$).

The Victim: An Ice Giant ($m_2 = 15$) is currently sitting at 3.2 AU ($a_2$).

Calculate Ratio ($\mu$): $\sqrt[3]{(300 + 15) / 999,000} \approx 0.068$

Calculate Average Distance ($a_{avg}$): $(3.0 + 3.2) / 2 = 3.1$ AU.

Calculate Hill Radius ($R_H$): $3.1 \times 0.068 \approx 0.21$ AU.

Calculate Required Space ($W_{req}$): $5 \times 0.21 = \mathbf{1.05 \text{ AU}}$.

The Test: The current gap is only $0.2$ AU. The required gap is $1.05$ AU. Conflict detected.

The Shove: The Ice Giant is pushed out to $3.0 \text{ (Giant Pos)} + 1.05 \text{ (Safety)} = \mathbf{4.05 \text{ AU}}$.