[SPACE] Tidal braking

[dataweaver]

Originally Posted by Agemegos

When calculating the height of the tide that a planet raises on its moon, or that a moon raises on its planet:

T = 1.6 million * (M * D-to-the-fourth) / (P * R-cubed)

Where M is the mass of the object raise the tide in Earth masses, D is the diameter of the object the tides are on in Earth diameters. P is the mass of the object the tides are on in Earth masses, and R is the distance between them in Earth diameters.

Note that P is used elsewhere to represent orbital periods; so you should probably choose another letter to represent the mass of the object that the tides are on. I'd be inclined to use a lowercase m. So:

Lunar tides: T = 1.6 million * (M * D-to-the-fourth) / (m * R-cubed)
Satellite Orbital Period: P = 0.0588 * square root of (R-cubed/(M+m))

T: tidal forces; 1 represents Earth-like tides.
P: orbital period of satellite in days
M: mass of satellite in Earth masses
m: mass of planet in Earth masses
D: diameter of planet in Earth diameters
R: radius of satellite's orbit in Earth diameters

Solar tides: T = 0.3 * (M * D-to-the-fourth) / (m * R-cubed)
Planetary Orbital Period: P = square root of (R-cubed/M)

T: tidal forces; 1 represents Earth-like tides.
P: orbital period in years
M: mass of star in solar units
D: diameter of planet in Earth diameters
m: mass of planet in Earth masses
R: radius of planet's orbit in AUs.

I've included the associated orbital period calculations here because they become important later on.

Originally Posted by Agemegos

In the case of an object under the tidal influence of more than one neighbour, add the T values together for a total (maximum) tidal range (ie. spring tides). But keep a record of the separate T values, because you will need them in the next step.

To determine the rotation rate of a planet or moon, roll 3d and divide by 120. The result is the initial rate of rotation of the body. Call this value I.

Now square each of the T values to which this world is subject, and add all of the squares together. Call this value S. Then calculate the present rate of rotation, w, as follows.

W = I - [(0.016 * A * P/D^5) * S]

You skimmed over the fact that planets generally have axial tilts, and thus the torque applied by tidal forces would tend to fight the axial tilt as well as the rotational speed. Because of this, planets that start with high axial tilts will tend to have their axial rotation slowed less than planets that start with low axial tilts. Also, it means that older systems will tend to have smaller axial tilts than younger systems will.

Likewise, the same tidal forces that tend to slow a planet's rotation will also tend to drive satellites into higher orbits, eventually letting them break away from the planet. Younger systems will tend to have more satellites and in tighter orbits than older systems will.

Originally Posted by Agemegos

If W is zero or less, the planet or moon is tide-locked, and its day-length is equal to its orbital period. Otherwise, the period of the planets' (or moon's) rotation is

Daylength =1/W,

in hours. If this is longer than its orbital period, the planet or moon is tidelocked. Change its period of rotation to be equal to its orbital period.

First: Tidal forces don't drive W toward zero. Rather, solar tides drive W toward 1/(8766 P), where P is the Planetary Orbital Period in years; and lunar tides drive W toward 1/(24 P), where P is the Satellite Orbital Period in days. If W is at or between these values, the various tidal forces will be in competition. In other words, a satellite will end up delaying or preventing a solar tide-lock, depending on how powerful the satellite's tides are.

Of course, the solar tides will be working to drive the satellite away from the planet at the same time that they're working to tide-lock the planet, with the twin side effects of lengthening the Satellite Orbital Period and weakening the lunar tidal effects. It may well be that planets never reach a point where lunar "tidal braking" puts up any significant opposition to solar tidal braking, because the solar tides will have driven the potential competition away first. If this is the case, it's likely that tide-locked worlds can't have satellites larger than asteroids, if that. Regardless, a tide-locked world will not be able to have a satellite with tidal forces equal to or stronger than the star's; the planet would have tide-locked to the satellite instead of the star.

Second, a nitpick: Daylength doesn't equal 1/W; Rotational Period equals 1/W. The length of a day is computed from the Rotational Period on page 118, under "Local Calendar".