[SPACE] Tidal braking

[dataweaver]

For the record, the means of determining how long it would take to get from one rotational period to another is:

A = 62.5 * [(P2 - P1)] / [P1 * P2 * M/D^5 * (S1 - S2)]
A: time needed in billions of years
P1: initial rotational period in hours
P2: final rotational period in hours
M: mass of planet in Earth masses
D: diameter of planet in Earth diameters
S1: sum of squares of tides from sun and satellites with orbital periods greater than P1
S2: sum of squares of tides from remaining satellites.

For multiple satellites:

First Step:
P1 is randomly determined
P2 is the longest satellite orbital period in hours (multiply days by 24)

Subsequent Steps (skip to end if S2 >= S1):
P1 is the previous step's P2
P2 is the next longest satellite orbital period in hours (multiply days by 24)

repeat until you run out of satellites

Last Step (skip if S2 >= S1):
P1 is the previous step's P2
P2 is the planet's orbital period in hours (multiply years by 8766)

For one satellite:

First Step:
P1 is randomly determined
P2 is the satellite orbital period in hours (multiply days by 24)

Last Step (skip if S2 >= S1):
P1 is the previous step's P2
P2 is the planet's orbital period in hours (multiply years by 8766)

For no satellites:

P1 is randomly determined
P2 is the planet's orbital period in hours (multiply years by 8766)

Again, keep a running total of the A's; stop and interpolate the current rotational period if the running total equals or exceeds the system's age.

(Oddly enough, the only equation I'm having trouble with is the interpolation of the current rotational period; and that's simple Algebra. Need sleep...)

[Agemegos]

Originally Posted by dataweaver

For the record, the means of determining how long it would take to get from one rotational period to another is:

A = 62.5 * [(P2 - P1)] / [P1 * P2 * M/D^5 * (S1 - S2)]

You have period increasing linearly, which seems odd. Does the torque decrease with rotation rate?

[dataweaver]

That formula was attained by taking the one that you provided (i.e., W = I - [(0.016 * A * P/D^5) * S]), solving it for A, and changing around several of the variables (e.g., I and W got replaced by 1/P1 and 1/P2, respectively). As such, it shares nearly all of the same properties that your equation has; and the differences come from the fact that some of the T-square terms are subtracted instead of added.

In particular, period does not increase linearly; note that the periods appear in the divisor as well as the dividend:

I - W = 1/ P1 - 1/P2 = (P2 - P1) / (P1 * P2)

The non-linear progression of the period is why I passed on providing the interpolation formula at the time. In effect, it starts with

(WC - WI) / (AC - AI) = (WF - WI) / (AF - AI)
WI: initial rotational speed
WF: final rotational speed
WC: current rotational speed
AI: age at initial rotational speed
AF: age at final rotational speed
AC: current age

The goal is to solve for the current rotational period (not speed) in terms of the initial and final rotational periods (not speeds) and the various ages, and to simplify the resulting equation to the point that the casual reader won't run screaming from the room.

--

I got a little lazy at the end and didn't bother providing different variations of the formula depending on the time units used for the various periods; instead, I only used one variation, where all periods are measured in hours. In that sense, this is an "almost-finished product" - the finished product would provide multiple formulas differing in terms of the constants used so that you could have 1) initial period measured in hours and a final period measured in days; 2) initial period measured in days and final period measured in years; 3) initial and final periods both measured in days; and 4) initial period measured in hours and final period measured in years.

The user should need only select the appropriate sequence of formulas and plug in the numbers from earlier in the design process, instead of having to reformat any of the previous numbers (e.g., changing their units) in order to use them in the current sequence of formulas.

[Agemegos]

For the benefit of users who are searching the forums for corrections to space errata, I re-post the following.

Six errata on p. 117

1. The equation for Satellite Orbital Period has been corrected by official errata.
   

   P = 0.166 * square root of (R^3/M)

2. Space p. 117 gives a formula for the height of the tide raised by a planet on its moon or by a moon on its planet as follows "T = (17.8 million * M * D) / R^3" (where M is the mass of the body raising the tides, D is the diameter of the world, and R is the radius of the relevant orbit). The published errata 'correct' this to "T= (2230000 * M * D)/R^3".
   Physics textbooks imply that the formula ought to be "T = 1.56 million * (M * D-to-the-fourth) / (P * R-cubed)". (Where P is the mass of the planet or moon on which the tide occurs.)
   

   <edit> T = 2.23 million * (M * D) / (density * R-cubed)

3. Space p. 117 gives a formula for the height of the tide raised by a star on a planet or moon "T = (0.46 * M * D) / R^3" (where M is the mass of the body raising the tide, D is the diameter of the body on which the tide is raised, and R is the radius of the orbit).
   Physics textbooks imply that it ought to be "T = 0.3 * (M * D-to-the-fourth) / (P * R-cubed)". (Where P is the mass of the body on which the tide occurs.)
   

   <edit> T = 0.455 * (M * D) / (density * R-cubed)

4. Space p. 117 provides a procedure for setting the primordial rate of rotation of a body (ie. its rate of rotation before tidal braking) in which small bodies rotate slower than large ones. This would seem to be based on an examination of the rotation rate of bodies in the Solar System without taking into account that Mercury and all the moons are tidally locked or tidally resonant. Considering that freely-rotating asteroids have periods of 2 to 40 hours, average about 12 hours it seems that small bodies rotate no slower than large ones unless tidally braked.
   
   
5. Space p. 117 provides that tidal braking increases day-length linearly, whereas physics textbooks suggest that it ought to decrease rate of rotation linearly. Space provides a formula for the amount by which day length is increased by tidal braking as follows: "E = (T * A)/ M", (where T is the sum of the heights of all tides on the world, A is the age of the system and M is the mass of the world).
   Physics textbooks suggest that tidal braking ought to reduce the rate of rotation by up to an amount as follows "E = (0.383 * A * M/D^5) * sum of squares of tide heights" (in rotations per day).
   

   <edit> E = 0.184 * sum of squares of tide heights * A * density/D^2

6. Space p. 117 provides that a tide-locked planet is always tide-locked to its innermost moon, if it has one, and to its star otherwise. It is better and just as simple to provide that a tide-locked planet is tide-locked to whatever body raises the highest tides on it.

[Agemegos]

There has been an update to the errata for Space! Yippee!

The formula for the period of moons has been corrected.

The formula for the height of lunar tides has been given a new coefficient, but its functional form is unchanged. The formula for solar tides has not been corrected, The formula for tidal braking has not been corrected. Initial rotation has not been altered. Planets' tide-locking to their innermost moon has not been affected. I don't know whether the story is that my suggestions about tides and tidal braking have not been considered yet or whether they have been considered and rejected.

I'd really, really like it if a few physicists, astronomers, and astrophysicists would check my maths in the posts above.

[thtraveller]

Algebraically the mass of the planet or moon (P) is proportional to the cube of its diameter (D) as volume =4/3 pi r^3.

So isn't D^4/P equivalent to D^4/D^3 = D
and then multiplied by a different factor?

[Agemegos]

Originally Posted by thtraveller

Algebraically the mass of the planet or moon (P) is proportional to the cube of its diameter (D) as volume =4/3 pi r^3.

So isn't D^4/P equivalent to D^4/D^3 = D
and then multiplied by a different factor?

Good point. P = 1/6 * pi * rho * D^3, so D^4/P = 6/pi * D/rho, so

T = 0.3 * (M * D-to-the-fourth) / (P * R-cubed)
= 0.3 * 6/pi * (M * D) / (rho * R-cubed)
= 0.573 * (M * D) / (rho * R-cubed)

where rho is the density of the object on which the tides are being raised.

Thanks very much for giving this attention.

[thtraveller]

Simplistically I thought it would be 2 / (4/3 * pi * density)
from
D=2R
and P = density * 4/3 * pi *R^3

So I thought the factor would be adjusted by about 0.48/density. But my algebra isn't that strong and I need to be fresh to work through yours (and it's late at night for me).


I also note that basic tidal height for Earth is only around half a meter assuming uniform water depth all over the world, but actual tides are significantly higher due to terrain.

Thanks very much for giving this attention.

I just stumbled across this thread today, though I don't have the latest Space (I have a previous version and First In).

[Agemegos]

Originally Posted by thtraveller

Simplistically I thought it would be 2 / (4/3 * pi * density)
from
D=2R
and P = density * 4/3 * pi *R^3

I reckoned that

P = density * 4/3 * pi * R^3

= density * 4/3 * pi * (D/2)^3

= density * 4/3 * pi * D^3 / (2^3)

= density * 4/3 * pi * D^3 / 8

= density * 4/24 * pi * D^3

= density * 1/6 * pi * D^3

So the factor is 6/(pi * density)

But now I realise that that is not right. we're doing density in Earth masses per Earth volume, not Earth masses per cubic Earth diameter. And that means that the mass of a sphere is M = rho * D^3. The pis and other constant factor are taken up in our daffy unit system.

:steamears:

Physics is a lot easier in consistent unit systems.

[Agemegos]

Originally Posted by thtraveller

I also note that basic tidal height for Earth is only around half a meter assuming uniform water depth all over the world, but actual tides are significantly higher due to terrain.

That's the amplitude of the equilibrium deep-water tide. The range is of course twice the amplitude.

About 8/11 of the tides on Earth are produced by the Moon, about 3/11 by the Sun.

<edit>About 11/16 of the tides on Earth are produced by the Moon, about 5/16 by the Sun.