08052006, 08:08 AM  #1 
Join Date: May 2005
Location: Oz

[SPACE] Tidal braking erratum
On p. 117 of Space, in Step 30 of the planet generation sequence, we estimate the tidal distortion of a planet as T = (0.46 * M * D) / R^3, (where M is the mass of the star, D is the diameter of the world, and R is the radius of the world's orbit). And the total tidal effect (amount of slowing) is E = (T * A)/ m (where A is the age of the system and m is the mass of the world). That expression expands to E = (0.46 * M * D * A) / (R^3 * m).
Now suppose that the system is only just old enough that the planet has tidelocked. Substituting E = 50 (the critical value for tidelock), we get (0.46 * M * D * A) / (R^3 * m) = 50, so A(lock) = (109 * R^3 * m) / (M * D) That's the time to tidelock in GURPS Space Look at the formula for time to tidelock given in Wikipedia. This is t(lock) = (w * R^6 * I * Q)/(3 * G * M^2 * k * (D/2)^5) (making the appropriate substitutions for the different notation.). I (moment of inertia of a spherical planet) is approximately = 0.1 m D^2. G, Q, and k are constants, the latter two having to do with the material of the planet. w is the initial rotation rate of the planet. Collect the constants t(lock) = (0.1 * Q)/(3 * G * k / 32) * (w * R^6 * D^2 * m)/(M^2 * D^5) = constant * (w * R^ 6 * m)/(D^3*M^2) Compare A(lock) = (109 * R^3 * m) / (M * D) Something is obviously wrong. I suspect that the authors of Space overlooked the following: that the tide height is proportional to teh tideraise force divided by the surface gravity; that the tidal torque is proportional not only to the height of the bulges, but to the gravity gradient (0.5*G*M/r^3) and the moment arm (D times the sin of the angle between the bulges and the starplanet line); and that the tendency of the planet to resist rotation, its moment of inertia, is proportion to its mass times the square of its radius. If I (and Wikipedia) are right, we ought ot replace the current expression for E with E = (0.46 * A * D^3 * M^2) / (R^6 * m) Which should make it a lot easier to get planets aroung type K stars that are not tidelocked. Which in turn should be greatly to Zorg's relief. Last edited by Agemegos; 10302010 at 08:33 PM. Reason: compacted the layout to improve readability 
08052006, 11:22 PM  #2 
Wielder of Smart Pants
Join Date: Aug 2004
Location: Ventura CA

Re: [SPACE] Tidal braking
You should PM (or email) Jon F. Zeigler.

08122006, 01:15 AM  #3 
Join Date: May 2005
Location: Oz

Re: [SPACE] Tidal braking
Here is a website that gives a comprehensible derivation for the equilibrium height of the tidal bulge. Its result is
h = (3 * M * r^4)/(m*R^3), where M is the mass of the tideinducing body, r is the radius of the world, m is the mass of the world, and R is the distance to the tideraising body. For simplicity, we want this in the form h = k * (M * D^4)/(m*R^3) To find k in terms of the relative units in Space we substitute the values for the EarthMoon system, noting that lunar tides produce about 70% of the observed tidal range: 0.7 = k * (0.0123 * 1^4)/(1*30.1^3) k = 1.55 million. A moon with mass M times Earth's mass circling a planet with mass P times Earth's mass and diameter D times Earth's diameter at distance R times Earth's diameter will raise a tide T times as high as the tides on Earth, where T = 1.55 million * (M * Dtothefourth) / (P * Rcubed) To find k in terms of the other set of relative units used in Space we subsitute the values in the EarthSun system, noting that solar tides make up about 30% of the tidal range on Earth: 0.3 = k * (1 * 1^4)/(1 * 1^3) k = 0.3 A planet with a mass P times Earth's mass and D times Earth's diameter orbiting a star of mass M solar masses at a distance of R astronomical units will experience a solar tide with a range T times as high as the total tidal range on Earth, where T = 0.3 * (M * Dtothefourth) / (P * Rcubed) Unless anyone has something to add I'll submit a corrigendum to errata@sjgames.com in a couple of days. Last edited by Agemegos; 10132007 at 11:47 PM. 
08122006, 02:33 AM  #4 
Join Date: May 2005
Location: Oz

Re: [SPACE] Tidal braking
I have been carefully through the derivation and got a result that is consistent with the result in Wikipedia.
The mass of a tidal bulge is proportional to the area of the bulge (D squared) times its height (h), times the density of the distorted body (P). That comes out to mass of tidal bulge = P * h/D The tidal acceleration is GMr/R^3, or since we are ignoring constants tidal acceleration = MD/R^3 F = ma, so Tidal braking force = h * M * P / R^3 Torque is force times moment arm, and for a given angular displacement of the tidal bulge the moment arm is proportional to D torque = h * M * P * D / R^3 The moment of inertia of a sphere is 0.4 * D^2 * m, so ignoring constants I = D^2 * P angular acceleration = torque/I = (h * M) / (R^3 * D) substituting h = (M * D^4)/(P * R^3) gives angular deceleration = (M^2 * D^3) / (P * R^6) E = A times angular acceleration E = (k * A * D^3 * M^2) / (R^6 * P) Woot! I love it when a plan comes together! Now there remains to chose the most convenient form for the equation and estimate the constant of proportionality. Given that the Ms and Rs are in different units according to whether we are calculating lunar or solar tides, but the D and P are always the same, the most convenient form might turn out to be angular deceleration = h^2 * P/D^5 Which gives E = (k * A * m/D^5) * sum of squares of tide heights. Now to estimate k. Last edited by Agemegos; 08122006 at 02:37 AM. 
08122006, 06:05 AM  #5  
Join Date: May 2005
Location: Oz

Re: [SPACE] Tidal braking
Quote:
1 = (k * 4.5 * 1/1^5) * (0.7^2 + 0.3^2) k = 0.383 revolutions per day. So, the reduction in spin rate E = (0.383 * A * P/D^5) * sum of squares of tide heights in REVOLUTIONS PER DAY lost from the original rotation speed of, say 3d/5 revolutions per day. Last edited by Agemegos; 10132007 at 11:59 PM. 

08142006, 05:36 AM  #6  
Experimental Subject
Join Date: Jan 2005
Location: saarbrücken, germany

Re: [SPACE] Tidal braking erratum
Quote:
A small Rock Planet on an inner orbital radius which got a Tidal Effect of 467,82 with the system in Space came out with a Tidal Effect of 1,37482470000e12 using your formula. I have no idea whether this is better, since I am unable to make sense of this number... Could you perhaps post the changed formulas together, in an easy to understand post, so that even a numberimpaired guy like me can use them? (iow, without your reasoning  just the formulas and a note what each variable is supposed to be, if it differs from Space p.117). I'm afraid I got something mixed up here... Before I tackle the remaining notes I have here, I would like to be sure that I understood/used your formulas correctly. I feel a little slow :
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Last edited by zorg; 08142006 at 06:17 AM. 

08142006, 07:44 AM  #7  
Join Date: May 2005
Location: Oz

Re: [SPACE] Tidal braking
Quote:
T = 1.56 million * (M * Dtothefourth) / (P * Rcubed) Where M is the mass of the object raising 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). T is the range of the tides in terms of the total tidal range on Earth, ie. if T is 1, tides will be about the same as on Earth; if T is 2 the tides will be about twice the size of the tides on Earth. When calculating the height of the tides raised by a star on a planet or the moon of a planet: T = 0.3 * (M * Dtothefourth) / (P * Rcubed) Where M is the mass of the star in solar masses, D is the diameter of the planet (or moon) in Earth diameters, P is the mass of the planet (or moon), and R is the distance between them in AU. T is the range of the tides in terms of the total tidal range on Earth, ie. if T is 1, tides will be about the same as on Earth; if T is 2 the tides will be about twice the size of the tides on Earth. 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 5. The result is the initial rate of rotation of the body (in revolutions per Earth day). 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 minimum rate of rotation, w, as follows. W = I  [(0.536 * A * P/D^5) * S] Now calculate the maximum daylength max_Daylength =1/W, in Earth days. Compare this to the world's orbital period around the body that raises the highest tide on it, or it's orbital period around the body that raises teh highest tide on it. If the orbital period is shorter or the rotational period is negative, then the world is tidelocked to that body. Set the rotational period equal to the orbital period. Last edited by Agemegos; 10132007 at 11:53 PM. Reason: updating figures 

12282006, 05:06 PM  #8 
Join Date: May 2005
Location: Oz

Re: [SPACE] Tidal braking
It looks as though noone cares to get this right. I suggested it as an erratum four months ago, and nothing has happened.
Last edited by Agemegos; 11142007 at 12:20 AM. 
12282006, 05:09 PM  #9 
Join Date: Aug 2004

Re: [SPACE] Tidal braking
I appreciate the effort put into this, and if I ever get a copy of Space I'll be using your erratum wether official or not.

12292006, 05:48 PM  #10 
Join Date: Nov 2005
Location: Seattle, Washington

Re: [SPACE] Tidal braking
For what it's worth, I put your formula into my PHP star system generator...
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Tags 
planets, space, system generation, tidal braking, tide, tidelocked, world generation 
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