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Old 05-20-2008, 11:33 PM   #31
Agemegos
 
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Default Re: [SPACE] Tidal braking

No, dammit, that's not right. The unit for tidal range/amplitude in Space is the lunar tide on Earth, not the total tide. So the figure for lunar tide has to work out to 1.

Now let's see. Earth's diameter and density are both 1 by definition. The mass of the Moon is 0.0123, and the radius of its orbit is 30.16 Earth diameters. So

T = x * (M * D)/(rho * R^3)

1 = x * (0.0123 * 1)/(1 * 30.16^3)

x = 30.16^3 / 0.0123

= 2.23 million.

T (lunar) = 2.23 million * (M * D)/(rho * R^3)

All the variables are 1 by definition in the Solar tide on Earth, but since the tidal range unit is the Lunar tide not the total tide the coefficient ought to be 0.38, not 0.3.
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Last edited by Agemegos; 05-20-2008 at 11:54 PM.
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Old 05-20-2008, 11:50 PM   #32
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Default Re: [SPACE] Tidal braking

As for tidal braking,my mistake with the unit of tidal height means that the sum of squares of tide heights on Earth is 1.14, not 0.58 as I thought.

<edit> Actually, it's 1.21.

Earth's initial day length was about 12 hours, so it has lost 1 rotations/day over 4.5 billion years.

E = (x * A * M/D^5) * sum of squares of tide heights

1 = (x * 4.5 * 1/1^5) * 1.14

x = 1 / (4.5 * 1.14)

= 0.195

<edit> Correct result is 0.184

E = (0.195 * A * M/D^5) * sum of squares of tide heights

<edit> Correct formula is
E = (0.184 * A * M/D^5) * sum of squares of tide heights.</edit>
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Last edited by Agemegos; 06-28-2008 at 02:57 AM.
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Old 05-21-2008, 02:26 AM   #33
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Default Re: [SPACE] Tidal braking

Quote:
Originally Posted by Agemegos
That's the amplitude of the equilibrium deep-water tide.
Isn't the drag increased depending on the amount of shallow water and topology of the bottom? The moon's current drag is higher than in some earlier periods. The 3.8cm per year Lunar recession is quite high.
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Old 05-21-2008, 06:28 AM   #34
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Default Re: [SPACE] Tidal braking

Quote:
Originally Posted by thtraveller
Isn't the drag increased depending on the amount of shallow water and topology of the bottom?
Yes, also depending on the Young modulus and dissipation function of the mantle, and something called the tidal Love number. I gather that these are poorly understood, and that deriving the tidal braking rate from first principles is challenging.
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Old 05-21-2008, 09:01 AM   #35
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Default Re: [SPACE] Tidal braking

Quote:
Originally Posted by Agemegos
Fair enough. I just have a paranoid fear of typographers' changing a lowercase m to an uppercase M for appearance's sake. It would be just like them to do so because the editor had declared the style of the work to use capitals in all formulas...
"Style guides" are the bane of communication.
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Old 05-21-2008, 11:22 AM   #36
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Default Re: [SPACE] Tidal braking

Quote:
Originally Posted by Agemegos

I'd really, really like it if a few physicists, astronomers, and astrophysicists would check my maths in the posts above.
This is probably old news for you that are whizzes at this, but I've seen this site posted on this BBS...

http://www.bautforum.com/

Not sure if that might be another place to get some feedback or not.






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Old 05-22-2008, 07:18 PM   #37
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Default Re: [SPACE] Tidal braking

Quote:
Originally Posted by Agemegos
E = (0.195 * A * M/D^5) * sum of squares of tide heights
Perhaps this is less confronting in the form
E = (0.195 * A * density/D^2) * sum of squares of tide heights
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Old 05-23-2008, 05:42 AM   #38
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Default Re: [SPACE] Tidal braking

Quote:
Originally Posted by Agemegos
E = (0.195 * A * density/D^2) * sum of squares of tide heights
I have no ideas whether the formulas are correct. But I do type them into a pocket calculator. So how about:

E = (0.195 * A * density/D^2) * (sum of squares of tide heights)

to make a little bit more obvious where the brackets go.

BTW, ease for typing into a pocket calculator is something that always interests me.
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Old 05-23-2008, 05:34 PM   #39
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Default Re: [SPACE] Tidal braking

Quote:
Originally Posted by Pomphis
I have no ideas whether the formulas are correct. But I do type them into a pocket calculator. So how about:

E = (0.195 * A * density/D^2) * (sum of squares of tide heights)

to make a little bit more obvious where the brackets go.
Well, since the sum of squares of tide heights is the result of a previous calculation, the best might be to put it first:
E = (sum of squares of tide heights) * 0.195 * A * density / (D^2)
Quote:
BTW, ease for typing into a pocket calculator is something that always interests me.
Commendable.
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Old 05-29-2008, 08:00 PM   #40
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Default Re: [SPACE] Tidal braking

For those of you who have been following this:

I've finally found the time to start re-evaluating . . . well, pretty much the entirety of p. 117 of Space. I've got most of Agemegos' math in front of me and have been trying to confirm it, with some success already. I think he's on the right track even if I'm not sure of all the details yet.

We may end up with a full-on rewrite of that page of the book. SJG errata coordinator tells me that this shouldn't present a problem, as long as we're satisfied that the results are correct.
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