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-   -   [SPACE] Tidal braking (https://forums.sjgames.com/showthread.php?t=18435)

Schedim 08-22-2009 01:24 PM
Re: [SPACE] Tidal braking
 
This thread is very interesting, aside from the problem to get the results to the powers that be ....

Has anybody an idea of how much the angle of the axis will reduce the tidal breaking?

Also the suggested forumula seems to give a lot of tidal effects to the moonlets in low orbit!

nick012000 08-22-2009 06:43 PM
Re: [SPACE] Tidal braking
 
Heh. Interesting thread. I'd help out, but astrophysics isn't my forte. ;)

nick012000 08-30-2009 09:40 AM
Re: [SPACE] Tidal braking
 
Okay, having used these rules, I've got to be a bit curious as to their tendency to produce extreme outcomes. Here's a gas giant and its moons I generated for Alpha Centauri B in a game I'm running:

Code:
Small Gas Giant, 3.6 AU
*Black Body Temperature=123
*Period = 7.2 years
*Mass: 20
*Density: .22
*Diameter: 4.49
*Surface Gravity: .98
*Age: 5 billion years
-3 Moonlets
-2 Major Moons
--Standard (Ice)
**Atmosphere: Suffocating, Thin
**Atmospheric Mass: 1.1
**Hydrographic Coverage: 18%
**Climate: Very Cold
**Density: 1.1
**Diameter: .54
**Surface Gravity: .59
**Mass: .17
**Atmospheric Pressure: .65
**Orbital Radius: 22.45
**Orbital Period: 1.39 days
**Tidal Height: 1,370
**Inital Days/Earth Day: 2.4
**Day Length: .44 days
**Moderate Volcanic Activity
**Light Tectonic Activity
**RVM: +0
**Habitability: 1
**Affinity: 1
--Small (Ice)
**Atmosphere: Suffocating, Mildly Toxic, Very Dense
**Atmospheric Mass: 1.1
**Hydrographic Coverage: 61%
**Climate: Very Cold
**Density: .5
**Diameter: .43
**Surface Gravity: .215
**Mass: .04
**Atmospheric Pressure: 2.34
**Orbital Radius: 28.93
**Orbital Period: 2.07 days
**Tidal Height: .0007
**Volcanic Activity: None
**Tectonic Activity: None
**RVM: +0
**Habitability: -1
**Affinity: -1
-5 Moonlets
As you can see, it tends to produce rather extreme tides. Two worlds, not too different, had five to six orders of magnitude difference in their tides- the bigger one had tides of over two kilometers, and only lost a fraction of a day to tidal effects!

Figleaf23 08-30-2009 02:45 PM
Re: [SPACE] Tidal braking
 
I'm curious -- do the tide height calculations assume there is surface water? Does there have to be surface liquids? What about other substances like a sea of amonia -- would it matter?

Agemegos 08-30-2009 07:48 PM
Re: [SPACE] Tidal braking
 
Quote:
Originally Posted by Figleaf23 (Post 842762)
I'm curious -- do the tide height calculations assume there is surface water? Does there have to be surface liquids? What about other substances like a sea of amonia -- would it matter?
What is calculated is the equilibrium tide height, which is the tidal height that the surface will tend to given indefinite time. A liquid surface will approach the equilibrium fairly quickly, mantle material will do so only over tens of thousands of years. If you have a mixed surface of solids and liquids the oceans will tend to conform much more rapidly than the land, and you will get the phenomena that we know as tides: the oceans rising and falling with respect to the land. That will involve sloshing and (in some bodies of water) resonance effects. These can result in local tides in some bays and straits much larger than the equilibrium tidal range. For example, the equilibrium tide on Earth has an amplitude of about 0.6 m, whereas some bodies of water such as the Bay of Fundy and King Sound have tidal amplitudes ten of fifteen times that.

Agemegos 08-30-2009 08:08 PM
Re: [SPACE] Tidal braking
 
Quote:
Originally Posted by nick012000 (Post 842656)
As you can see, it tends to produce rather extreme tides. Two worlds, not too different, had five to six orders of magnitude difference in their tides- the bigger one had tides of over two kilometers, and only lost a fraction of a day to tidal effects!
One or the other of us is making arithmetic errors. I get a tide height of 1 935 for the inner moon and 1 584 for the outer moon. Even ignoring the solar tides 'E' would be 24 063 631 for the inner moon and 1 248 421 for the outer one, implying that they are rigidly tide-locked as you would expect.

Quote:
Originally Posted by Brett
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)....)
<edit> T = 2.23 million * (M * D) / (density * R-cubed)
M = 20, D1 = 0.54, density1 = 1.1, R1 = 22.45

T1 = 2.23 million * (20 * 0.54) / (1.1 * 22.45 ^ 3)

My calculator gives

T1 = 1 935.026 413

Considering the outer moon

M = 20, D2 = 0.43, density2 = 0.5, R2 = 28.93

T2 = 2.23 million * (20 * 0.43) / (0.5 * 28.93 ^ 3)
= 1 584.119 752
Quote:
E = 0.184 * sum of squares of tide heights * A * density/D^2[/indent]
You can ignore the solar tides since the planetary tides are enormous and will obviously dominate the sum of squares.

E1 = 0.184 * T1^2 * A * density1 / D1 ^ 2
= 0.184 * 1 935 ^ 2 * 5 * 1.1 / 0.54 ^ 3

= 24 063 363.69
E2 = 0.184 * T2^2 * A * density2 / D2 ^ 2
= 0.184 * 1584 ^ 2 * 5 * 0.5 / 0.43 ^ 3

= 2 903 306.023

Figleaf23 08-30-2009 08:39 PM
Re: [SPACE] Tidal braking
 
Quote:
Originally Posted by Brett (Post 842878)
What is calculated is the equilibrium tide height, which is the tidal height that the surface will tend to given indefinite time. A liquid surface will approach the equilibrium fairly quickly, mantle material will do so only over tens of thousands of years. If you have a mixed surface of solids and liquids the oceans will tend to conform much more rapidly than the land, and you will get the phenomena that we know as tides: the oceans rising and falling with respect to the land. That will involve sloshing and (in some bodies of water) resonance effects. These can result in local tides in some bays and straits much larger than the equilibrium tidal range. For example, the equilibrium tide on Earth has an amplitude of about 0.6 m, whereas some bodies of water such as the Bay of Fundy and King Sound have tidal amplitudes ten of fifteen times that.
Thank you, that's informative. Just to be sure -- does this add up to saying that it does not matter whether or what oceans actually are there?

Agemegos 08-30-2009 08:52 PM
Re: [SPACE] Tidal braking
 
Quote:
Originally Posted by Figleaf23 (Post 842895)
Thank you, that's informative. Just to be sure -- does this add up to saying that it does not matter whether or what oceans actually are there?
For the purposes of tidal locking? Well, really it does. But the effects are very difficult to calculate from first principles. Tidal braking depends on how squishy the mantle of a planet is, and that depends on what material it is made of, how hot it is, et. And in fact a planet shallow or partial oceans brakes more rapidly than a planet with deep oceans or none. The configuration of the ocean bottoms matters, and how many continents straddle the equator matters too.

A planet with no oceans will tide-lock. All the major moons in the solar system have done so. It will do so more slowly than one with oceans. No-one is really sure how much more slowly. The formula I gave for tidal slowing is based on an estimate of how much Earth has slowed It should be approximately correct for a habitable planet to the extent that habitable planets are more-or-less Earth-like. It might be off for a world with no oceans. How much? I don't know. A factor of two? An order of magnitude? I don't know.

Figleaf23 08-30-2009 09:29 PM
Re: [SPACE] Tidal braking
 
yEeeRg!

I'm adding GURPS SPACE: Lite to my personal pdf wishlist.

Pomphis 08-31-2009 04:08 AM
Re: [SPACE] Tidal braking
 
Ger a copy of Space 2e. It had nice and simple rules to roll up systems. Of course, they were probably considerably less realistic.


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