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Why Mars has no Plate Tectonics ?


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1. Simple Model of Terrestrial-Planetary Structure

 

Advanced Computer Simulations of Planetary Structures give the relationship, between Planet Radius vs. Planet Mass, for Terrestrial Planets*. For Terrestrial Planets, of sufficiently small Planetary Mass (M [math]\leq 10 \; M_{\oplus}[/math]), this result can be linearly approximated accurately. For such Rocky Worlds, of Earth-like Composition, the Average Bulk Density ([math]\rho[/math]) increases nearly linearly with Surface Gravity (g):

[math]\rho = \rho_{0} + k \times g[/math]
(eq. 1)

where, using Earth-normalized units:

[math]\rho_{0} \approx 0.536[/math]
[ 2.96 g cm
-3
]

[math]k \approx 0.464[/math]
[ 1 - p0 ]

Note that [math]\rho_{0}[/math] represents the natural "uncompressed density", of Terrestrial-type rocks, at zero Surface Gravity, consistent with the observed density of Asteroids**.

*
Carroll & Ostlie.
Introduction to Modern Astrophysics
(
1996 ed
.), pg. 825.
See also
:
Attachment 1
.

**
Carroll & Ostlie.
ibid
., pg. 878.
Asteroid
Ida has a density of 2.2 - 2.9 g cm
-3
. Our "uncompressed density" is slightly higher, as we are here dealing with solid bodies, whereas
Asteroids
are comparatively uncompacted
Rubble Piles. See also
:

 

 

2. Estimation of Terrestrial-Planetary Raleigh Convection Numbers

 

Earth's Plate Tectonics are driven by massive Convective Currents in the Earth's Mantle*. Such Convective Systems are characterized by the dimensionless quantity called the Raleigh Convection Number (Ra)**, defined as:

[math]Ra \equiv \frac{\Delta \rho \; g \; L^{3}}{D \; \mu}[/math]

where:

[math]\Delta \rho[/math] is the density contrast across the convective region

g is the local gravitational acceleration

L is the characteristic length-scale of convection

D is the
Diffusivity

[math]\mu[/math] is the
Dynamic Viscosity

The Dynamic Viscosity ([math]\mu[/math]), being complicated function of Temperature, will be discussed in an Appendix to this post. The Diffusivity (D) is a function of the Thermal Conductivity & Specific Heat Capacity of the convecting material, assumed to be constant for Terrestrial-type Bulk Compositions, and inversely proportional to the density of said material***.

*
History Channel
How the Earth was Made -- The Alps
[TV]

**

***

Thus, for our model of Planetary Convection, for a presumed planet of density [math]\rho[/math], radius R, Surface Gravity g, and mass M, we adopt:

[math]D \propto \rho^{-1}[/math]

 

[math]L \propto R[/math]

 

[math]\Delta \rho \equiv \rho - \rho_{0} = k \times g[/math]

so that:

[math]Ra \propto \rho \; \Delta \rho \; g \; R^{3} \propto \rho \; g^{2} \; R^{3} \propto M \; g^{2}[/math]

where we have used the basic identity that [math]M = \rho \times R^{3}[/math] (Earth-normalized units). Or, noting that [math]g = \rho \times R[/math] (Earth-normalized units), we have that:

[math]Ra \propto \frac{g^{5}}{\rho^{2}} = \frac{g^{5}}{\left( \rho_{0} + k \times g \right)^{2}} \approx 4 \; \frac{g^{5}}{\left( 1 + g \right)^{2}}[/math]
(eq. 2)

where we have observed that both [math]\rho_{0}[/math] and k are each roughly equal to one-half*.

*
M is cubic in g, making impossible a simple closed-form solution, for Ra, in terms of M alone.

 

 

3. Analysis

 

Eq. 2 is a steeply increasing function of Surface Gravity. For Mars (M = 0.1, g = 0.4), [math]Ra \approx 0.016[/math] (Earth-normalized units). Thus, Mars' lower density ([math]\rho[/math]), lower density contrast ([math]\Delta \rho[/math]), lower Surface Gravity (g), and smaller size ®, all combine to make Mars two Orders-of-Magnitude less convective than Earth. This is consistent with the observed lack of Plate Tectonics upon that planet.

 

Note that a hypothetical Super-Earth ([math]g = 2, M \approx 3.6[/math]) would have a Raleigh Convection Number of [math]Ra \approx 14[/math]. Thus, such Super-Earths could quite conceivably be much more convective than Earth.

Planet_Radii_vs_Mass_medium.jpg

Edited by Widdekind
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Venus is thought to lack plate tectonics, lack of water as a lubricant between plates and possibly lack of a large moon have been suggested as possible reasons for this lack. Is it possible to have a large diameter planet with a similar density to Mars? If so would it be likely to have plate tectonics?

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Judging from my attached figure (from Carroll & Ostlie), such a large-but-low-density Planet would probably be composed of Ices. To wit, it would be a "Super Pluto".

 

Perhaps, then, you could have Convection Currents of water, or various Ices. To wit, like Europa. But, according to Eq. 2 ([math]Ra \propto M \; g^{2}[/math]), Europa's Raleigh Convection Number is 0.00015, roughly 1% of that of Mars.

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so a planet made of exactly the same elements [as] Mars in the same proportions would be dense as the earth if it was much bigger than mars?

 

Yes -- the rock would compress under its own Self Gravity. Look at Earth itself -- our Iron Core has a density of ~11, yet, Iron on the Surface only has a denisty of ~8. Why is core Iron so much more dense? B/c it's compressed, by incomprehensible pressures.

 

The Moon, Mars, Venus, and Earth are all of (basically) comparable compositions -- b/c they all formed in the same liquid water "Climate Zone" around the Sun -- but their Bulk (Average) Densities increase in the same order as their Masses, and Self-Gravities.

 

Mercury is comparatively over-dense... but it formed so close to the Sun, that liquid water turned to steam vapor. Thus, Mercury lacks allot of Volatiles compared to the other Rocky Planets, so that it is composed of greater concentrations of heavy, dense elements like Nickel & Iron.

 

In the other direction, the Gas Giants formed out past where liquid water turns to ice. This increased the density of solid materials in the Proto-Solar Disk, allowing the Gas Giant proto-planets to get much bigger -- big enough to keep their Primordial H/He Atmospheres.

 

To brazenly coin terms, then, the Sun is surrounded by "Climate Zones" -- "steam zone" (Mercury), "water zone" (Venus, Earth, Moon, Mars), "ice zone" (Jupiter, Saturn Gas Giants), "methane-ice zone" (Uranus, Neptune Ice Giants). Within a single such "Climate Zone", the planets all tend to have similar overall average Bulk Compositions. And, all common materials compress under applied pressures.


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TANGENT: Looking again at the attached figure, it seems that Uranus & Neptune are only slightly larger, in Radius, than purely solid Pluto-like Ice Planets of the same Mass. This suggests that Uranus & Nepture are "almost" giant Plutos -- to wit, that their Atmospheres are relatively thin, and that "most" of those Ice Giants are actually solid Icy Cores. Conversely, Jupiter & Saturn are accurately described by the Gaseous H/He curve, suggesting that they are, in fact, true Gas Giants possessing only comparatively small solid cores.

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So a huge planet with a small iron core composed of mostly granite like materials as proposed by Asimov would still be as dense as the earth? He seemed to think it would maybe 12,000 miles in diameter with a surface gravity close to earth normal but with a much denser atmosphere due to the less steep gravity gradient.


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I thought the general consensus was that Uranus and Neptune were worlds of extremely hot water oceans thousands of miles deep over Earth sized rocky cores. Far to hot to contain ice of any kind.

Edited by Moontanman
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OBSERVATION:

 

The Relative Raleigh Convection Score of Venus (M ~ 0.8, g ~ 0.9) is only 0.66. Thus, even "Earth's twin", being only slightly smaller, already drops down to ~2/3rds of the "convective vigour" of Earth.

 

According to Wikipedia, Plate Tectonics on Venus ended between 1,200 to 500 million years ago. Therefore, Venusian Plate Tectonics lasted for 3.4 to 4.1 billion years, from the formation of the Solar System. If Terrestrial Plate Tectonics last 3/2 as long, we should expect them to cease after 5-6 billion years (~0.5 to 1.5 billion years from now).

 

 

 

ANSWERING QUESTION:

I thought the general consensus was that Uranus and Neptune were worlds of extremely hot water oceans thousands of miles deep over Earth sized rocky cores. Far to hot to contain ice of any kind.

 

Uranus' "Solid Core" is composed of Rocks (~Silicates) & Ices (~Water). Uranus' "Solid Core" is differentiated, into chemically distinct layers, analogous to the Core & Mantle of the Earth. And, even as Earth's Core & Mantle are quite hot, so, too, is the central "Solid Core" of Uranus. Thus, the Ices melt, forming a vast High-Pressure Ocean (even as Earth's Inner Core is liquid, albeit, liquid Iron -- at high-pressure, to wit, a "High-Pressure Iron Ocean").

 

Never-the-less, the above figure strongly suggests that the bulks of Uranus & Neptune are comprised of Rocks & Ices, w/ comparatively thin Atmospheric Envelopes.

 

Earth is a Rocky World, w/ merely a salting of Ices. Uranus & Neptune are Rock-Ice Worlds, having Rocks & Ices in nearly equal measures.

 

Uranus is made up of a swirl of gases, liquids, and solids. It has no solid surface to stand on. The solids have settled to the center of the planet. These include silicon, iron, and a basalt like rock. The core of Uranus is about the size of Earth. At the outer edge of the core... is an ocean that is about 5,000 miles deep. It is thought to be mostly water and a mixture of liquid ammonia and methane. An interesting thing about Uranus' ocean, is that it is extremely hot-almost 8,000 degrees Fahrenheit. At the surface of this ocean, there is a gaseous layer that is 4,000 to 5,000 miles deep. This layer of atmosphere is mostly liquid hydrogen, methane and some helium. The methane gas gives Uranus its blue-green color. Above the gas layer is another layer made of hydrogen and helium.

 

Most planets orbit the sun by rotating and creating days and nights as Earth does. Uranus does not. Instead, its poles are on its sides. The planet spins from north to south instead of west to east. It takes it 17.25 hours to spin around completely. It is tilted so far on its side that it actually lies almost level with its orbital path around the sun. Most planets axis' tilt less than 30 degrees, but Uranus tilts at 98 degrees.

 

Traveling on an ovular path around the sun, it speeds through space at 4.25 miles per second. It takes 84 years for it to complete one orbit or one Uranus year. The mass of this planet is 14.5 times larger than that of Earth. Its density is about 1.27 grams per cubic centimeter [cf. Water ~ 1, Rock ~ 3] and the force of gravity is ninety percent of that on Earth. So if one were to weigh 100 pounds on Earth, he would weigh only 90 pounds on Uranus.

 

The atmosphere is composed of many gases. It is 83% hydrogen, 15% helium, 2% methane, and it has small amounts of ethane as well as other gases. The temperature of Uranus [Atmosphere] is about –355 degrees Fahrenheit. The interior temperature can rise fast and even reach up to 4,200 degrees in the ocean and 12,600 degrees at the core.

 

 

http://74.125.155.132/search?q=cache:zNM0K0PAfSYJ:library.thinkquest.org/C002416/uranus/index.htm+temperature+uranus&cd=3&hl=en&ct=clnk&gl=us


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This Journal Article deals explicitly with this topic.

Edited by Widdekind
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So the the surface of the ocean on Uranus is covered by ice? Even at the pressures involved the temps would seem to preclude ices. The temp of -355 is the top of the atmosphere, the top of earths atmosphere is less the the freezing point of water but you don't consider the earth to be an ice planet. or that ice is a significant part of it it's make up. be that as it may. I my main interest here is the possibility of a large planet with gravity close to earths IE a 12,000 mile planet with a surface gravity of less than 1.25X of Earths, for a story I am writing. I'd like for it to be plausible at least.

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"Rock" is the "frozen ice" form of Magma.

 

Likewise, "Ice" is the frozen form of Water.

 

So, just as Earth is a "Rocky World", even though its interior is molten Magma...

 

so, too, Uranus is a "Rock-Ice World", even though its interior is molten Magmas & liquid Water.

 

B/c of Uranus' super-thick, "ultra-Venusian" Atmosphere, its "Surface Temperature", down in at the base of its Atmosphere, is probably so high that the whole "Ice Mantle" is "molten" (ie, that it's a High-Pressure, High-Temperature, liquid water ocean).

 

As an analogy, consider Venus. Imagine that Venus' Atmosphere was even thicker, and its Green-House Effect even higher. You could imagine, then, that such a "super Venus" would not have a solid Rocky Crust... but, be covered entirely in a molten Magma Ocean (like the early Earth). Now, just replace the word "Rocky" for "Icy" -- the "surface" of Uranus is a "molten Ice (= water) ocean".

 

Ice is to Water

as

Rock is to Magma.

 

Earth's interior is melted Rock...

Uranus' interior is melted Ice

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I am pretty sure that in astronomy ice means simple compounds of Hydrogen and oxygen or carbon or nitrogen or sulfur when they freeze. Granite is not considered an ice.

 

http://en.wikipedia.org/wiki/Ice

So, any normally liquid or gaseous covalent compounds. Rocks and minerals are generally ionic compounds and thus are out of the equation.

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I am pretty sure that in astronomy ice means simple compounds of Hydrogen and oxygen or carbon or nitrogen or sulfur when they freeze. Granite is not considered an ice.

 

In astronomy, everything with an atomic number higher than 2 is considered a "metal." :eek::rolleyes::D

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  • 3 weeks later...
http://en.wikipedia.org/wiki/Ice

So, any normally liquid or gaseous covalent compounds. Rocks and minerals are generally ionic compounds and thus are out of the equation.

 

That's true, and those Chemical Properties determine specific details, like Melting Points & Heats of Fusion. However, I'm only speaking in generalities -- water is "molten ice", much as magma is "molten rock". And, icy worlds will have "molten ice", much as rocky worlds will have "molten rock".

 

NOTE: From the above figure, you can estimate the size of Uranus & Neptune's Rock-Ice Cores, by simply projecting them, straight down, onto the Rock-Ice Curve. That amounts to keeping the mass constant, but reducing the radius. This is a "zeroth order" approximation, implicitly assuming that the atmosphere doesn't mass much.

 

Visually, it looks like both Uranus & Neptune's cores are roughly 1/2 [math]R_{\oplus}[/math] smaller in radius that the whole planets. Since both of these "hidden planets" are quite big (~ 10 [math]M_{\oplus}[/math], ~ 3 [math]R_{\oplus}[/math]), perhaps they are still Geologically Active, deep beneath their thick protective Green House Atmospheres. Perhaps they still undergo "Rock-Ice Volcanism", and the like. Perhaps they could be used to study, and model, Geological Activity on Terrestrial-type planets too.

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Eq. 2 is a steeply increasing function of Surface Gravity. For Mars (M = 0.1, g = 0.4), [math]Ra \approx 0.016[/math] (Earth-normalized units). Thus, Mars' lower density ([math]\rho[/math]), lower density contrast ([math]\Delta \rho[/math]), lower Surface Gravity (g), and smaller size ®, all combine to make Mars two Orders-of-Magnitude less convective than Earth. This is consistent with the observed lack of Plate Tectonics upon that planet.

There are two problems with this paragraph.

 

For the moment I'll accept that your simplified treatment produces a valid result. Schubert et al in Mantle Convection in the Earth and Planets, page 297, observe "Thus the ratio of the Rayleigh number Raq to the minimum critical Rayleigh number for upper mantle convection Raq,cr(min)is between r= Raq/Raq,cr(min) =3.6×10^3 and 1.2×10^4, depending upon the applicable velocity boundary conditions. If, as you say, the Martian condition will be two orders of magnitude less than the Earth, this is still at least one order of magnitude more than the value required to initiate convection. Therefore, based upon your own figures, Martian mantle convection is probable.

 

Secondly, you state there are no observed plate tectonics on Mars. This would be disputed by several authors.

 

During the aero-braking maneuver to place it in orbit, Mars Global Surveyor detected linear magnetic anomalies in the southern hemisphere of Mars. Researchers, using the data from two full Mars years of polar orbital data and with an improved technique to eliminate the effects of external fields, constructed a global map of remnant crustal magnetization (I note in passing that the strength of this magnetism is more than an order of magnitude greater than comparable magnetism on the Earth.)

 

The similarity between these parallel patterns of magnetic reversal on Mars and those flanking mid-ocean ridges on Earth, immediately suggested a Martian equivalent to sea floor spreading. The patterns are absent from large impact basins, such as Hellas and Argyre, and from the volcanic terrain of the Tharsis bulge. This is consistent with the loss of the Martian dynamo early in history, before the final Heavy Bombardment Phase and Tharsis volcanism, but after crustal formation.

 

The researchers provisionally identified two major faults from offsets in the magnetic patterns. The character of these faults shows them to be transform faults, rather than simple strike slip faults. “The great faults in Meridiani are consistent with the properties of transform faults and define an axis of rotation (23°S and 80.5°E) describing the relative motion of two plates, north and south of the equator. The separation of the faults (1,200km) and offset of the putative ridge axis (+/-240km) in Meridiani are comparable with what is observed along ocean ridges on Earth."

 

There is also work by Sleep interpreting the Northern Plains/Southern Highland intersection as arising from plate tectonics, while Zhong and others have published widely on the proposal that “degree-1 mantle convection induced by a layered viscosity structure may be responsible for the formation of the crustal dichotomy.” and that this requires a mobile lid, rather than stagnant lid scenario. (Reference for these last two hypothesis can be provided if required.)

 

(1) Connerney, J.E.P., et al. (1999) Global distribution of crustal magnetization discovered by the Mars Global Surveyor MAG/ER experiment. Science 284, 794–798.

 

(2) Connerney, J.E.P., et al (2005) Tectonic implications of Mars crustal magnetism. P.N.A.S. , 102, 14970–14975

Edited by Ophiolite
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Mars' Planetary Magnetic Dynamo, generated by Core Convection, apparently dissipated by about 3.8 billion years ago:

Astronomers believe that these craters formed when chunks of debris pelted the inner solar system some 3.8 billion years ago. As old as these craters are, the global magnetic field that once existed on Mars must have vanished before they formed, Acuna asserts.

And, the collapse of Core Convection is likely linked to the collapse of Plate Tectonics (from Mantle Convection), at about the same time:

The younger, northern lowlands of Mars show no evidence of striping, and much less of the crust appears to be magnetized, Connerney says. These observations suggest that both the Martian magnetic field and plate tectonics had died away before volcanic activity melted and resurfaced this vast region of the planet.

 

It's possible that the fading of the global magnetic field and plate tectonics are intimately linked, Connerney speculates. When the interior lost so much heat that it could no longer power the dynamo, it may also have had too little energy to drive plate tectonics.

Thus, on Mars, all Convection Currents completely stopped by about 3.8 billion years ago, when Mars was merely ~3/4ths of a billion years old.

 

 

REVISION of THEORY:

 

Mars being some ~60 times less Convective than Earth (Ra ~ 0.016), Convection-driven processes, and in particular Plate Tectonics, stopped on Mars much sooner than upon this planet. To wit, the Martian Mantle did once Convect, but only in the deep past.

 

 

REFERENCES:

Edited by Widdekind
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...

During the aero-braking maneuver to place it in orbit, Mars Global Surveyor detected linear magnetic anomalies in the southern hemisphere of Mars. Researchers, using the data from two full Mars years of polar orbital data and with an improved technique to eliminate the effects of external fields, constructed a global map of remnant crustal magnetization (I note in passing that the strength of this magnetism is more than an order of magnitude greater than comparable magnetism on the Earth.)

 

...

 

According to the Science News article I cited:

Surveyor detected several magnetized patches of terrain, some with fields as strong as 400 nanoteslas, or 1.3 percent of Earth's field (SN: 10/18/97, p. 246).

According to a diagram from the NASA article I cited, the maximum Magnetic Field Strength the MGS detected was 1200 nT (roughly 4% [math]B_{\oplus}[/math]).

 

CONCLUSION (?): While Mars' Remnant Magnetism is much weaker than Earth's Dynamo-driven Planetary Magnetic Field, it's still much greater than Earth's own Remnant Magnetism.

 

Does this imply, that Earth's Planetary Magnetic Field inhibits Crustal Magnetism upon this planet ?

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  • 1 month later...

Evidence of ancient Oceans & Plate Tectonics on Venus:

 

http://news.yahoo.com/s/space/20090714/sc_space/oceansonancientvenusstudysuggests

 

The new map shows that the rocks on the Phoebe and Alpha Regio plateaus are lighter in color and look old compared to the majority of the planet. On Earth, such light-colored rocks are usually granite and form continents.

 

Granite is formed when ancient rocks, made of basalt, are driven down into the planet by shifting continents, a process known as plate tectonics. The water combines with the basalt to form granite and the mixture is reborn through volcanic eruptions.

 

"If there is granite on Venus, there must have been an ocean and plate tectonics in the past," Müller said.

 

See also Intelligent Life in the Universe by Peter Ulmschneider, pp. 71-72:

VenusTectonics.jpg

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  • 3 years later...

According to the definition of Bulk Modulus (K):

 

 

[math] \frac{\rho - \rho_0}{\rho_0} = \frac{P}{K}[/math]

 

 

For crude estimate, assuming a constant K, and static equilibrium:

 

 

[math] \frac{dP}{dr} = - g® \rho[/math]

[math] \frac{K}{\rho_0} \frac{d\rho}{dr} = - \frac{G M_{<r}®}{r^2} \rho®[/math]

 

But

 

 

[math] M_{<r}® \equiv \int_0^r 4 \pi r'^2 \rho(r') dr'[/math]

 

So, re-arranging terms, and taking the derivative, of the integral, to get the integrand:

 

 

[math] \frac{K}{\rho_0} \frac{d}{dr} \left( \frac{r^2}{\rho} \frac{d\rho}{dr} \right) = - 4 \pi G r^2 \; \rho[/math]

[math] \left( \frac{K}{4 \pi G \left(\rho_0 \; R \right)^2} \right) \frac{d}{dx} \left( \frac{x^2}{y} \frac{dy}{dx} \right) = - x^2 \; y[/math]

 

where we have normalized the radius coordinate value, by the radius of the world ( R ); and the density value, by the uncompressed "natural" density ( [math]\rho_0 \approx [/math] 3000 kg m-3 ). In words, that equation states:

 

 

(scaling factor) x (increase in density) = RHS

 

 

The equation is similar, for all worlds (of the same bulk composition, defining K, [math]\rho_0[/math]); but bigger worlds (larger R) have a much smaller scaling factor ( [math]\propto[/math] R-2 ), requiring much larger increases in density. Now, note:

 

[math] \boxed{ \alpha \equiv \left( \frac{K}{4 \pi G \left(\rho_0 \; R \right)^2} \right) }[/math]

[math]\alpha_{\oplus} \approx \frac{1}{3}[/math]

[math]\alpha_{moon} \approx 5[/math]

[math]\alpha_{mars} \approx 1[/math]

 

So, the planet Mars seems to be "transitory", between the "low-compression" regime vs. "high-compression" regime, i.e. of "moons" vs. "worlds" (for want of worthier words). And, Mars also seems transitory, between the inert Moon, and geologically active Earth. Perhaps plate tectonics somehow results, from "high compression" of rocky material ??

 

Note, that resembles supra-adiabatic compression, inside stars, which generates convection, i.e. material in the center is so compressed, that were it to expand and rise, it would still have more heat energy, i.e. temperature, than surrounding material, so being less dense, and so buoyantly rising. So, by analogy to stars, compression could conceivably cause convection, i.e. plate tectonics. On a P-T diagram, with adiabats over-drawn ( P [math]\propto[/math] T5/2 ); increasing the size ( R ) of the planetoid, and so decreasing its scale-factor ( [math]\alpha[/math] ), results in the world's "geotherms" ( P( r ) vs. T( r ) plotted parametrically, from the surface at R where P( R ) = T( R ) = 0 ) rising towards much higher densities (and, so, Pressures), until, with worlds bigger than Mars, their "geotherms" rise up above the adiabatic curve, so that central material tends to begin convecting.

 

The above equation, when numerically solved & plotted (with the "Wolfram Alpha" website, using initial values y'(1) = 0, y(1) = 1, i.e. at the surface (x = 1), material has its natural density (y=1), and is initially non-compressing (y'=0) ), appears perfectly plausible

Edited by Widdekind
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What promotes convection is when the densest material is on top and the lightest at the bottom - when material at the bottom expands from heat, be it through Sunlight (our atmosphere), radioactivity or maybe crystallisation (our Earth's core)...

 

The density difference to be injected in Rayleigh's equation, point 2 of the first post of this discussion, is measured horizontally across the convection cell and results from temperature.

 

Though, point 1 of the first post of this discussion injects a density difference computed vertically that results from pressure.

 

I disagree with this use, because hydrostatic pressure alone is not the cause sought for plate tectonics, and because a higher density at the bottom would kill convection instead of promoting it.

 

---------

 

The situation is slightly more complicated, especially for gases like our atmosphere. Convection needs that an air bubble that ascends continues to do so because it keeps warmer during its trip than the surrounding air, despite the dropping pressure of the bubble cools it as it rises.

 

So the proper condition is that the temperature profile of the surrounding atmosphere drops quicker with altitude than does the temperature of an ascending bubble - where this bubble follows with a good accuracy an adiabatic expansion, telling that P varies as T high Cp/R. This is the case of a hot bottom during a Summer afternoon.

 

Similarly in a planet's interior, the contribution to density resulting from depth cancels out as a convection bubble ascends and keeps the same pressure as its surrounds, with which it must be compared.

 

What stands is the contribution by the temperature gradient within the planet. Solids (and pastes...) are simpler here than gases, as compressing them heats them little; neglecting that completely, the criterium would just be "center warmer than surface" to make convection possible - where "possible" does not mean "efficient" nor "existing". Rayleigh's coefficient just compares how efficiently heat would be conducted away from a hot bubble, impeding its ascent.

 

You also have antagonist forces that block convection, especially the segregation among rocks and between rocks and metals, which have already put deeper the densest materials (resulting from the composition, not pressure nor temperature).

Edited by Enthalpy
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First, convection occurs in stars, and dredges dense material, from the deep interior, out to the rarified surface. That is the opposite of what Enthalpy stated, "convection is densest matter on top" (paraphrase).

 

Second, a one-zone model:

 

[math] \frac{dP}{dr} = - g® \rho[/math]

 

[math] \frac{K}{R} \frac{\Delta \rho}{\rho_0} \approx - \frac{G M}{R^2} \rho[/math]

 

[math]\frac{\rho}{\rho_0} \approx 1 + \left( \frac{4 \pi G}{3 K} \left( \rho_0 R \right)^2 \right) \rho^2[/math]

 

has the solution, via Pythagoras' theorem:

 

[math] \frac{\rho}{\rho_0} \approx \frac{ 1 - \sqrt{1 - 4 \alpha} }{2 \alpha} [/math]

 

[math] \boxed{ \alpha \equiv \left( \frac{K}{4 \pi G \left(\rho_0 \; R \right)^2} \right) }[/math]

 

For earth, the relative density is nearly 2. And, if you plot that function, then you observe, that high relative densities only occur, near [math]1 - 4 \alpha \approx 0[/math]. So, if [math]\alpha_{\oplus} \approx 1/4[/math], then [math]K_{\oplus} \approx 500 GPa[/math], comparable to the bulk modulus of diamond. Ergo, perhaps the high-pressure phases, of material, in earth's mantle & core, have collapsed into super-dense configurations, with diamond-high bulk moduli.

Edited by Widdekind
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First, oops on "alpha":

has the solution, via Pythagoras' theorem:

 

[math] \frac{\rho}{\rho_0} \approx \frac{ 1 - \sqrt{1 - 4 \alpha} }{2 \alpha} [/math]

 

[math] \boxed{ \alpha \equiv \left( \frac{4 \pi G \left(\rho_0 \; R \right)^2}{K} \right) }[/math]

 

For earth, the relative density is nearly 2. And, if you plot that function, then you observe, that high relative densities only occur, near [math]1 - 4 \alpha \approx 0[/math]. So, if [math]\alpha_{\oplus} \approx 1/4[/math], then [math]K_{\oplus} \approx 500 GPa[/math], comparable to the bulk modulus of diamond. Ergo, perhaps the high-pressure phases, of material, in earth's mantle & core, have collapsed into super-dense configurations, with diamond-high bulk moduli.

 

Second, if you plot that function, you get (relative) density, as a function of the (corrected) alpha. That alpha is a function, of R2, and 1/K. So, you can then estimate the planet's "bulk bulk-modulus", K ~ R2 / a.

 

E.g. for Earth, a ~ 0.25, R ~ 6400km, K ~ 450 GPa; for Mars, a ~ 0.18 (= 3/4 earth value), and R ~ 3200km (= 1/2 earth value), K ~ 1/3 earth value ~ 150GPa. So, qualitatively, perhaps Mars has allot less material, compacted into high-density, high-bulk-modulus phases ? Is that not plausible / probable ? Earth's estimated bulk bulk-modulus resembles diamond, which does in fact form at depth, down in the mantle, at high pressure, as a high-pressure phase (of carbon). Mars' estimated bulk bulk-modulus resembles regular rock, as known, from earth's crust, at low pressure.

Edited by Widdekind
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i employed Mathematica, to numerically solve the ODE. (For a given world, and its [math]\alpha[/math], there exists some central density, for which the curve of the density drops to one, at the surface (x=1, y=1).)

 

The equation works remarkably well, for the Moon, Mars, Earth, with a plausible "bulk bulk modulus" comparable to concrete. The simple model accurately, if not precisely, reproduces the average & central densities, of those bodies. But, world's cannot get much bigger than Earth, with this model. Larger worlds have lower [math]\alpha[/math], and worlds larger than ~7000km cannot satisfy (x=1,y=1); the equation demands too much slope (to generate the pressures offsetting gravity), so the densities drop well below uncompressed (y=1), well before reaching the surface (x=1). Interpreted, rocky worlds don't get (much) bigger than earth; more mass merely becomes more density; the solutions all have earth-sized worlds, with super-dense cores.

 

So, whilst a simple bulk modulus model can account for moon-to-earth-sized worlds, "super earths" are either earth-sized, but super-dense; or else something else happens, e.g. high-pressure high-density matter states have higher bulk moduli.

 

...


[math] \alpha \frac{d}{dx} \left( \frac{x^2}{y} \frac{dy}{dx} \right) = - x^2 \; y[/math]

 

... Now, note:

 

[math] \boxed{ \alpha \equiv \left( \frac{K}{4 \pi G \left(\rho_0 \; R \right)^2} \right) }[/math]

[math]\alpha_{\oplus} \approx \frac{1}{3}[/math]

[math]\alpha_{moon} \approx 5[/math]

[math]\alpha_{mars} \approx 1[/math]

 

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