How do I explain why the Tibetan plateau is colder than lowlands at similar latitudes?

The answer is because the Earth is not a static system.

Due to the ideal gas law, air cools as it rises.  This is referred to as the dry adiabatic lapse rate.  However, you are curious why every location on earth is not the same temperature at the same latitude.  We know this is not true. But why is not true?  Weather.

The earth, as with most natural phenomenon, attempts to achieve static equilibrium.  The differences in incoming and outgoing radiation leave the earth with net energy reserves that need to be balanced out.  Weather is the mechanism that the Earth uses to attempt to achieve balance.

Okay, so we know that weather is disrupting static equilibrium on earth.  But how does that work?  Wind.  Or better yet, advection.  Advection is simply the transport of an atmospheric parameter (such as moisture, temperature, or rotation) from one location to another.  So, wind can 'transport' temperature, moisture, and rotation.

You don't necessarily need a storm to advect any of these parameters. In fact, they are always in motion.  Cold air is always moving away from the poles.

Still, that doesn't quite answer it, does it?  Well, imagine you are hundreds or thousands of miles away from Tibet.  You aren't in a mountainous region anymore but instead a flatlands at sea level. The environmental lapse rate is 6.5°C/1000m above this location (because we are not dealing with an air parcel).  Tibet is 4500m - 8850m high.   That would be around approximately 32.5°C of cooling at 5000m.  The air that advects into Tibet from elsewhere is on average of 32.5-50°C cooler than at sea level.

So, it isn't so much that Tibet is receiving less radiation than other points at the same latitude (its not). Its more that, the air at other locations at the elevation of Tibet is much cooler than the air at sea level. Remember, we measure air temperature!


How can scientists measure chemical composition on other planets?

Scientists can measure the chemical composition of the atmosphere of other planets using spectroscopy.  Each molecule in the atmosphere absorbs and re-radiates at specific frequencies of light that are a function of temperature and unique to the specific type of molecule.  For more information see https://en.wikipedia.org/wiki/Extraterrestrial_atmosphere and http://www.physics.uu.se/research/astronomy-and-space-physics/research/planets/exoplanet-atmospheres/ .


Volcanic activity in Taiwan, ever in recorded history, much chance in the near future?

There are at least 9 old centres of volcanic activity in Taiwan:

The Tatun Volcano Group 15km north of Taipei, they think there are active magmatic systems there but the last eruption was 200,000 years ago.

The Chilung Volcano Group in the north is extinct, with no magmatic activity and hasn't erupted in 800,000 years.

Mount Guanyin is listed as extinct but not a lot of work seems to have been published on it as a volcano.

Caoling Mountain is similarly extinct and no-one seems to have done any real work about it's eruptive history at all.

Guishan Island is still active and last erupted in 1795, it could start a new eruption at any stage.

The Hai'an Range on the Eastern Coast is tectonically active and has experienced volcanism in the past which could recur at any time depending on the movement along that plate boundary.

Green Island and Orchid Island off the East Coast are old volcanoes of the Luzon volcanic Arc and haven't erupted in hundreds of millions of years.

Lastly there are the Pescadores out west, 90 odd volcanic atolls and small islands probably all monogenetic in origin (the individual volcanoes erupted only once in a given location but the area has experienced repeated eruptive cycles over geological time) none of the existing members is likely to erupt ever again but new volcanism in the region is likely at some stage.

So the long and short is that while it's been pretty quiet in Taiwan in the last couple of centuries there are areas where new eruptive cycles could start at any time at existing sites or new ones.

Please note any dates given are pretty approximate.


Why does wind blowing against a strong current create way bigger waves?

There are two physical processes at play here:

Wind speed relative to the ocean surface
Wave focusing and blocking in opposing current

Wind speed relative to the ocean surface

As you describe in your question, wind speed relative to the ocean surface varies depending on the current orientation and magnitude. The rate of wave growth is proportional to the wind speed $U$ in direction of the wave relative to the phase speed of the wave $c_p$, squared:

$$
S_{in} \sim \left( U - c_p \right)^2
$$

However, the wave phase speed is modulated by the current:

$$
c_p = c_{p0} + u
$$
where $c_{p0}$ is the intrinsic phase speed (phase speed in absence of currents) and $u$ is in the direction of the wave. Negative value of $u$ is then an opposing current. Wave growth is then:

$$
S_{in} \sim \left( U - c_{p0} - u \right)^2
$$

Let's plug in some numbers here. Let's say $U = 10\ m/s$, $c_{p0} = 3\ m/s$, and $u = \pm 1\ m/s$:

In the case of following current, $u = 1\ m/s$, $S_{in} \sim 36\ m^2/s^2$ ;
In the case of opposing current, $u = -1\ m/s$, $S_{in} \sim 64\ m^2/s^2$ ;

Thus, by changing the orientation of a moderate current in characteristic wind and wave conditions, we increased the rate of wave growth by a factor of 1.78! This effect will be more powerful the smaller the relative wind speed $U - c_{p0}$ is.

Wave focusing and blocking in opposing currents

Wave energy is advected (transported) by the waves' own group speed $c_{g0}$ and also by currents. If for simplocity we drop all the other terms in the wave energy balance equation, we get:

$$
\dfrac{\partial E}{\partial t} = -\dfrac{\partial \left[\left(c_{g0} + u \right)E\right]}{\partial x}
$$

The crux here is in the modulated group speed $c_{g0} + u$. If the waves propagate into an area of increasingly stronger opposing current, there will be a convergence of wave energy in that space, which will manifest itself in increase of high and steep waves. At the point where $c_{g0} = -u$, that is, the current is exactly countering the group speed of the waves, they can no longer propagate in that direction! Wave energy accumulates locally and vigorous wave breaking ensues.

Summary

Wave blocking is a different process from the wave growth dependence on relative wind speed described above, but, they often correlate because young windsea (short waves locally generated by wind) are almost always aligned with wind direction, so all you need is the currents to oppose wind for both of these effects to come into play at the same time.

Both are likely to occur in strong and steady western boundary currents such as the Gulf Stream, Kuroshio, Aghulas, and Brazil current, and also in any region with strong tidal currents.

More on this topic in the answer to this question.


Using Enhanced Rock Weathering in the Sahel Region of Africa

OK I did a paper on mitigating climate change by enhanced weathering of rock on agricultural land for my masters (2014) it's not brilliant, but it does include some useful references if you want to research this further.
I would imagine the main limiting factor in the Sahel region would be that the process needs to be wet. So you're possibly only talking about being able to add material to irrigated areas. There are also the limitations you've already discussed of getting the material on site. In terms of crop yield, I would suggest that it will only increase plant growth in areas which do not already have alkaline geology (and I have no idea what the geology looks like in the Sahel, I'm afraid) and adding too much to soils may well actually decrease site fertility without the addition of something else (manure is the traditional amendment)
As far as I'm aware, enhanced weathering is perfectly technologically feasible, but it does have a limited application - due to the sheer quantity of it we'd need.
Hope this helps.


in picture 11 there is erosion by water and wind.

it is sandstone and this is easily eroded by water each time it rains the water takes particles of the sandstone and moves them down the mountain.

sorry for this short and bad answer,it was a simple question so the answer is simple too.


What is the average cholorphyll concentration for the ocean?

Behrenfeld and Falkowski, 1997 have maps of cholorphyll concentration for the entire world on pages 11-13 of the linked paper. The maps are divided by into yearly concentration and seasonal concentration, with the maps filled out from experimental data by two different modeling methods.

These maps are in mg C / m$^2$ / year, but they offer the normalization of $$\frac{4.6 \text{mg C / hour}}{1\text{ mg Chlorophyll}}.$$ Chlorophyll production varies by optical depth, which is the distance that light can penetrate into the water. Optical depth depends on various characteristics of the seawater like sediment node, so it doesn't have an exact analogue to water depth. However, Fig 1A supports a conclusion that plankton production is reduced by an order of magnitude at depth of 20 meters, and by 2 orders of magnitude by 60 meters.

If we confine the chlorophyll to entirely the top 60m of the ocean, then we can convert $$\frac{1 \text{ mg C}}{\text{m}^2\cdot\text{year}}\cdot\frac{1 \text{ mg Chlorophyll}}{4.6\text{ mg C / hour}}\cdot \frac{1 \text{year}}{8760 \text{ hours}} \cdot \frac{1}{60 \text{m}}= \frac{0.41 \text{ ng Chlorophyll}}{\text{m}^3}.$$

Using this convertion specifically on map A on page 11 of the linked paper, I get for your answers:

Open ocean in general: 75-150 g C/m$^2$/year $\rightarrow$ 30-60 $\mu$g/m$^3$
N. Atlantic Ocean: 150-250 g C/m$^2$/year $\rightarrow$ 60-100 $\mu$g/m$^3$
Coastal waters around the UK: 300-450 g C/m$^2$/year $\rightarrow$ 125-185 $\mu$g/m$^3$
Equatorial waters: 100-200 g C/m$^2$/year $\rightarrow$ 40-80 $\mu$g/m$^3$


Why are pole reversals not more detrimental to life?

There are several factors to consider. The main one is the atmosphere (especially if you want to compare Mars with the Earth's during magnetic reversals). Earth's atmosphere is a formidable shield against solar wind and cosmic radiation. Each type of radiation have a different penetration, but in general the radiation dose associated to each type of radiation decays exponentially at lower elevations. For example for ionizing radiation the relationship is

$E(z)=E_0 \left(0.21 e^{-1.649 z} + 0.79 e^{-0.4528 z}\right)$

Where $E$ is the dose, $E_0$ is the dose ate sea level and z is the altitude (reference). The following figure illustrate how the radiation dose drop at lower elevations. But if you extend that curve to 100 km you will be able to grasp how much of the incoming radiation is filtered by the atmosphere.

Figure: Cosmic radiation exposure as a function of altitude, latitude 50° north, taken from the European Nuclear Society.

Therefore even in the absence of magnetic field, only a small fraction of the radiation would make it to the surface. And note that reversals only means a weakening of the magnetic field, not a complete disappearance. What can disappear is the dipole component of the magnetic field, not the magnetic field as a whole.

It is important to note also that the magnetic field is very important to prevent atmospheric erosion. Therefore, a permanent lack of magnetic field can generate the loss of the atmosphere (likely what happened to Mars). Conversely, Earth's magnetic field is one of the reasons we have a very thick atmosphere compared to Mars, and during the brief periods of magnetic field weakening (due to pole reversals), our thick atmosphere still provided a quite effective protection.

The fact that space radiation doesn't penetrate much can be observed in many different ways, one of them is noting that auroras (a.k.a. northern lights) usually never happen below 90 km of elevation.

Moreover, the reason why auroras happen only at high latitudes (i.e. near the poles), is due to the "geomagnetic latitude effect" that describe how radiations dose rates are minimal at the equator and maximal at the geomagnetic poles. Nevertheless, the atmosphere absorbs most of that extra radiation and life thrives in the arctic forest without any problems.

Also, the atmosphere would not be just a passive filter, it could create its own magnetic field as some studies suggest. Such studies argue that as long as we have atmosphere, the solar wind can induced a magnetic field around the Earth even in the absence of internal magnetic field (reference)

The other important point is: Would the paleontologic record of migratory birds populations and magnetic-dependant microbial life be good enough to detect a catastrophic population reduction on migratory birds and some microbial life that happened 780,000 years ago? I'm not an expert but I would be skeptical. In fact, 780,000 years seems to be enough time for the evolution whole new species and the rise of the migratory behaviour from scratch. More recent reversal events like the Laschamp event 41,000 years ago were much quicker than 1,000 years, and even so, I wonder how well we could reconstruct migratory birds populations that far back (considering that we have trouble reconstructing human populations back then).

Regarding migratory birds, you say that after the reversal the birds would migrate in the wrong direction. And that is quite questionable. In the lack of any point of reference, using an improvised compass to know the north-south direction is easy, but figuring out which side is north and which is south is much more difficult. Therefore, maybe birds use other means to figure that out which side of their compasses is north, means that don't depend on the magnetic field. For example, the side at which sunrise happen or something like that.

Finally, microbes evolve and adapt very quickly, so even assuming that some (or many) of them would die during reversals, I'm sure others would take their place. At the end the ocean microbiota is nutrient-limited, and all the nutrients not used by the dying species will be used by other species that can thrive with a weaker magnetic field. So I wouldn't be so quick to think that such event would lead to a food chain collapse.


Is there any evidence for higher air pressures in the geological past?

Yes, there is.
But the data is still very sparse and errors are large.

Past atmospherics pressures have been estimated by at least three different methods:

Isotopic composition of fluid inclusions trapped hydrothermal quartz ( Nishizawa ey al 2007; Goldblatt et al 2009; Marty et al, 2013)

Image from Fig. 2 of Nishizawa et al 2007.

Size distribution of gas bubbles in basaltic lava flows (Som et al 2016)

Figure 3c of Som et al (2016): Beasley River geologic context and flow detail (scale bar, 1 cm)

Size distribution of fossilised raindrop imprints (Som et al, 2012; Kavanagh & Goldblat, 2015)

Figure 1 of Som et al (2012) The 2.7-billion-year-old Ventersdorp Supergroup raindrop imprints lithified in tuff at Omdraaivlei, South Africa.

Each study is based on samples that capture the conditions at a fairly specific point in time. Therefore, different results not necessarily contradict each other, but offer a sense of how variable has been the atmospheric pressure over geological time.

The size distribution of fossilised raindrop imprints might have a large range of error due to the many factors that influence drop size beside atmospheric pressure. However, some studies suggest that the Archean (4 to 2.5 billion years ago) atmosphere was almost ten times denser than it is today (Kavanagh & Goldblat, 2015). That figure is based on fossilized raindrop imprints dated about 2.7 billion year ago, that is after the evolution of photosynthesis, but still in its early stages, when most oxygen was absorbed by the oceans and there was very little of it in the atmosphere.
On the other hand, studies based in isotopic composition of fluid inclusions and bubbles in basaltic lava have found that the atmosphere was less dense than it is today.

These studies are of great interest, as a denser atmosphere even if poor in greenhouse gasses can produce warmer surface condition, helping to resolve the faint young Sun paradox. In a nutshell, this paradox refers to how could liquid water exist on Earth in the past when the sun was much fainter than it is today.

It is important to note that the additional temperature in a thick atmosphere doesn't come from adiabatic warming as some people have suggested here. A good treatment of the phenomena is presented by Chemke et al (2016) in the paper "The thermodynamic effect of atmospheric mass on early Earth's temperature", there they say:

  We find that higher atmospheric mass tends to increase the
  near-surface temperature mostly due to an increase in the heat
  capacity of the atmosphere, which decreases the net radiative cooling
  effect in the lower layers of the atmosphere. Additionally, the
  vertical advection of heat by eddies decreases with increasing
  atmospheric mass, resulting in further near-surface warming.


Does a table summarizing basic details of significant operational weather models exist?

MetEd from NCAR has a good encyclopedia of models that summarizes some weather models. Of course, when models update, such as the GFS dynamical core being converted to the FV3, or changes in data assimilation systems, they may not be up-to-date. This also doesn't have ALL weather models, but some of the more common ones.

Of course, there are models missing from the encyclopedia, such as the SREF, GEFS, the JMA model, the Panasonic weather model, any iteration of the WRF model, statistical post-processed models, etc.


Hypothetically, would digging a big trench from the ocean and flooding desert areas be a bad solution to combatting sea level rise?

Yes, it would be a bad solution.

Aside from the construction and maintenance cost, the impact to the populations affected, and the impact on the environment around the destination, the benefits would be minimal, as roughly calculated in the answer to Volume of Land Below Sea Level

  So to reduce the sea level by 1m the total area of these depressions
  needs to be around 5,000,000 km^2. This is around half the total land
  area of the USA. I don't have the figures to hand but given that these
  area would probably all fit into an area considerably smaller than
  Texas - I estimate that flooding land which is below sea level will
  reduce the sea level by exactly 'not very much at all' (give or take a
  millimeter - and very much less than 1m in any event).

There is also a rate of flow issue. You basically want to create rivers flowing inland at a rate to offset or reverse sea level rise, which means the flow rate of those rivers has to approach the volume of global sea rise. Global sea level is rising by an average of about 3.4mm per year, or 1,230 cubic kilometers per year according to NASA (and the rate is increasing). That amount is about the flow of the Congo River (the second largest river in the world) or roughly 19 percent of the flow of the Amazon River.

Fed by gravity, these are among the widest and deepest rivers of the world; if you're going to move that volume of water, you need to build channels to each basin with a net flow similar to that capacity, with that gradient, that won't leach saltwater into the groundwater of surrounding communities. (The engineering issues likely pale in comparison to the political challenges.)

And keep in mind that transferring water inland is only a temporary solution. Water evaporates and by spreading water over a larger area you're increasing its surface area; Death Valley has an evaporation rate of 140 inches or 3.5 meters per year. (In Southern California, the Salton Sea is drying out and causing a variety of salinity and dust issues due to lack of inflows.) So you're transferring ~60-degree seawater into a salt pan where it's going to heat up and evaporate and leave behind a toxic brine, then fall or flow back into the ocean, contributing in some small way to the thermal expansion of the ocean.

(Also, the net volume of CO2 released by building and maintaining these massive projects would also contribute to global warming in some non-negligible way.)


Where can I obtain ocean wave data of the Southeast Asian region?

The only open and ongoing data source for in-situ ocean wave measurements I am aware of is the National Data Buoy Center. Though NDBC manages data service from plenty of moored buoys in the Gulf of Mexico and the Atlantic and Pacific coasts of North America, unfortunately there isn't much in your region of interest.

The only buoys I have found that are somewhat near the Southeast Asian region are 2 buoys on the east and west side of Guam, and one right by Northern Mariana Islands:

http://www.ndbc.noaa.gov/station_page.php?station=52200
http://www.ndbc.noaa.gov/station_page.php?station=52202
http://www.ndbc.noaa.gov/station_page.php?station=52211

These buoys provide directional energy spectrum data, as well as integrated quantities like significant wave height and mean and dominant wave period. I've written a Python library that lets you parse NDBC server and access data from buoys from command line or your Python programs. Once I tidy it up and make it public, I will update this answer so that other people can use it.

Your other option, as suggested by @DeerHunter, is to use altimetry data from polar-orbiting satellites. Although these are sparse and provide only wave heights (no period or direction information), they may be your only bet depending on the region of interest.


Connection between radiative imbalance and radiative forcing

That +2.2 W/m2 is the effective radiative forcing, which is the difference in net top of atmosphere radiation in response to a changed set of conditions (e.g., greenhouse gas and aerosol concentrations, land use) after allowing fast system feedbacks to equilibrate (e.g., tropospheric humidity, stratospheric cooling) but not allowing for slow system feedbacks (ocean heat uptake, land- and sea-ice change).  That latter point is important because relatively little energy is stored in the atmosphere and land, so it’s the slow components that would adjust the system state to remove any net radiative imbalance.
But note that it’s a theoretical, diagnostic property of the system; +2.2 W/m2 is not an energy imbalance that was ever actually expressed between the pre-industrial period and the present.  In reality, the forcing conditions and radiation imbalance occurred gradually, and the system responded gradually along with them, mainly by storing heat in the ocean and increasing the ocean surface temperature.  That has kept the expressed, measurable imbalance down to much lower values, giving us the +0.6 W/m2 that we have today.
As for the pre-industrial imbalance, well, to quote Andrews and Forster (2020):

Since heat uptake was not observed in the nineteenth century, climate and energy balance models must be used to infer the pre-industrial energy imbalance. Forster summarized this to be around 0.10 W/m2 …

My experience from climate models is that 0.1 W/m2 is the typical variability in the decadal mean radiation balance in pre-industrial control simulations, so that observed estimate is probably not distinguishable from noise for a system that’s basically in radiative balance.  In the absence of any solid evidence to the contrary, we normally use a target of radiative balance and no drift in long term model state when spinning-up pre-industrial model climates.


Do desalination plants affect the ground water table?

With desalination plants the import thing to consider is from where such plants are taking water and where they are disposing of their waste.

The waste from desalination plants is either salt of very salty water. If the plant you are referring to is just dumping its waste on the ground nearby or into a nearby river or water channel it is possible that the salt in the discharge water from the plant will enter the ground water, making it salty.

Generally this should not happen and in most situation it does not. With desalination plants near the seashore, the waste water is discharged back into the sea, but far from the intake for the plant.


Why is the ice on a frozen lake sometimes cracked?

Ice does crack completely naturally without human intervention. There are many reasons it could do so, one is as mentioned before mechanical stress from in- and outflow that lifts/settles the ice cover.

Another one is temperature stress from density changes, because ice changes its density e.g. depending on diurnal temperature changes or simply by cooling as it thickens (or warming as it thaws).

... and more.

Changing density:
https://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html


Ozone gas (O₃) endurance in the atmosphere

$\mathrm{O}_3$ decomposes to $\mathrm{O}_2$.

The rate of decomposition increases with temperature.

At 20 degrees C, the half-life is about 3 days in air.

$\mathrm{NO}$ can catalyze the formation and decomposition of ozone.


A model is a simplified representation of a system.
Rather than try to model everything down to the microscopic level, or as vectors, we often aggregate phenomena across geographic areas, to simplify the computation a lot, and to lose only a little accuracy and precision.
So if I want to model the UK, which roughly spans 8°W - 2°E, 50°N - 60°N, that's an area of 10° x 10°. I can split that up into 400 cells, each 0.5°x0.5°. So, starting in the south-west corner, I model the area 8°W-7.5°W x 50°N-50.5°N as a single homogenous area, with no variation within it. And the same for the other 399 cells (8°W-7.5°W x 50.5°N-51°N and so on all the way through to 1.5°E-2°E x 59.5°N-60°N). That would be a 0.5° grid.
Note that the gridcells do not overlap at all, and they completely cover the area of interest. (in mathematical terms, they are exclusive and exhaustive - every point of interest is in exactly one gridcell).
That can create vast amounts of data - particularly if I'm doing this for the whole world, at second-by-second resolution for ten years, for a few hundred variables of interest. And no one will ever need that amount of precision from my results. So I aggregate up to a more coarse grid: so that I take 25 grid squares that form a larger square, and aggregate each of my statistics of interest in them. So now my area 8°W-7.5°W x 50°N-50.5°N is just one of 25 cells in a square in the coarser grid which spans 8°W-5.5°W x 50°N-52.5°N, and I report statistics for 16 cells in a 2.5° grid, rather than 400 cells in a 0.5° grid.
Exactly how that aggregation is done, will vary for each statistic:

If it's a minimum overnight temperature, then I'll take the minimum across all 25 0.5° gridcells, to give the minimum overnight temperature for the 2.5° grid cell that consists of those 25.

If it's an average rainfall, I'll take an average.

If it's insolation per unit area, and I'm interested in insolation per unit land area, then I'll take a weighted average of insolation, weighting by the proportion of each 0.5° gridcell that contains land.

And then I'll have modelled on a 0.5° grid, and archived on a 2.5° grid
Here, for illustration, is Britain in a 100km grid:
 source
If I wanted to aggregate that to a 200km grid, then the four squares in the bottom-left corner, SQ, SR, SV, SW, would be combined into a single grid cell. And SS, ST, SX, SY would be combined into another grid cell. And so on.


Why are cobalt deposits so unevenly distributed?

Part 1

The Democratic Republic of the Congo's Katanga Province contains
almost 40% of the world's reserves of cobalt [1].
Why are deposits concentrated so strongly in such a small portion of
the earth?

Cobalt isn't as unevenly distributed as it seems. It is correct that most of the world's cobalt reserves are in DRC, but most of the cobalt resources are not in DRC. And this is a very important point.
Resources are known and estimated quantities of economically extractable materials. Reserves are resources for which detailed plans for extraction have been made. This means that to turn a resource into a reserve you need to have plans for the extraction and refining plants, have environmental approval, have the workforce figured out, have authorisation from the owners of the land, and more. As you can guess, making a resource into a reserve is a time consuming and expensive process.
The reason why most of the Earth's reserves are in DRC is simply because it's easier and cheaper. The infrastructure already exists, and the expertise exists. At current technology levels and cobalt prices, the investment in defining a reserve from a resource elsewhere is simply not economical.
This does not mean that DRC has most of the cobalt. It only means that getting cobalt out of the ground and making it into a product is cheaper in DRC than it is in other countries.
Part 2

Do the geological processes causing this uneven distribution significantly differ from those that cause the uneven distribution of gold?

The reasons are the same, in principle. Every element behaves differently and will separate from other elements and concentrate because of various geological processes. Cobalt tends to follow copper and nickel, and in most cases it is mined as a by-product of those elements. That's why cobalt is extracted from the African Copperbelt, and why it's extracted from nickel deposits in Australia. Not much point going into the chemistry and thermodynamics of why this happens, but the point is that elements which are concentrated by rare geological processes will be found in fewer places, and elements that are concentrated by common geological processes will be found in more places.


It's biotite. Not muscovite. That's why you have the pleochroic halos in it.

I don't understand why you are saying that the interference colours are like muscovite. First of all, the lack of cleavage suggests you are looking down (or close to) the x axis, so this means the interference colours you see are going to be very low. Let's say 1st order white or brown. And then, it's going to get the anomalous brown colour of the biotite together with it, which is exactly what you see here. Also, muscovite is not brown in ppl. This one is. Therefore, it's biotite.


The ‘Oligo-Miocene’ part

The Oligocene and the Miocene are epochs of geological time. The Oligocene lasted from ~33.9 million years ago to ~23 million years ago; the Miocene followed immediately after the Oligocene, and lasted until ~5 million years ago.

Oligo-Miocene refers to events that happened around the boundary between the two epochs at ~23 million years, or that happened throughout the Oligocene and Miocene epochs.

The ‘radiation’ part

‘Radiation’ here is evolutionary radiation, a biological term usually referring to an increase in the number of species or in the differences between them. The Palaeontology Online glossary defines ‘evolutionary radiation’ as follows.

  An evolutionary radiation is an increase in either taxonomic diversity (i.e. the number of species) or morphological disparity (how they differ in anatomy). This can result  from an adaptive change or the opening of ecospace.

The combination

Putting the two parts back into the contexts where you found them: the first article mentions an increase in the diversity or marine mammals, and the second in the diversity of dry-adapted shrubs. Both these increases occurred between the start of the Oligocene and the end of the Miocene.


Mg C is a tonne of Carbon.

An Mg - a megagramme - is a tonne. It's a silly way of writing a tonne for a general readership, but it's the strict SI equivalent of a tonne. So an "Mg C" is a tonne of Carbon - the standard unit when it comes to talking about pricing Carbon.

Normally you wouldn't see anyone writing about a million Mg of anything; (in this case, of carbon). You'd either write about a Teragramme, or a million tonnes, but not a million Mg.

However, if you draft a paper referring to millions of tonnes, and then the editor or a peer reviewer comes back and says that the tonne is not a strict SI unit but the Mg is, then someone in the chain of authors and editors may just do a search and replace on all the "tonnes" to replace them with "Mg". And that's one way that you might end up with lots of references to "million Mg C", instead of the more usual "million tonnes of carbon"


How it's possible to measure temperature 2000 years ago with such precision?

How it's possible to measure temperature 2000 years ago?

Sans the technology used by Bill and Ted ("Bill and Ted's Excellent Adventure"), it obviously is not possible to directly measure the temperature from yesterday, let alone 2000 years ago, or longer. What is used are "proxies", things that can be measured today that serve as stand-ins for things such as temperature in the past.
One example of a proxy is the amount of oxygen-18 in the ice in Greenland and Antarctica. Water with two hydrogen atoms and one oxygen-16 atom has a boiling point of about 100° C. Water with the oxygen-16 atom replaced by an oxygen-18 atom has a slightly higher boiling point. This means that heavy oxygen water evaporates less readily but precipitates more readily than does normal water.
This in turn means that the fraction of oxygen 18 versus oxygen 16 in the water and air bubbles in the ancient ice in Greenland and Antarctica are indicative of the climate at the time that that ice and those air bubbles formed. Despite being a half a world apart, the ratios of oxygen-18 to oxygen-16 over time are highly consistent between Greenland and Antarctica. The consistency of these measures at half a world apart are almost universally taken as being a proxy for something else, and that something else is climate.
There are many other proxies for past climate. There are other isotopes that serve as proxies. The amounts and kinds of pollen in mountain glaciers and ice sheets form yet another kind of proxy. The kinds of plants that do grow in some locales and how fast they do grow is highly temperature dependent. Like ice, buried muds in the oceans also show variations in various proxies. Climate scientists put these proxy measurements together to arrive at the temperatures from 2000 years ago, and even further into the past.


converting segd data to SU format using seismic unix

SEG-D is an uncommon format. From my experience, here is how you need to approach this problem. You can't easily jump straight from SEG-D to SU. I've had extensive experience with SU and the best methodology is to first take your current seismic data type (SEG-D) and convert to SEG-Y. Only after converting to SEG-Y format can you then jump to SU.

So, here is an outline and workflow of a solution:

  1.) Download and use this tool for SEG-D to SEG-Y format: http://users.chariot.net.au/~witek/segd2segy.htm

  2.) Now that you have all your SEG-D data converted to SEG-Y data, use the following BASH  script to convert a single SEG-Y file to SU:

#!/bin/bash
echo Provide Input File: ;
read sdata;
segyread tape=${sdata} verbose=1 endian=0 segyclean > ${sdata}.su
echo "New File: ${sdata}.su";
rm -rf binary header;
echo Done.;

***Keep in mind that - based on the type of machine you have - endian may need to be set to 1.


Are there General Circulation Models that can simulate the atmosphere of Venus?

I'm not sure of their current status, but I read some papers based on these models a few years ago:

Oxford Planetary Unified Model System for Venus (OPUS-V), which is based on an old version of the UK Met Office climate model, HadCM3 (paper from 2016).
ROCKE-3D at Nasa GISS, which is based on an old version of the GISS ModelE2‐R climate model (blog article and paper from 2016).

It looks like the introduction of Yamamoto et al (2019) has a more up-to-date summary of the current set of models, and there's quite a healthy selection from groups around the world.


Do I need a permit for putting retrievable sensors in ocean (Atlantic)?

It depends where you live as laws verry from country to country. Just ask your local lawyer or law enforcement official. It may also depend on what data you are recording


CMIP5 Historical experiments, what do they mean?

The CMIP5 experimental protocol is documented at https://cmip.llnl.gov/cmip5/docs/Taylor_CMIP5_design.pdf

The historical experiment for a model spans model years 1850-2005 and is initialized from a control experiment for that model (forcings from 1850 are used repeatedly).

For models that are not coupled to a modeled carbon cycle, greenhouse gas forcing are based on the observed record (including ice cores) of atmospheric concentrations. For models that are coupled to a modeled carbon cycle, atmospheric CO2 concentration is predicted, and the model is forced with anthropogenic CO2 emissions (and land use change).

RCPs are extensions into the 21st Century. There are different RCPs, based on different assumptions about future greenhouse gas concentrations and/or emissions, land use, population, etc.

Reanalysis data is from a model this is constrained to be close to observations of the atmospheric state (e.g., air temp, winds, pressure, etc.) This differs from CMIP models that run unconstrained.


It depends on which pole.
The south pole is in the middle of Antarctica, with mean temperatures of -57°C, liquid water doesn't exist there and the air is extremely dry. Therefore, I would say that it is not possible to create more ice due to the lack of liquid water. And if there were any liquid water it would freeze naturally. This will make sense if we have a look at the antarctic geography. This map will help

[image from geology.com]

One possible action following your line of thought, would be to bring liquid water to the south pole from the coast and let it freeze there. However, the more than 500 km that separates the South pole to the nearest ocean shore (that is actually under hundreds of meters of ice) or 1,300 km to open water, makes this idea very impractical. A similar approach but more efficient would be to pump water from the coast to a nearby antarctic basin, and that would indeed help reducing sea level. This idea have been proposed by scientists. However, to make this effect significant you would need to pump incredible amounts of water, something that makes this geoengineering idea very costly and not very appealing.

Other scientist have proposed that dumping piles of rock in the ocean floor at specific points could be the most cost-effective way to slowdown sea level rise. The idea is to redirect warm ocean currents away from glaciers, something that could significantly reduce the rate at which they are melting down. This approach would have a much lower (but still very high) cost.

This paper explain this idea and the following figure summarizes it:

The North pole in contrast is just in the middle of the Arctic ocean. You could indeed take ocean water there and turn it into icebergs. However, as others have said in the comments, using any kind of cooling system would produce a lot of heat so it would not be a good idea. But you can just enhance the heat transfer with the atmosphere as ski resorts do to make artificial snow. Their approach is to wait for a cold day and spray water into the air. If it is cold enough, the water droplets will turn into ice before falling back to the ground.

However, even if you succeed in making icebergs, that would not make any change in sea level, because they are floating. It would be equivalent of melting of ice cubes in a glass, something that doesn't change the level of the liquid in the glass. Here a graphical example of that

Nevertheless, your feat of engineering would have a positive effect in slowing global warming by covering the ocean with a layer of highly reflective ice. That ice would reflect back to space a lot of sunlight that would have been absorbed by the ocean otherwise. However, when you freeze sea water, most of the salt is expelled, so you would also put a lot of salt into the water, that would make sea water more dense and it would sink. That could have a undesired effect on ocean circulation, so you have to be very careful in studying the side effects of such approach.

Similar to the antarctic approach. You might get better results if you pump water from the Arctic ocean and let it freeze on top of a desolate arctic island. However, it still doesn't seem to be the most cost effective approach.

Said that, there is a long list of geoengineering ideas to battle against climate change and sea level rise. However, the most cost-effective way to slowdown sea level rise, is to reduce the rate at which we are consuming stuff. So just buy less and reduce/repair more. Also very important is reducing the use of fossil fuels (something that would also be a side effect of consuming less). Having less children or adopting instead of having your own is also a way to make big contributions to slow down sea level rise.


Why does chlorophyll only come in green?

chlorophyll is green because that was the part of the spectrum that was left when plants evolved. The bulk of the spectrum was already being harvested by other photosynthetic life. there is actually a wide variety of photosynthetic pigments chlorophyll is just one of many. You may want to check out the purple earth hypothesis.

Sauce


What factors determine the number of Hadley cells for a planet?

Well for one, this will surely be the speed of planetary rotation and the wind speeds your atmosphere can generate in North-South-direction.
In my understanding those cells form as N-S winds flow and become deflected from this direction. So the number of cells will be determined by the winds that start either at the equator or the pole and the distance that they can cover before being deflected into completely zonal direction.

To have a rough idea about this distance I tried to estimate this (in a very quick, dirty and most likely extremely wrong way) by assuming the simplified Navier-Stokes equation in planetary coordinates on synoptic scales:
$$\partial_t \vec v = 2 \Omega \sin\Theta (\vec v \times \vec z)$$
leading to a harmonic equation for the meridional speed
$$\partial_{tt} v_{\phi} = -(2\Omega \sin\Theta)^2 v_{\phi}$$
which with the usual arguments of scale-analysis gives us a timescale $\tau_{\phi}$ for the periodic movement involved in solutions of that equation, of
$$\tau_{\phi} \sim \frac{1}{2\Omega \sin\Theta}$$
corresponding to a distance
$$ d \sim \frac{v_0}{2\Omega \sin\Theta_0}$$
covered by the vortex. Of course, I cheated here by assuming the initial speed $v_0 = \text{const}$ and the initial Coriolis-factor constant with $\Theta = \Theta_0$.
As now such a "structure" (solution of the above eqs.) will have a size of $2d$ in the meridional direction, having space of $\pi R_{\text{planet}}$ our number of Hadley cells per hemisphere becomes simply
$$N = \frac{\pi R_{\text{planet}} \Omega \sin\Theta_0}{v_0}$$
Where we'd need to estimate this number starting at the poles, as the Coriolis-parameter vanishes at the equator.

I will later (if I don't forget it) pluck in some numbers and see how bad this estimate is, or perhaps someone can try that. Anyway, I'm open for improvements of those arguments from anyone who knows something, but a quick look into Holton, An Introduction to Dynamic Meteorology, unfortunately didn't reveal much.

But to also address your question a little bit further: Of course, we'd need to have some climatic model to be able to estimate $v_0$ and know the speed of planetary rotation. Independent of how bad my formula is, the initial arguments of deflection vs. 'climbing the meridian' should hold for the formation of a Hadley cell.
Returning winds from higher atmospheric layers, surface friction in the atmospheric boundary layer will also play a role for an exact calculation.

Addendum: Rhines (1975) has however found a relation $N \sim R_{planet} \sqrt{\frac{\Omega}{v_0}}$, by considering how turbulent eddies will break up energy-limited by zonal jets. Also turbulent eddies can provide a good mechanism for justifying a locally constant $v_0$, that I simply assumed in my calculation above.

On the other hand I just noticed a trivial error in my above calculation and cleared that. The fraction that is N was the other way round. Sorry for that.

I wanted to note one other thing that seems important to me (and is perhaps also to anyone on this planet..):
While this rule of thumb-calculations can work out, there is more physics that play a role in shaping the exact morphology of circulation-cells on a planet. Namely the radiative equilibrium in the atmosphere, which links the exact positions of those cells to a changing climate.

As a study suggests by measuring Tropical to Polar-Jet distances and Tropopause heights, those cells can expand and reshape the environment people live in.


Sedimentology (Particle size distribution parameters)

Folk and Ward (1957) were the ones who created a measure of sorting using graphical moments of particle size distribution based upon φ values obtained graphically from the cumulative frequency curve at specific percentile levels.

I wasn't aware that negative values were possible based on the equation, but I am not up on my sedimentology..

Assuming your value is correct, it would signify extremely well sorted material:  the Inclusive Graphic Standard Deviation (Folk and Ward, 1957) has a phi range of <0.3 (extremely well sorted) to >3.0 (extremely poorly sorted).

Folk, Robert L., and William C. Ward. "Brazos River bar: a study in the significance of grain size parameters." Journal of Sedimentary Research 27.1 (1957).


The easiest way this is measured is via Argo floats. They're probes which periodically dive into the sea, and measure the ocean heat content.

This is done by calculating the formula. According to the European Environmental Agency, this formula is:

  Ocean heat content is defined as the integrated temperature change times the density of seawater, times specific heat capacity from the surface down to the deep ocean.

This could also be done manually by calcuating out the formula, and obtaining the temperature of the seawater by using a device such as a Nansen Bottle.


Magma chambers can exist at a range of depths from very deep to very shallow.  Only requirement that is really necessary is that there be enough pressure from overlying rock strata to keep the magma chamber from reaching the surface and erupting/exploding.  Volcanoes typically represent a slow leak of molten magma.
Even large Volcanic eruptions will only expel a fraction of the magma available.

There are several important geological factors that will control the size. location, and depth of a magma chamber:

Magma buoyancy: Density of the magma (Dissolved gases/liquid and chemical composition vs density of the surrounding host rock.
Integrity of the overlying host rock: tectonic settings, fault and fracture patterns, how seismically active is the local region.
Rate of melting: dependent upon source rock composition, heat flow, depth of source material.

Back in the early 1990s, Japanese researchers imagined magma plumes near the lower mantle  / outer core boundary.

Here is an article discussion shallow and deep Magma sources for the kilauea, Hawaii. As this article suggests, magma can originate at depth and migrate upward and supply shallow chambers.

Facts on the Yellowstone volcanic complex.


What is the CO2 concentration by altitude and why?

The volume of carbon dioxide in the atmospheric column becomes uniform at altitude because the sources and sinks for CO2 are generally at ground level and, as it rises, CO2 becomes well-mixed into the atmosphere.

This is why, for example, David Keeling began the measurements for what would become known as the Keeling Curve at a station 3.3 kilometers above ground level at Mauna Loa.

At ground level, CO2 levels will be heavily influenced by local and seasonal sources such as forests, grasslands, and human activity. That is, if you have a city or industrial area, CO2 emissions will be high due to the burning of fossil fuels. In the case of nature, forests and grasslands will absorb CO2 during the local spring and summer when they grow and emit CO2 in fall and winter as they decompose. (Seen in the annual wiggles in the generally rising Keeling Curve.)

The specifics will vary by location. In the case of your graph, this late-fall test proofing lidar against balloon measurements over Oklahoma on Dec. 7, 2008 shows CO2 at about 395 ppm at about a half a kilometer, declining with altitude so that at about 3.3km, the measurement is around 385-386ppm.

At Mauna Loa, at 3.3km, the NOAA monthly CO2 data average for December 2008 was 385.56ppm.


How does liquid iron create a magnetic field?

It seems your question assumes that Earth magnetic field is caused by natural magnets inside the Earth. Instead, the magnetic field comes from the movement of charged particles.
Any electrical charged particle in movement generates a magnetic field (see Biot-Savart law and Ampere law), hence, thanks to the movement of particles in a liquid mantle we have a magnetic field. Obviously, particles must have electrical charge, which is more common in metals.
If mantle turns solid, movement is gone, and the magnetic field reduces its strengh.


Mostly stratocumulus with a scattering of emergent cumulus towers which have been lofted by strong internal convection cells.


Wind profile below the aerodynamic roughness length

The key text here is "for $z>z_0$".  It's telling you that, while you can evaluate the equation for other values of $z$, outside of that range the equation is not a valid description of the physical system.  The equation could be written piece-wise to be complete:

$u(z) = \begin{cases} (u_*/k) \ln(z/z_0)& z>z_0 \\ 0 & z\le z_0\end{cases}$

But this doesn't really add anything useful.  In practice, the "log-law" is used to describe the wind profile over 10s of metres and values of $z_0$ range from 1 mm to 2 m, so values of $z$ are likely to be in the valid region.  If you do need to make calculations that close to the surface (in the interfacial sublayer) then you'll need a different equation anyway.


How to start studying dinosaurs and pre-historic mammals/sea creatures

Here's what it takes to really study this: you need to go through the whole bachelor program for geoscientists, that includes fundamental geodynamics like plate tectonics, magmatism, volcanism, volcanic and metamorphic rocks and generally the cycles that make up earth's internal dynamics.
Then there is the huge field of external factors, like sediment geology (that's really complicated stuff), weathering and transport and how soils come to being, diagenesis and the structures sediments can form and their classifications. Role of the ocean (that's where it starts, before all) and the atmosphere, of course.
When through that, usually 4 semesters or so, you can start to specialize. For paleontolgy you need knowledge of earth history, of course, it's subdivision, and the conditions at certain times as far as they are known. Once that's done, then comes real paleontology: Animals (invertebrates and vertebrates), plants, and their development, biological evolution (that's frequently underrated, I find), taphonomy, ... For a sturdy base count another 2-4 semesters.
You may see that even a bunch of websites, maybe all of them together, cannot replace actual study. I am not aware of any site that even gives a reasonable overview of the field. Geoscience, and thus paleontology, touch many fields of natural science.
That said, when asked "How to learn about animal paleontology ?" I allways mention Micheal Benton, Vertebrate Paleontology. It needs a basic understanding of geoscience, evolution and skeleton anatomy. Functional morphology, phylogeny and an overview over sediment geology and earth history also won't harm, but you could give it a try. Some things are explained in between.
I'd recommend to at least get a good introduction and overview into geology, in form of one or two books on the matter. Maybe a colleague knows other good books on the subject. Actually, I'd like to update my bookshelf myself :-)
Also, Museums of Natural History have publications that could interest you, but best would probably to sacrifice some time, visit them and then browse their book shops. Depending on your time and where you are.


How far downwind of a covid-19 emitter has a 1000-fold reduction in PPM?

As I said in my comment, viruses aren't my area (concerning units). But the answer is actually unitless. The exact answer you seek is dependent on a couple of different variables:

The speed of the exhalation (cough or breathing)+speed of wind
The atmospheric stability

Of course, the viral load of the exhalation is also important, but since you're asking for a ratio, I won't list that.
The Gaussian Plume model can be used to describe pollutants. The Gaussian Plume model is: $$\frac{C}{Q}=\frac{1}{2\pi U \sigma_y \sigma_z}\exp\left(-\frac{1}{2}\left[\frac{y}{\sigma_y}\right]^2-\frac{1}{2}\left[\frac{z-H}{\sigma_z}\right]^2\right)$$,
where $C$ is the concentration, $Q$ is the emission rate, $U$ is the wind speed (or worst case-scenario is cough speed), $x$ is the downwind distance, $z$ is the distance from the ground, and $\sigma_x, \sigma_z$ are the horizontal and vertical diffusion constants.
$\sigma_y, \sigma_z$ are functions of stability and wind speed. The actual formulas are... iffy. Theoretically they can be described as $$\sigma_{y,z}=\sqrt{\frac{2K_{y,z}x}{U}}$$, where $K$ is the eddy viscosity. There are empirical formulas based on stability classifications too.
Lets assume that the receiver is directly downwind and at the same height (z=H and y=0). Therefore the equation collapses to: $$\frac{C}{Q}=\frac{1}{2\pi U \sigma_y \sigma_z}$$. Since the distance downwind is expressed only by $\sigma_{y,z}$, then to answer your question, we need to isolate $\sigma_{y,z}$. $$\sigma_y \sigma_z=\frac{1000}{2\pi U}$$
Let's just say, for the sake of argument, consider two different scenarios:

Empirical $\sigma_{y,z}$ (refer to this for values of constants)

$$(ax^b)\times (cx^d+f)=acx^{b+d}+afx^b=\frac{1000}{2\pi U}$$
And this is complicated to solve. So I'll leave solving for $x$ as an exercise to the reader.

Constant $K_{y,z}$

$$\sqrt{\frac{2K_{z}x}{U}}\sqrt{\frac{2K_{y}x}{U}}=\frac{1000}{2\pi U}$$
$$\frac{4K_{z}K_{y}x^2}{U^2}=\left(\frac{1000}{2\pi U}\right)^2=\frac{10^6}{4\pi^2U^2}$$
$$x=\sqrt{\frac{10^6 K_{z}K_{y}}{16\pi^2}}$$
$$x=\frac{10^3}{4\pi}\sqrt{K_{z}K_{y}}$$
If we assume that $K_z=K_y=K$, then the equation simplifies further. $$x=\frac{K10^3}{4\pi}$$
There are many ways on how to calculate $K$. One simple formula for $K$ can be used if mixing length theory is invoked
$$K=\frac{ku_*z}{\phi(\frac{z}{L})}$$, where $k$ is the von-Karman constant, $u_*$ is the friction velocity, $\phi_M(\frac{z}{L})$ is the Businger-Dyer function for momentum, and $L$ is the Monin-Obukhov length. If we assume neutral atmospheric stability,$\phi_M(\frac{z}{L})=1$ and the resultant equation can be written as:
$$x=\frac{ku_*z10^3}{4\pi}$$
More assumptions! Let's assume z=1.5 m and $u_*=1.11$ m s$^{-1}$ (corresponds to a 5 kt wind speed at 10 m with a roughness length of 0.1 m under neutral stability). Plugging those numbers in gets
$x=35$ m
Of course, there were a LOT of assumptions to get to this point. But this is what I have to offer.


how to know the correct results in watershed delineation in ArcGIS?

I would recommend looking at automatic stream network delineation and/or known stream locations features, and overlaying those as a check on the final watersheds created through whatever procedure you have. Watersheds can be merged or lumped together, which is fine, you just want to make sure it is consistent with the other information you have and what your goal is. There will always be some variation in the exact extents of the watersheds depending on what algorithm you use, but you should definitely make sure the general areas and drainage patterns are consistent with what you expect. Beyond that, I am not sure there is a correct answer.


Some mountains form from the collision of tectonic plates; volcanoes are different, and there are some other mechanisms too.

IANAgeologist, and perhaps one will be along to give a more technical explanation, but intuitively, it's about viscosity:

If you push an object very slowly into a bowl of water, nothing much happens - the water flows around the object.

If you push the object equally slowly into a tray of treacle, it's harder work, and you can form a "hill" in front of the object. The treacle flows too slowly to get out of the way.

At very high pressures, rock behaves a bit like the treacle[1] - only much more so. It takes a huge force to push one lump of rock into another, but because the rock flows very slowly, by pushing very slowly you can still create a mountain.

[1] I realise it's not exactly like treacle. If you're an expert here, feel free to post another answer getting it right :-)


Is moonlight a significant condition for making life possible on earth?

One significant biological event that requires a specific "type" of moonlight is the reproduction of corals in Great Barrier Reef, in Australia.

'One week each year in spring, after a full moon, millions of corals release eggs and sperm in what Bill Leggat, a co-author of the new study, called "a slow symphony."'

'An ancient light-sensitive gene has been isolated that appears to act as a trigger for the annual mass spawning of corals across a third of a million square kilometres of Australia’s Great Barrier Reef, shortly after a full moon.'


Website that contains the maps of atmospheric pressure and wind speeds

Check out this beta data portal from the Cooperative Institute for Meteorological Satellite Studies / University of Wisconsin-Madison
CIMSS Tropical Cyclones
Data Archive http://tropic.ssec.wisc.edu/archive/  .

However, my favorite source of information is the Regional and Mesoscale Meteorology Branch (RAMMB) of NOAA/NESDIS http://rammb.cira.colostate.edu/products/tc_realtime/

And there are good summaries here:
https://www.tropicalstormrisk.com/hislink.html

and another good general source of information is from the National Hurricane Center http://www.nhc.noaa.gov/data/


There's no official distinction between a river and a creek, both of which are categories of streams. As an example, consider Pine Creek in Pennsylvania and Po River in Spotsylvania County, Virginia. Pine Creek is 87.2 miles long and has a flow rate that ranges between 40 to 30000 cubic feet per second. Po River on the other hand is 1.9 miles long and has a row rate that ranges between less than 2 to 4000 cubic feet per second. The longer and larger one is called a creek while the shorter and smaller one is called a river.


GOES Visible Channel - asymmetry between morning and afternoon

The absolute accuracy for GOES-13 visible channel is 5%, so 2% is well within the noise. http://noaasis.noaa.gov/NOAASIS/ml/imager.html


Did the Geothermal Power Plant contribute to Pahoa's Lava Fissures?

I think we can confidently enough answer "no".

From this report (p. 23):

  Kilauea eruptions occur either at its summit or within two well-defined swaths (called rift zones) that radiate from the summit.

In the same report, you can see those rift zones on the map page 3, or on the sketch of the plumbing system page 24. Most lava flows are emplaced along these rift zones. For instance, look at this map of the 1955 eruption (public, USGS):

You can see that the flows were emplaced in a zone very similar to the 2018 eruption. Apparently the power plant opened in 1993, and the first exploratory wells were drilled in 1961-62, so after this event.

Actually, the rift zones are in the lava flow hazard zone 1, i.e. the most severe. The zone is defined as follow by Mullineaux et al. (1987):

  Zone 1 consists of the summit areas and active parts of the rift zones of Kilauea and Mauna Loa; in those areas, 25 percent or more of the land surface has been covered by lava within historical time, during the 19th and 20th centuries. These areas contain the sites of most historical eruptions, and a large majority of the lava flows that will affect other zones on Kilauea and Mauna Loa in the near future probably will originate in zone 1. [emphasis added]

So the real question is: why was the plant (and estates...) built in an area well known for lava flow hazards?!

To conclude, by researching the plant I found that it is actually where people accidentality drilled through a magma pocket for the first time (Teplow et al. 2009)! They hit a pocket of dacite; it did not trigger an eruption. This happened later in Iceland: this time, they even used it as an experimental well (there is a whole issue of Geothermics on this). Again, no eruption.


Why do forecast areas on weather.gov appear as twisted squares?

The equal-area projection used to create the continental U.S. grid for weather forecasting does not represent lines of latitude and longitude as straight lines.  Instead, they are curved (note the Canada-USA border curvature below, which is actually along the 49th latitude parallel).

An equal-area projection is often used in regional weather modeling because it is important for each cell to represent the same amount of physical space during computation.  The map images posted in the question are using the Mercator (or Mercator for web) projection, where lines of latitude are horizontal and lines of longitude are vertical, but area is distorted.

The grid-cells from the regional model are also designed in a way that maximizes cost-benefit of computational power.  You want to minimize the number of grid cells modeled while maximizing the number of grid-cells in your domain that are relevant to your study area.  This ensures that you aren't wasting computational power and is often achieved by changing the orientation of the domain a little.  So, the grid cells do not align nicely on the mercator map used on web-app maps like google, bing, etc.  However, the grid-cells line up perfectly if visualized using the native projection for the image frame (e.g. below).

The reason that Hawaii does not follow the pattern is that the data for Hawaii is obtained from a model simulation for a different grid.  The nesting of grids can be seen in the image below.

I suggest you read Grids in Numerical Weather and Climate Models
 for a nice look at the factors involved when choosing a grid definition.
You could also look here for more info about NOAA prediction.

North American Mesoscale (NSM) forecast model domains used by NOAA.


When did people first realize that hurricanes are a particular type of storm system?

Trade winds from a general easterly direction occur almost all of the time in the Caribbean and tropical Atlantic. A hurricane approaching from any direction except from the south would change the wind direction because winds spiral counterclockwise into the hurricane. For example, a hurricane approaching from the east would cause the wind direction to back to northerly, while one approaching from the west (very rare in the tropics before recurvature occurs) would cause the wind direction to veer to southerly. If a hurricane approached from a general southerly direction, the wind would remain generally easterly but would gradually increase in speed and gustiness. Hurricanes are often accompanied by anticlyclonic outflow at upper levels; this could be observed via the motion of cirrus clouds.


Is there any place on Earth that is permanently a desert?

UserLTK, above, has mentioned the Dry Valleys of Antarctica and the Atacama desert as two notable examples of "permanent" deserts. It's likely that these two locations have been deserts for at lease several millions of years despite changes in global weather patterns from multiple glacial events. The Atacama is well shielded by a very old mountain range to the west, and the Dry Valleys are also well protected by both an inland location and by global ocean currents that isolate the continent from much of the world's weather.


The short answer: cubic cleavage is a specific example of 3D cleavage.

A cleavage plane is a direction along a crystal splits preferentially when exposed to mechanical stress.

3D crystals are objects where the spatial arrangement of atoms is described in  a parallelepiped (a unitcell) with three outer dimensions $\vec{a}$, $\vec{b}$, and $\vec{c}$.  The crystal literally builds by translating this unit cell along $\vec{a}$, $\vec{b}$, and $\vec{c}$.

The relative lengths of $\vec{a}$, $\vec{b}$, and $\vec{c}$ and the angles these vectors enclose define the outer symmetry of this unit cell.  Any of the following seven crystal classes is possible: triclinic, monoclinic, orthorhombic, tetragonal, rhombohedral, hexagonal, and cubic.  In this set, starting from triclinic and moving to cubic, the inherent symmetry increases.

With $\vec{a}$, $\vec{b}$, and $\vec{c}$, equally crystallographic planes may be defined; some of them parallel to / coincide with the cleavage planes observed at macroscopic scale.

Thus, as listed e.g., in Wikipedia you may discern:

Basal or pinacoidal cleavage occurs when there is only one cleavage plane. Graphite has basal cleavage. Mica (like muscovite or biotite) also has basal cleavage; this is why mica can be peeled into thin sheets.

Cubic cleavage occurs when there are three cleavage planes intersecting at 90 degrees. Halite (or salt) has cubic cleavage, and therefore, when halite crystals are broken, they will form more cubes.

Octahedral cleavage occurs when there are four cleavage planes in a crystal. Fluorite exhibits perfect octahedral cleavage. Octahedral cleavage is common for semiconductors. Diamond also has octahedral cleavage.

Rhombohedral cleavage occurs when there are three cleavage planes intersecting at angles that are not 90 degrees. Calcite has rhombohedral cleavage.

Prismatic cleavage occurs when there are two cleavage planes in a crystal. Spodumene exhibits prismatic cleavage.

Dodecahedral cleavage occurs when there are six cleavage planes in a crystal. Sphalerite has dodecahedral cleavage.

Intentionally, the description above was constrained to 3D crystals which eventually are (typically) described in one of 230 3D space groups. Note, there equally are seven planar space groups, and a growing body of work about 2D crystals, too.


Are Richter-magnitude 10 earthquakes possible?

Adding a little bit of practical data to the above answer about the Gutenberg-Richter relationship, here is a plot of the per-year cumulative probability of earthquakes in a particular province in Japan, based on observed frequencies over many hundreds of years:

Source: http://topo.earth.chiba-u.jp/afr/backnumber/no24/24%E5%8F%B702kumamoto.pdf

The relationship is plausibly log-linear (in accordance with G-R); if you accept that the relationship will hold up to higher magnitudes, you would estimate the probability of a M10 event in this location at once every 30,000 years.

To get an estimate for "anywhere in the world" you would need the cumulative data for all. A good place to start is the USGS website - they have a handy table with data since 1900.

Taking these data and plotting them on a log-linear graph, then extrapolating a linear fit, gives the following chart:

This is rather scary because it suggests that there is about a 1:100 chance of a M10 earthquake anywhere in the world, in any given year. Note that I plotted the data for magnitude x to x.9 at the location of x - which tends to underestimate things a little bit. Note also that at the extreme end of very large earthquakes (8 and above), the data looks like it might be deviating from the straight line - but there are not enough data points to draw any firm conclusions about the shape.

There are several other caveats. First - the assumption that the model can be extrapolated can be challenged: a particular fault may not be constructed in such a way as to store the energy required for an M10 event, as it will always release the energy before it gets there (and there may be a "stress shadow" effect, which says that after a big earthquake, the chance of another big one is temporarily reduced because stresses have been relieved; this is why this model can only be used "over a long period", and does not accurately reflect the risk of earthquake in the next five years).

Still - one percent.


Is the representation of Earth oriented correctly in this image?

The way "I see the diagram" it looks weird, but it is probably correct.

My perception is that it is about frames of reference. We are used to seeing the Earth with Earth's north at the top, not at 23.5 deg from the vertical.

This picture has the ecliptic pole at the top. As you state, Earth's equator matches the celestial equator. The Earth's poles would be at right angles to it and the north pole would be in the middle of the white mass at the top of the Earth - representing the Arctic ice sheet

If you imagine a line from this point, perpendicular with the celestial equator it matches zero degrees latitude and is sort of sub parallel with the west coast of Africa.

The other thing that might be confusing our perception is the representation of the land masses is not accurate.

Looking at the representation for Europe, Italy is not present, neither is the Balkan Peninsular or the islands of Britain & Ireland. Scandinavia just looks wrong! Iceland is present and the Arctic ice extends too far southwards. The last time I looked there wasn't an ice sheet between Iceland & Greenland. Also, the Red Sea isn't represented.


Given a 3 box model of the atmosphere,determine fraction of mass transport from lower layer

My results:

If an $\alpha$ sized chunk was moved up, then what remains is $(1-\alpha)$.  Because each layer is 300mb thick, they have equal parts mass.  What goes up must be compensated, and due to conservation of mass, there is no gain or loss of mass in my system.  Potential temperature is a conserved quantity, so I can say:

$$350K(1-\alpha)+325K\alpha = 325K(1-\alpha)+375K\alpha$$
$$\alpha = \frac{1}{3}$$

Further,
$\frac{1}{3}$ mass movements equates to $\frac{1}{3}$ of a $300mb$ layer, which is $100mb$ subsidence.


Observations of sea ice typology in the Arctic

Concerning datasets, there are several, depends on Your final purpose.
One of the simplest is output from OSI-SAF, which has info about Sea Ice edge, concentration, type, emissivity and drift. It is derived mostly from polar orbiting meteorological satellites, I guess.
More detailed info can be found on Univ of Illinois pages
Visual images from  MODIS satellitescan be handy
But, most detailed info about sea ice conditions can be derived with radar satellites, with commercial options like radarsat being available. From free sources, data from Sentinel-1 is available since late 2014.
So the best info is available from satellites, in-situ measurements do not cover most of the area.
Our knowledge of Arctic sea ice is pretty good IMO, at least since 1979 when satellites started to measure the extent. Understanding the age of sea ice is a bit more tricky, as it can be up to ten, maybe even few more years
old.


Can overshooting tops enter the stratosphere, or will they rather push the tropopause upward?

This is an interesting question and I feel the best answer is "a little bit of both".  Consider your picture without the overshooting top and just the anvil.  The tropopause is clearly defined as the anvil spreads out horizontally against the static stability of the stratosphere.

Now let us consider a single parcel in the updraft, accelerating upward due to CAPE and hitting the tropopause with significant vertical momentum.  This parcel is full of tropospheric air but makes an excursion into the stratosphere until the negative buoyancy it experiences reduces its vertical velocity to 0 (the top of the overshooting top) and then continues to accelerate downward to eventually oscillate about its equilibrium level (level of neutral buoyancy).  For this single parcel I would argue that it is in the stratosphere.

Now consider all of the updraft parcels doing this.  They are all transporting tropospheric air upward into the stratosphere -- inside that cloud it is all tropospheric air and toward the edges it is a mix of tropospheric and stratospheric air due to entrainment.  The tropospheric air transported upward that mixes with the stratospheric air will reduce the static stability locally and may eventually modify the temperature gradient such that you consider the tropopause lifted.

I may run a simulation later with many grid levels around the tropopause and see what a model has to say.  I'll edit with my results.


How much of the Earth's surface was covered with water during the Mesozoic?

At that time there were large areas of shallow seas, so small differences in elevation could lead to large differences in water area. I doubt that the elevations are well enough known to estimate the area precisely.
https://dinosaurpictures.org/ancient-earth/#90
The coefficient of expansion of water is only $207 \times 10^{-6}/K$. 15 degrees would give 0.3%.


Are Southern Hemisphere 'Polar Blasts' caused by global warming?

Looking at the southern hemisphere polar jet, there is a large trough (meander) in the jet stream bringing cold air a long way north from the Antarctic to south eastern Australia at present. The relevance of the position and shape of the jet stream is thought to be important for keeping cold polar air in place above the antarctic/arctic, with a strong jet stream keeping cold air hemmed in above the poles while a weak jet stream meanders allowing cold waves/blasts to escape into the midlatitudes. This previous Stack Exchange question sets out how the theory of arctic/polar amplification which is thought to affect the position and shape of the jet stream is being studied in connection to climate change (Generally it has been proposed that global warming will make the jet stream more wobbly/meandering, and so increase instances of cold weather in the mid latitudes).

Single weather events such as this are not caused by global warming, but it has been hypothesised that their frequency of occurrence may change due to warming, though as you can read in the linked stack exchange this theory is not universally accepted.


Was Judea as desertified 2000 years ago as it is now?

There has always been a strong north-south rainfall gradient from approximately the Golan Heights (high rainfall), through Judaea and southwards into the arid Sinai desert. We know that this gradient once had a higher rainfall because there are abundant rainfall harvesting and groundwater collection structures, built by the Nabatean civilization about 400 BC to 100 AD. They are now long defunct because of climatic desiccation, but are characteristic of dry regions with marginal cultivation. Similar working structures can now be found in Oman, Iran and parts of north Africa. So it is reasonable to deduce that the region was once significantly wetter.

In addition to this overarching climatic gradient there is a very strong local rainfall gradient from the Israeli Highlands in the west, about 1100 mm, to the Dead Sea-Rift Valley in the east which receives about 100 mm per year. This is a rain shadow effect from the prevailing Mediterranean airflow to further east into the Arabian desert - obvious from the vegetation density on Google Earth. Masada currently receives about 200 mm per year, but it is very erratic, with huge inter-annual variability. One study of the Negev and adjacent dry areas concluded that the spotty distribution of annual rainfall doesn't average out - at least, not on any time-scale useful to humans. Because of its position in an intense rain-shadow, Masada will always have been dry.

There are two potential sources of water supply to places like Masada; springs, and ephemeral (very unreliable) flash floods. The springs are groundwater which has percolated through the limestone massif from the Israeli highlands around Jerusalem,. Hebron, etc. There are currently no springs in the immediate vicinity of Masada. There may once have been a spring under higher water-table conditions, but none that we know of historically.

The solution used to supply Masada was to build two dams in canyons draining the Israeli highlands. Ephemeral flash floods were thereby intercepted and conveyed by gravity-drained aqueducts to tanks dug deep inside the Masada citadel. These tanks were of about 30 to 35 cubic metres capacity, so even allowing for evaporative and seepage losses in the aqueduct, the water supply could be topped up - provided the frequency of flash flood was sufficient. Even today it is likely that flash-floods would have been of adequate frequency. Under the slightly wetter regime of 2000 years ago, it is near certain that there would have been more than sufficient water every year.

Unfortunately, there are no rainfall records from so far back, but some interesting general climatic inferences may be found in Issar A.S. (1990), 'Water Shall Flow from the Rock'.


What is the residue after plant matter is completely decomposed?

From Mary, et al., 1996, Figure 1 on the second page shows a nice breakdown of there the Carbon and Nitrogen go when a plant decomposes. Hadas, et al., 2002 has experimental data from plants with C:N ratios from 11:1 to 136:1.

In summary, most carbon ends up in microbial biomass or as CO$_2$. If the process is allowed to proceed to infinity, then presumably, the microbial biomass will itself be decomposed by other microbes until all of it in the form of CO$_2$.

The nitrogen is mineralized, first into NH$_4^+$, then NO$_2^-$ and eventually all into NO$_3^-$ by various decomposing microbes. Normally these ions are attached to something else that they got from the soil and thus enter the soil. In the regular nitrogen cycle, the ions that eventually end up as NO$_3^-$ will be processed by other bacteria into atmospheric nitrogen (N$_2$). So, if decomposition proceeds to infinity, all the nitrogen would be expected to return to the atmosphere eventually too.

That leaves us with the small amounts of other biologically common elements such as phosoprous, calcium, etc. This will be the only stuff left. From Latshaw, 1934 (side note, not the oldest paper reference I've seen on Stack Exchange, but close!), and the chemical composition of a corn plant, considering all parts of the plant together, there is on average 44.9% oxygen, 44.0% carbon, 6.3% hydrogen, and 1.5% nitrogen by mass; all of these would be expected to leave as as gasses.

Of the remainder, representing 3.3% of the mass, probably most of that would stay. Whatever elements are left are most likely to form molecules with oxygen, since it is freely available in the air and likes to react with just about anything. Silicon, potassium and calcium don't seem to have any likely gaseous forms they could take, while sulfur (SO$_2$) and phosphorous (possibly P$_2$H$_4$) might take some more volatile forms. In the end, little will be left of the former plant, just a scattered residue of oxidized minerals, around 3% of the original mass.


Coal, oil and gas can be found anywhere on Earth: on land and on sea.

The Japanese once mined coal from beneath the sea floor, using Hashima Island as the base of operations. The island is only 6.3 ha in size.

The British also mined coal from under the sea, in the north east of England.

As for oil, the following picture from USGS shows the current oil-prone petroleum basins in the world, outside of the USA.

The following oil basins are all on land:

Russia
Arabia, Kuwait, Iraq and north Africa
China
Eastern Europe
The Canadian oil sands and other plays in the province of Alberta, Canada
Colombia
The oil fields of east Texas, USA
The former oil fields in Pennsylvania

When oil & coal deposits formed they each had different mechanisms of formation. Coal is derived of plant material that once lived on land, died, formed peat, which then got buried.

Oil however, owes its origins to algae & plankton that lived in the oceans. When they died they sank to the bottom of the ocean and deposited in waters with very low, if any oxygen. Eventually, these deposits were covered by sediments. If the deposit was subject to a significant amount of geothermal heating gas deposits formed, otherwise oil deposits formed. Movements of the Earth's crust then repositioned some of the deposits which is why some coal is now under the sea and some oil is now on land.


I have found an archive of small-scale charts encompassing part of Africa here.


Would the climate still rise when we reduce our carbon footprint down to zero?

While I can't comment on whether the global temperature would still rise if humanity's carbon footprint were reduced to zero, I do know that the heat our machines produce would not contribute any net warming effect. The issue is not the amount of heat which is entering the atmosphere -- it's where and when that heat came from.
Assuming that we are successful in reaching 100% wind, water, and solar energy, the energy consumed at the end use would be from only seconds to at most a year ago:

Seconds: Solar power from photovoltaic panels is collected, converted to electricity, transmitted to the grid, and used almost instantaneously
Hours to days: Wind is primarily generated by daily fluctuations in surface temperature caused by the sun
Months: The water in rivers which power hydroelectric dams is generated by seasonal patterns in precipitation, which is a function of changes in solar intensity over the course of the year

In contrast, the energy we're using today from fossil fuels is tens to hundreds of millions of years old, containing energy harvested from the sun over a period of over a hundred million years (source).
The energy from wind, water, and solar sources is being added to our atmosphere anyway -- we just convert it into different forms to use it for other purposes in between its transit from the sun to the planet.
By contrast, the energy from fossil fuels had been locked away, safely stored for 100 million years or more. Until we started drilling and mining, it was no longer having any significant impact on the planet.
This also applies to the food we consume -- the energy that fruits and vegetables store is at most a year old -- for animal products, it's at most two to three years old.


Utah rock pattern identification - looks like small craters

Relief features made by microbial mats which were gelatinous. The ridges and fins are thought to be caused by small currents that flowed through the umbrella like bodies of these ancient bacteria.


Why do Tsunamis travel slower than sound?

Tsunamis and sound waves are different types of wave - one is a transverse wave and the other is a longitudinal one. Let's look at the factors that influence the speed of each one.
Tsunami - transverse wave in shallow water
A transverse wave is one of the type that we think of from day to day - where the direction of oscillation is perpendicular to the direction of travel. The speed that a transverse wave travels at depends on different factors depending on the depth of the water. For this purpose, "shallow water" is usually defined as existing where depth < wavelength/20. The wavelength of a tsunami is very large - of the order of hundreds of kilometres - so for a tsunami, any part of the world's oceans counts as "shallow water".
In shallow water, the speed of a transverse wave can be described by,
$$V = \sqrt{gD}$$
where $V$ is the wave's speed, $D$ is the depth, and $g$ is the acceleration due to gravity (9.81 m/s2). In the case of a tsunami in the deep ocean, then, if we assume a depth of 4 km we can estimate a speed of 198 m/s, or 713 kph. That's a back-of-an-envelope calculation, but it's sufficiently similar to the 800kph that you quoted in the question that I'm happy with it.
Sound - longitudinal wave
In a longitudinal wave, the direction of the oscillation is parallel to the direction of travel - i.e. it's an oscillation in the density of the material. We don't see many of these in everyday life, but one good example is the wave that moves down a slinky if you jerk the ends towards or away from each other.

(Image source)
Sound, in a liquid or a gas, is an example of a longitudinal wave. The speed of a longitudinal wave depends on the stiffness and the density of the material that it travels through, in the following way:
$$c = \sqrt{\frac{K}{\rho}}$$
where $c$ is the speed of the wave, $K$ is the bulk modulus of the fluid, and $\rho$ is its density. Note that there is no dependence on the depth of the fluid (in this case the sea) - sound one metre below the surface of a salt water swimming pool would move at roughly the same speed as sound one meter below the surface of an ocean.
Summary
Sound waves and tsunami waves propagate through different mechanisms, and thus different things influence their speeds.


Why Earth's magnetic poles are (and were) in their positions?

Actually, all of the three questions are directly related.

As you noted, the Earth's magnetic field is generated by flow of molten conductive materials (probably mostly iron) in the outer core. (Contrary to popular belief, only the outer core and the oceans are liquid. The crust, mantle, and inner core are all solid. Solids can flow, but they have a shear strength. Liquids have no shear strength.)

However, this flow is turbulent.

Therefore, Earth's magnetic field can be described by having two poles, but in reality, it's much more complex.  Part of this is due to crustal structure (i.e. the magnetic anomalies we typically work with in geophysics to, say, locate an igneous intrusion or understand the evolution of the oceanic crust). However, the part that's due to crustal structure doesn't change rapidly and only influences a very small region.  The "longer wavelength" (i.e. broader) differences are due to variations in the flow of the outer core.

Every "eddy" (i.e. any vorticity) in the flow produces its own magnetic field.  The sum of these is what produces the global magnetic field.  At the surface, we're a significant distance from the flow that produces the magnetic field.  Therefore, it gets "smoothed out" and these small perturbations are less apparent. It looks fairly close to a single magnetic dipole because the bulk of the flow is approximately around Earth's rotational axes.

This is the answer to your first and second questions.

The magnetic field is not the result of a single "current" of laminar flow, but rather the sum of much more complex turbulent flow.  Therefore, it doesn't exactly match the rotation axes, and the position of the magnetic poles changes over human timescales.

I'll hold off on answer the third question, as you split it into a separate question, but basically, the flow tends to self-organize, producing a strong magnetic field that's close to a single dipole.  Then, for reasons that aren't entirely clear, it episodically de-stabilizes (basically, you'd have multiple equally strong poles that somewhat cancel each other out), and then re-stabilizes (sometimes in the opposite direction).

If you're interested in a "pop-sci" style writeup, there was a very nice Scientific American piece on this awhile back written by Glatzmaier (a very big name in the field).  Unfortunately, the web version is missing  the figures that were in the print version, but they're mostly from Glatzmaier & Roberts, 1995 or one of the follow-up papers by the same authors.


Is there an estimate for the length of the American Cordillera?

A difficulty in finding a resource that provides an estimate of the length of the entire American Cordillera is as, @fre0n states in the comments

  there is very few standard mention regarding this unit as a whole, instead it is more often localized

As can be read in the Wikipedia article American Cordillera, where they state that

  The American Cordillera is a chain of mountain ranges (cordillera) that consists of an almost continuous sequence of mountain ranges that form the western "backbone" of North America, Central America, South America and Antarctica. It is also the backbone of the volcanic arc that forms the eastern half of the Pacific Ring of Fire.

The Wikipedia article lists the constituent mountain ranges as being a line of overlapping mountain ranges from the Alaska and Brooks Range right through the Andes to the southern tip of South America, also suggesting that the Cordillera could

  possibly be followed through the arcuate South Georgia Ridge across the Southern Ocean to the mountains of Graham Land on the Antarctic Peninsula.

Given the earlier definition of the Cordillera being

  an almost continuous sequence of mountain ranges that form the western "backbone" of North America, Central America, South America and Antarctica.

according to Google Maps, where I used this definition and traced the distance (right-click on map -> Measure distance -> select points along the 'spine' of the Americas and Antarctica), I determined an estimate of the total length as being approximately 17,000 km (or about 21,500 km if the South Georgia Ridge and Graham Land Mountains are included).


Where can I get good raw meteorological data for Middle East?

This work intend to consolidate climate data series from 79 stations in North Africa and the Middle East, between the 19th century to present day. Distribution of the stations are shown in Figure 1:

One challenge they had was an amount of incomplete records in the available dataset during that studied lapse. Table 1 and 2 show the name and location of the sources for the datasets, which could be useful for you to get started if your research need recent (2008-2018) records

Further they provided as supplemental material their raw datasets used in their research here, as used in their paper (-> 2008).


How can a starting point south of the north pole to an endpoint north of the south pole be halfway around the world?

I think you are too into the numbers.   When talking about longitude and latitude they are two different things.  Longitude goes east and west from a point on the earth from 0 to 180 degrees east and 0 to 180 degrees west and describe semicircles of equal length running from the north pole to the south pole.  Latitude however is in degrees north or south of the equator and describe circular lines around the earth that get smaller as you get closer to the poles.  See this image:

Slicing along the lines of longitude would give you wedge shapes like when you think of orange slices.  Slicing along lines of latitude would give you circular slices of varying sizes.
To get an 'opposite' longitude you spin the earth on it's axis 180 degrees.  The opposite of 1 degree east is 179 degrees west.  The opposite of 10 degrees east is 170 degrees west.  So in your example it will be from 121.895 degrees west (longitude -121.895) to 58.105 degrees east (longitude 58.105).
The opposite latitude is also 180 degrees away.  However there are only 180 total degrees of latitude because their shapes are different.  Longitude goes from 0 to 180 degrees east and 0 to 180 degrees west giving a total of 360 degrees.  Since latitude represents a circle, there are only 180 degrees, from 0 to 90 degrees north and from 0 to 90 degrees south.  But you need to go a full 180 degrees.  The way it works is that it is just the same degrees,  just flip north and south.  Think about two points on opposite sides of the planet on the equator.  They are both at 0 degrees latitude.  If you move north 1 degree, the opposite will now be one degree south.  If you continue to 90 degrees north, the opposite is now 90 degrees south, which makes sense since the north and south pole are opposite.
So let's take Monterey, CA at 36.6 (positive is north), -121.895 (negative is west).  You cannot get to the opposite side of the earth by walking along a line of latitude because all lines of latitude except for the equator are not the right size.  If you divided the earth at that line of latitude you would not get two equal halves.  You can however get to the opposite side of the earth by walking along lines of longitude.  So say you start walking north.  After 53.4 degrees of latitude you arrive at the north pole.  Keep going in a straight line and you will start walking south on longitude 58.105 (positive is east).  After you reach the equator you will have travelled 143.4 degrees.  You have 36.6 degrees of walking south to go before you end up at -36.6 degrees latitude (negative is south).  You've travelled 180 degrees in a straight line to get to the opposite side of the earth and now you're in the Indian Ocean.


Have these optical satellite ground station locations been chosen for clear skies?

Yes, mean cloud cover is routinely measured from satellites.  Like all satellite data (and in fact all measurements), it does have an uncertainty, but for the purpose of this question the satellite product of mean cloud coverage is good enough.  Personally, I would hesitate to trust the data at very high latitudes with frequent cover of snow or ice, because those are hard to distinguish from clouds with any form of passive down-looking remote sensing.

The figure below is released by ESA and derived from the MERIS and AATSR instruments on the former ENVISAT satellite.  Although the caption does not specify, I suspect that it uses visible channels and that it therefore only relates to daytime cloudiness. It would be interesting to see how it compares to 24-hour cloud cover!

Full image caption and credit:

  Global annual mean cloud cover derived from three years (2007–09) of Envisat data. The map shows areas with little to no cloud coverage (blue) as well as areas that are almost always cloudy (red). Clear skies are immediately visible over deserts (such as the Sahara, Namib and over the Atacama). Regions with constant high-pressure systems are detectable, such as the Pacific. On the other hand, the North Atlantic Ocean in the mid-latitudes is mostly covered with clouds, where storms develop. The UK is often cloudy, whereas the Mediterranean shows sunny skies.
  Data from both the MERIS and AATSR instruments on Envisat were used.

  Credits: ESA/Cloud–CCI

(On behalf of ESA and the Cloud-CCI I would like to apoligise for the poor choice of colourmap, but this figure predates the time at which many of us were educated on its poor suitability)


What is are the benefits of map 500 hPa geopotential for weather forecasting?

The 500 hPa geopotential is usually called the Level of non divergence. It is understood that beneath that level there is a level of convergence and above that level there is a level of divergence so that this surface(500 hPa) is the surface to look for vertical motions. To  a first approximation one can assume the time derivative of density of a particular surface to be a constant and then set the continuity equation to be zero.

$$\nabla. u = 0  $$

We can then expand on the terms on the LHS and separate the horizontal components from the vertical components.

This turns out to be(in isobaric coordinates)

$$ \nabla_H. u_H = -\frac{ \partial \omega}{\partial p}$$

One can then conclude from this equation the following

1) If the LHS is less than zero then there is an increase in vertical velocity with decreasing pressure

2) If the LHS is greater than zero then there is a decrease in vertical velocity with decreasing pressure

Since the atmosphere is bounded at two levels (the surface and tropopause) the vertical velocity is close to zero at these two levels. However in between these two boundaries the vertical velocity starts to increase and achieves it's maximum value before again tapering off at the tropopause. The level of maximum vertical motion is usually the 500 hPa and this is also called the level of non divergence.


Need help identifying this rock that my grandfather found in or around Hot Springs, AR 75 years ago. PLEASE

From the looks of it, I would suggest that it is more likely pyrite than gold. However, it is difficult to say without inspecting it in person.


Can you have a low pressure with no clouds?

Low pressure areas are simply areas where the air is less dense than the surrounding areas.  The scale of these differences can be small, size of a small farm field, or large, covering expansive areas of the earth's surface.

Clouds form around low pressure systems typically because of the way low pressure systems develop and the air moves around them.  Typically low pressure systems start forming over relatively warmer sections of the earth, this is gross over simplification, as the air warms it becomes less dense and will tend to rise.  This rising of warm often moist air leads to cloud formation because as air rises it also becomes cooler, see Lapse Rate.  Typically at some time the temperature drops below the dew point in the air moving upwards and clouds form.  But if the relative humidity of the air is not high it is possible that the air mass in the low pressure system never rises high enough for condensation of the water vapor to occur, in this case no clouds will form.

In conclusion it is possible to read a lower air pressure at a site and still not have cloud formation.

This site has a good graphic on how low and high pressure systems relate to each other. High and Low weather systems described

Please note that I am not meteorologist.  This answer does not consider all the aspects associated with cloud formation such as how airborne particulates can affect condensation


How is the radiative efficiency of a given gas (like a given CFC) analytically calculated?

Radiative efficiency is the radiative forcing per unit change in halocarbon mixing ratio.  Radiative Efficiency assumes that the Radiative Forcing is linear in mixing ratio—this is normally appropriate in the case of small perturbations in mixing ratio about current concentrations and is also appropriate for CFCs.  To calculate it requires a radiative transfer model, which includes both analytical and theoretical contributions.

The diagram shown in Hodnebrog et al.: "Global warming potentials and radiative efficiencies of halocarbons and related compounds: A comprehensive review" in Reviews of Geophysics
( http://dx.doi.org/10.1002/rog.20013 ) explains how radiative efficiency is calculated and is pasted below:


Why isn't the acceleration of an object in Earth's gravity related to the object's mass?

This is clearly related to Newton's 2nd law, but since you have asked about the phenomenon rather than the formulas, a Physics SE user explained much better than I can possibly can:

  (..) the gravitational force is different on different masses. Half
  the mass only has half the weight, so it is also only pulled in half
  as much by gravity.

  What gravity does is to pull in every single "particle" equally. If
  there are double as many "particles", then the pull in each is still
  the same and each accelerates the same amount.