how can a complete novice find ore? I live near Sydney, Australia and I really like the idea of sourcing my own ore from the wilderness. I want to give my kids a really hands on education into how primitive industries actually kicked off.

Is there some source of public information explaining exactly where you would find surface deposits of ore? Is this going to be achievable for a lay person?

Edit:
I'm thinking of copper ore, for example, which can be smelted into pea sized droplets relatively easily using a makeshift clay furnace with bellows. I didn't specify exactly which ore because I realise that ore is regional. I realise that it is unlawful to steal materials out of the environment, but I'm assuming that the likelihood of being prosecuted for conducting a small scale science experiment for your kids is relatively low. taking a couple buckets of gravel is not going to cause an environmental disaster. If the land is privately owned then it may be a matter of simply asking the landowner for permission.

how can a complete novice find ore?

You don't.

In addition to Fred's excellent answer regarding the legalities of the matter, I would add the following:

All of the easy stuff has already been found, and finding any more ore requires efforts beyond what you can do on your own.
Stuff that is easy to find now is commonly in old mines or old mine tailings. These are dangerous places to be in. There is risk of collapse, acid mine drainage, and a bunch of other nasty things.
Learning how to find "ore" regardless of the first two points requires proper education in geological sciences.
Smelting copper ore, as per your idea, is a bad bad bad idea. Copper ore is usually in the form of copper sulfides. You really don't want to breath sulfur-bearing gases while you're smelting the ore.

If you are interested in looking for minerals and such, as a resident of New South Wales I recommend you talk a look at this website:

https://www.resourcesandgeoscience.nsw.gov.au/miners-and-explorers/fossicking-in-nsw

It contains information and resources that might be useful to you.


How to obtain the saturation point of a soil layer from the volumetric soil water content data? Would you please let me know if the following is accurate as an answer to the question:
Suppose we have sensors which measure the volumetric soil water content of a soil layer for a long period (more than 6 months). Could I attribute the maximum (percentage) in this dataset to the saturation point?
I understand that we do need to have a lengthy rainfall event to have a saturated soil condition. How long the event should be so that the above proposal works?
And if there is any other way that I can get to the saturation point of a soil layer from the volumetric soil water content data/sensors, please let me know.
Mostly focused on the topsoil and preferably only using the dataset.
Thank you all.

I assume you mean sensors that measure volumetric soil water content. The answer will depend on the amount of precipitation and the permeability of the soil. Soils don't reach full saturation in some climates - especially coarse texture soils. I would say that if the soil water content reaches a maximum value that persists for a period of time after large rainfalls (giving a flat top) and ideally reaches the same maximum level during multiple wetting-drying cycles that it is reasonable to infer that is the saturation value.
Ideally you would collect undisturbed soil samples of known volume (e.g in by driving a cylinder into the soil (gently as possible) to determine the pore volume and saturated water content.
Also be aware that many soil moisture sensors give readings that are dependent on soil properties and are not linear up to 100% saturation so they may not give good values at high water contents. It is best to calibrate those types on your soils.


How is global weather model skill measured? The models I have in mind are the major global models such as the Global Forecast System (GFS) and the European Center for Medium range Weather Forecasting (ECMWF).

These models produce numerous output and as such there are numerous ways to validate the forecast they produce. I've seen many plots that focus on specific heights on the 500 mb surface to gauge forecast accuracy, but that is all I'm familiar with.

How is model skill and forecast correctness measured?  Is it an overall global skill, or are various skills in specific regions of the globe also evaluated?  Are these validation results freely available on the internet?

There are lots of articles that assess global models both locally and globally. They usually focus on a few parameters that are easily comparable with observations (e.g., temperature, wind, surface pressure). A couple of examples of the many available:

for ECMWF:

Roberto Buizza, 1997: Potential Forecast Skill of Ensemble Prediction and Spread and Skill Distributions of the ECMWF Ensemble Prediction System. Mon. Wea. Rev., 125, 99–119, doi: 10.1175/1520-0493(1997)125<0099:PFSOEP>2.0.CO;2.
for both only looking at precipitation:

Thomas M. Hamill, Renate Hagedorn, and Jeffrey S. Whitaker, 2008: Probabilistic Forecast Calibration Using ECMWF and GFS Ensemble Reforecasts. Part II: Precipitation. Mon. Wea. Rev., 136, 2620–2632, doi: 10.1175/2007MWR2411.1.
For the climate model world, check out: http://www-pcmdi.llnl.gov


How do I convert specific humidity to relative humidity? How do I convert specific humidity to relative humidity? What variables are needed (e.g. air temperature, pressure, etc.)?

Relative humidity is just $e/e_s$, the ratio of the vapor pressure to saturation vapor pressure or or $w/w_s$, the ratio of mass mixing ratios of water vapor at actual and saturation values.  If you have specific humidity, which is the mass mixing ratio of water vapor in air, defined as:

$$ q \equiv \dfrac{m_v}{m_v + m_d} = \dfrac{w}{w+1} \approx w.$$

Relative humidity can be expressed as the ratio of water vapor mixing ratio to saturation water vapor mixing ratio, $w/w_s$, where:

$$ w_s \equiv \dfrac{m_{vs}}{m_d} = \dfrac{e_s R_d}{R_v(p-e_s)} \approx 0.622\dfrac{e_s}{p},$$

and from Clausius-Clapeyron:

$$ e_s(T) = e_{s0}\exp\left[\left(\dfrac{L_v(T)}{R_v}\right)\left(\dfrac{1}{T_0} -\dfrac{1}{T} \right)\right] \approx 611\exp\left(\dfrac{17.67(T-T_0)}{T-29.65}\right).$$

Once you have calculated $w$ and $w_s$ you can obtain the relative humidity as:

$$ RH = 100\dfrac{w}{w_s} \approx 0.263pq\left[\exp\left(\dfrac{17.67(T-T_0)}{T-29.65}\right)\right]^{-1}. $$

You could also calculate $RH = 100(e/e_s)$, but I think since you are starting with $q$ it isn't as straightforward as doing it this way.

Variables used:

$q$ specific humidity or the mass mixing ratio of water vapor to total air (dimensionless)
$m_v$ specific mass of water vapor (kg)
$m_{vs}$ specific mass of water vapor at equilibrium (kg)
$m_d$ specific mass of dry air (kg)
$w$ mass mixing ratio of water vapor to dry air (dimensionless)
$w_s$ mass mixing ratio of water vapor to dry air at equilibrium (dimensionless)
$e_s(T)$ saturation vapor pressure (Pa)
$e_{s0}$ saturation vapor pressure at $T_0$ (Pa)
$R_d$ specific gas constant for dry air (J kg$^{-1}$ K$^{-1}$)
$R_v$ specific gas constant for water vapor (J kg$^{-1}$ K$^{-1}$)
$p$ pressure (Pa)
$L_v(T)$ specific enthalpy of vaporization (J kg$^{-1}$)
$T$ temperature (K)
$T_0$ reference temperature (typically 273.16 K) (K)


Can spraying sulphate particulates in the upper atmosphere to lower Earth's temperature have side effects? According to Geoengineering treatment stratospheric aerosol injection climate change study

  Planes spraying tiny sulphate particulates into the lower
  stratosphere, around 60,000 feet up. The idea is to help shield the
  Earth from just enough sunlight to help keep temperatures low.

  The researchers examined how practical and costly a hypothetical solar
  geoengineering project would be beginning 15 years from now. The aim
  would be to half the temperature increase caused by heat-trapping
  greenhouse gases.

Can spraying these sulphate particulates in the upper atmosphere to lower Earth's temperature have side effects?

Short Answer yes, there may be side effects.

Long answer. They are looking at more than just sulfates. Initial research has shown an optimistic lack of side effects for sulfates but we won't know until we actually test it and testing over a 1km area, something attempted multiple times, has been blocked for environmental concerns. Concerns exist because we can't test it because it's blocked for environmental concerns.

The thing to keep in mind is that the original idea was

proposed to shock the scientific community into researching multiple avenues by Lowell Wood in 1999 (I could not find the source but it was a Wikipedia article so take it as you will.)
is based off the cooling effects of volcanoes.

The second point is the most important because we have a natural phenomenon to work off of. there was a volcanic eruption in the 90s that lowered global temperatures by ~1.5 degrees (I believe it was C) over 18 months before things returned to normal. This means we can study the effects of volcanoes and predict the effects of the aerosol injection. Side effects of volcanoes include large impacts on vegetation and cooling but little else is cited in the articles I've seen. Fortunately, they believe that aerosolized injection into the upper atmosphere will avoid affecting plants very much if at all as the particles will mostly stay up there. However, there is concern that it might modify our atmosphere in some unknown way.

I have seen some articles saying sulfur emissions in China/asia might actually already be helping lower global temperatures a little but the idea was theoretical and the articles mostly seemed like clickbait.

Really we aren't sure and need to do some testing to find out. However, this will be small scale testing that will ramp up slowly.


Could an earthquake on flat land still kill someone without any buildings/boulders or such near by? Suppose you were sitting near the epicenter of an earthquake, sitting down on flat land without much else near by, maybe some grass. Is it still possible that say, a magnitude 8 or 9 earthquake would be lethal, assuming you can spread out enough to avoid tipping over and hitting your head?

It seems like the biggest damage in earthquakes comes from buildings, not the ground splitting apart, but it is also possible the terrain itself could warp like an ocean wave several meters I guess. And if a fissure was opening up near someone from the quake's epicenter, would they have time to react and avoid being dropped into it?

Yes, a man in the open sitting on grass could be killed by a magnitude 8 or 9 earthquake, but in the circumstances you describe it would be very unlikely. Not all large earthquakes produce huge rifts on land, but if you were unlucky enough to have one open up beneath you then the chances are you would fall into it. Another danger would be liquefaction. In some places the moisture content of the land is such that although in normal circumstances it is perfectly solid and stable, when subjected to the violent shaking of an earthquake a phenomenon called liquefaction occurs. The ground becomes a sort of quicksand into which buildings can collapse and humans disappear. It is quite rare, however, and fatalities among people are unlikely if they are out in the open where nothing can fall on them, though I dare say a few unlucky ones might have a heart attack. On hard ground,though, even a fit and healthy young man might be severely bruised and shaken in a magnitude 9 quake.


Why is Venezuela so dry? I'm trying to understand why the summer (January) precipitations in Venezuela are so low in the central and southern parts of the country as well as in eastern Colombia. The most plausible explanation I could find was a rain-shadow effect but unless I'm wrong, the mountains are on the northern side while the winds are blowing from the ocean, east to west. This should not result in a dry season.

What have I missed?

My references are :

http://research.jisao.washington.edu/data_sets/ud/samerica/

http://kids.britannica.com/comptons/art-166714/The-map-at-left-shows-the-wind-belts-of-the

In January the Intertropical Convergence Zone shifts to south of the equator. And the Subtropical High Pressure Belt moves into areas of Venezuela, which leaves it dry. I am guessing this from a study of principles of world geography and climatology; I don't live in Venezuela though.


How can I learn to spot fossils and fossil-bearing rocks on the Jurassic coast? Due to a pleasant change in my circumstances, I have been to Lyme Regis several times recently, and it seems as though I'm going to be a regular visitor over the next few years.

As a former biologist I found the experience of combing the shore of the famous Jurassic coast for fossils to be delightfully addictive. Having spent about six hours over three days doing so, I've amassed a decent collection of ammonite fragments. Having got the bug for doing this, I'm eager to try and find some more impressive specimens.

Chatting to some of the local palaeontologists on the beach it became apparent that to do so you need to split rocks, and that you need to know which rocks to split. It's also clear to me that while I've become attuned to watching for the shapes and patterns indicative of ammonites, I don't yet have the eye for the other colours and shapes of fossil shells and bones.

I tried to do some reading about the geology of the region, but most of what I found online was either too basic (i.e. aimed at non-scientists) or too advanced (i.e. aimed at actual geologists) for me.

So: how can I learn to better spot fossils and fossil bearing rocks in this region? I'd be very interested to learn some geology if it would help, but if that's the case, how can I find material pitched to the right level?

You are correct when you say it is mostly a matter of getting your eye in and getting used to the shapes and types of rock you need to be looking at. Hopefully however the below information will help you find a few more fossils or at least be of interest to you.

The main fossil bearing beds in Lyme Regis are from the Jurassic period and formed in a warm sea, hence the marine fossils.The Jurassic strata (layers of rock) in this area are mainly composed of grey limestones, shales and marls (lime rich mudstones).  These are the rocks you want to focus on when searching for fossils (see image below taken facing church cliffs).

The majority of the very well preserved ammonites are found in the limestone and can be found inside limestone nodules or lying loose on the beach. This rock is very hard and will require a good geological hammer and maybe a chisel to split. This is often easiest if you find a nodule with lines of weakness or cracks that you can split it along. It is also a good idea to look for signs that the rock contains a fossil before trying to break it, part of a fossil may be visible on the surface of the rock.

You can identify the limestone by it's lighter grey colour and hardness, it should be quite hard to break without a hammer.

It is however possible to find ammonites in the shale and mudstones, the ammonites in these however tend to be less well preserved. These rocks are typically a darker grey and very easy to break. However there are other fossils that can be found in these shales, including belemnite rostrum (belemnite guards). These bullet shaped fossils are from an ancient relative of the squid and the cuttlefish and are very abundant in certain areas.

Finally there are very soft mudstone beds out near Church Cliffs that you can walk across along the foreshore and are rich in belemnites, as well as pyrite (fools gold) and pyritised ammonites. These are formed when the original shell is replaced by pyrite in low oxygen conditions such as deep on a muddy sea floor and can be recognised by their golden colour and shine.

All the best with your fossil hunting!


ID on a potential flint fossil I was out in Swinley Forest in Bracknell(United Kingdom) last weekend. The area is a large pine forest but the ground is full of flint and is very sandy.

I stumbled across a piece of flint and noticed that there were some very interesting marks on it.

As you can see on the right edge of the rock there is an indentation with some vertical lines and some circular patterning. I've included a zoomed in image below.

My question is whether this could be plant fossil or even part of the fin of a fish with scales attached.

Many thanks!

Assuming that the identification of the rock as Flint/Chert and not Limestone/Dolomite is correct I'd say it's a formation artifact rather than a fossil. What I mean is that at some point as it was forming, flint can form in two ways one is due to silica saturated water acting on limestone and the other is by direct precipitation from mineral saturated water in shallow environments, the small bead-like structures were created within or on the surface of the rock. It may be a surface impression from a fossil that was once part of a larger piece but as it is now there's no evidence of a material difference between "fossil" and the country rock of the flint. This opinion is coloured by the fact that here in New Zealand fossils in our flint are vanishingly rare, I'm not sure if that holds in England though.


Negative values of relative humidity in CFSR I recently started using the Climate Forecast System Reanalysis (CFSR) 6-hourly products (http://rda.ucar.edu/datasets/ds093.0/), and I noticed many values of relative humidity are negative. Is this an error in the dataset? If not, how should I interpret negative values?

I am also getting negative relative humidity from the GFS (Global Forecast System): 36-hour predicted atmosphere dataset (Provider: NOAA/NCEP/EMC, their website: http://www.emc.ncep.noaa.gov/index.php?branch=GFS)

I found this link that answers as to why relative humidity is possible in their dataset:

http://ftp.cpc.ncep.noaa.gov/wd51we/reanal/random_notes/negative.h2o

This might apply to the one you have access too.

Let me know if this helps.


Does dirt compact itself over time? If so, how does this happen? If I were to bury something 10 feet (~3 metres) underground, with loose soil on top, would the ground naturally compact itself over time, until whatever I had buried has dirt tightly pressing against it on all sides?

What if I buried it 50 feet (~15 metres) underground?

If it exists, what is this compaction process called and how does it happen?

Soil is a collection of various sized minerals grains, of various types of minerals produced by the weathering of rock. Typical soil minerals are clays, silts and sands.

The properties and behavior of different soil types depends of the composition of the soil: the proportion of clays, silts and sand in a soil. Sandy soils are well draining and clayey soils are sticky.

Between the grains of minerals that comprise a soil are spaces, called pores or pore spaces. The pores can be filled with either water or air, depending the location of water tables and wetting events like rain, snow melts or other forms of water inundation.

The density of a soil is dependent on the degree of compaction of the soil. For to a soil to be compacted, a stress has to be applied to the soil to realign the grains of soil which reduces the total volume of the pores and reduces the amount of air within the pores.

Consolidation of a soil occurs when pore space is reduced and water in a soil is displaced due to an applied stress.

Regarding having something buried and soil compacting around it over time, yes that will occur but it is a question of how much stress the soil experiences, the duration of time and the nature of the soil - sandy or clayey. Something buried for a day without any stresses not much will happen. But, something buried for thousands of years with people and animals walking over it, rain falling on the soil, vibrations from nearby human activity and an occasional earthquake all add to the stresses the soil will experience and increases the degree of compaction or consolidation over time.


Why is thunder and lightning rare, even though it rains frequently? Why in some places (for example Dublin, Ireland) the thunder and lightning is such a rare thing even though it rains so frequently?

Rain in a climate such as Ireland is most often stratiform. That is, it is created when moist air is lifted by a front, or by divergence in the upper troposphere. Thunderstorms involve buoyant air, which rises on its own due to its being warmer than its surroundings. Typically solar radiation is involved in heating the surface such that the air near it warms and becomes buoyant. That is more common in lower latitudes and locations that are typically sunny.


El Nino in the Middle East According to reports and charts meteorological phenomenon El Niño is close to Iran, but the question is, El Niño phenomenon in Iran is likely to achieve?
What are the consequences?

A very quick and general answer, if anyone wants to get more specific, feel free:

  the question is, El Niño phenomenon in Iran is likely to achieve? What
  are the consequences?

Iran's been through several El Nino's, so its not like this year you'll see things you've never seen before.   I mean, you could.  There's always a chance of record breaking weather, but have a look at El Nino's and how often they occur, here and more numerical and monthly specifics here.

If we assume this year will be a strong El Nino, then look back at weather in Iran in winter 97-98 through Summer 1998, and Winter 82-83 into 1983.  Those were the 2 most recent big El Ninos.

I found this article from last month.  You probably have better access to local updates than I do, but it seems that frost and flooding are the primary concerns.    As Gordon Stranger points out, that's not a guarantee, only an increased chance that it happens.

  El Niño is close to Iran

El Nino is in the Pacific Ocean.   The El Nino effect is all over the globe, so it's, in a sense, close to everywhere on earth.  Here's a kind of cool picture of the effects of the 1982-83 El Nino.  Those were just the effects for that one year, and no two El Nino's will affect the globe in the same way, though they can have some similar patterns.

Source

  However, you should be afraid of the possible occurrence of El Niño in
  Iran?

When I first read this, I thought you were suggesting that he be afraid, but reading it again, I think what you're asking is whether you should be afraid.

El Nino probably isn't something to be afraid of cause there's no guarantee the effects will be bad where you are.   It's silly to worry about a 30% increased chance of something happening - which is basically what this is.

If you live in a flood prone area try to prepare for flood.  Have food and drinking water on hand, maybe sandbags to put in-front of your doorway - which might help for a smaller flood.  If you're a farmer, as much as possible, try to prepare for Frost.   Fear isn't helpful, but planning ahead for possible bad weather events is always a good idea.


Was Judea as desertified 2000 years ago as it is now? The ancient fortress of Masada is currently so dry and forbidding that is hard to imagine anyone living there. Was the climate in the surrounding area more wet 2000 years ago, or was it as much a desert 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'.


Limitations on mineral precipitation from solution If the solubility constant for a mineral is exceeded and, thermodynamically, a mineral is favored to precipitate in solution, are there any factors that may limit this from happening? Perhaps kinetics? Is it known if calcite precipitates out faster than dolomite does?

Also what will happen once a mineral precipitates out? Will concentrations decrease until the solution is saturated with respect to the mineral?

Thank you

This is really a wide open question on which books could be written. First let's stipulate that we are considering aqueous solutions rather than say magma.

In aqueous solution the solubility of a particular mineral depends on the temperature, pressure and chemical composition of the liquid. Generally as a solution would rise towards the surface of the earth the pressure and temperature drop which tends to decrease the solubility. Once the solution becomes saturated with respect to a certain mineral then it becomes thermodynamically favorable for the mineral to deposit. However it is possible for a solution to become supersaturated.

Also you have to remember that homogeneous conditions wouldn't really exist. There would be temperature and pressure differences in water flowing through a crevice. The water wouldn't flow at a constant rate over long periods of time. So the whole aspect of deposition and dissolution would be very dynamic.

It is also possible to dissolve minerals A and B, yet deposit some mineral C.

Also for minerals there are often many different crystal formations. Calcite deposits in hundreds of different types of crystals.


How to calculate cumulative moment magnitude? What is the proper way to calculate cumuluative moment magnitude if I know the moment magnitude of several earthquakes?

The moment magnitude ${\displaystyle M_{\mathrm {w} }}$  is a dimensionless number defined by Hiroo Kanamori as:

$$M_{\mathrm {w} }={\frac {2}{3}}\log _{10}(M_{0})-10.7$$

To sum several moment magnitudes we need to first convert them to  ${\displaystyle M_{\mathrm {0} }}$ (seismic moment) which you can simply sum and then convert it back

for each of the magnitudes you have - insert the magnitude into ${\displaystyle M_{\mathrm {w} }}$, solve for ${\displaystyle M_{\mathrm {0} }}$ and write down the answer.
Sum all the answers
Insert the sum back into ${\displaystyle M_{\mathrm {0} }}$ and solve for ${\displaystyle M_{\mathrm {w} }}$


What factors determine the number of Hadley cells for a planet? We know (think?) that Earth has three Hadley cells per hemisphere, but from observing gas giants such as Jupiter, we see that they have many more cells. According to a link from a comment in this question, Titan might have one cell going from north to south poles.

What factors affect the number of cells a planet has, and how? Or to put it another way, given a hypothetical planet with atmosphere, what facts would you need to know in order to guess at how many cells it has?

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.


Why does part of the P-wave convert into a S-wave with an increase of incidence angle? With the increase of incidence angle, a portion of a P-wave converts into a S-wave, how does this behaviour occur?

The way I imagine this phenomenon is illustrated in the picture below. I have to warn you that it lacks scientific accuracy and is just a simple way to imagine the effects we observe on a reflective boundary. However, it can be useful as a simple model you can easily imagine.
In the upper picture you can see an illustration of a reflection event at a horizontal boundary.
The lower picture illustrates how the vector of particle motion in a down-going P-wave is 'decomposed' into two vectors. First component is a vector of particle motion in an up-going S-wave, the second corresponds to particle motion in an up-going P-wave.

Once again, this is not how the rays exactly behave in a reflection event. Here we do not take into account transmitted waves and lots of other factors. Nevertheless, sometimes you can use this 'approximation' because it satisfies two conditions:

At normal incidence, there is no mode conversion (shear wave does not appear)
As the incident angle increases, the amount of shear-wave energy increases too.

These effects are also observed in real life.
Exact description of a reflection event on a plane boundary, as kwinkunks has  mentioned above, is given by Zoeppritz equations. They take into account all details and are widely-used. No doubt, they are much more difficult (but physically correct!) than a simple sketch I suggest.

.


P wave to S wave conversion While passing through layers inside the earth some P waves get converted to S waves and then back to P waves while returning towards the surface. Is this statement true?

If yes, then why? (Considering that they are both fundamentally different in terms of wave theory - longitudinal vs. transverse)

P and S waves are fundamentally different, when it comes to properties of the wave. An example might be that P waves can travel through fluids while S waves cannot. However, when it comes down to wave theory, these two are just different polarizations of a mechanical wave.

Seismology

In seismics this concept may be puzzling as we make some very fundamental assumptions. However, seismologists are very well aware that these polarizations (they call them phases) usually change throughout their travelpath. When you look at the image below, you can see different travelpaths and the same source will trigger P and S waves. This P wave will travel to the point P where it can be recorded as direct P wave. Upon reflection it will be partially converted to a S wave, which you can also see on the nomination PP is the P part of the reflected P wave and PS is the S part of the reflected and converted P wave. (Actually you can go on happily converting back and forth.)

Seismics

In seismics we get this conversion on interfaces or layer boundaries (whichever name you use). So everytime a wave is reflected it may also be partially converted into a different polarization. So essentially, your statement is wrong. The wave doesn't just convert passing through a layer, it is converted upon interaction with an interface.
So while this wasn't a yes directly to your question let me elaborate why this conversion happens.

This picture basically shows the rate of conversion depending on the incident angle. At 30° the rate of conversion is maximized. (The scales are terrible to read sorry.) At this interface we can see the incident wave as excitor of a new wave package. So we have to take a detour into wave excitation.

Exctiting waves

In marine seismics we can excite pure P waves, because we're operating in a fluid. On land you will always excite a mix of P and S waves (and variations thereupon, namely surface waves). The excitation of these waves is direction dependent. Starting a P wave in one direction will always start an S wave component perpendicular to this direction. Of course we have some energy distribution so that we get something like this:

P waves in the + direction and S waves in the - direction. (This image was vor conductivity but it serves the point.)

At the interface

We have to look at the incident wave as an excitor at the interface, which will give us a P and a S wave response. In seismics, wave theory and other solutions you can usually see the S wave solution stands perpendicular on a P wave solution (except when we're dealing with anisotropy). So yes in the end your incident wave will start two P waves and two S waves that look like this:


how to know the correct results in watershed delineation in ArcGIS? When I use one DEM in my study area to delineate watersheds, the results are different from when I mosaic 4 DEM layers adjacent to each other.
After the mosaic process, 3 of the basins were merged in one large basin, and 2 other basins got larger extent.

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.


which are the two countries with the flight time between them and time zone differences are equal I have been flying to different countries so often and this question came to my mind out of nowhere! Which are the two countries with the flight time between them and time zone differences are equal?

For example, considering current time differences between time zones on all over the world, let's say that hypothetical i will take off at 16:00 from Istanbul and i will land to Munich. And time difference between these cities is 2 hours and flight duration is 2 hours. So, when i will land, the local time in Munich will be 16:00.

My question basically is that is there any countries/cities in the world like the example above?

I'm not going to look for examples, but there will be some. More interesting is to consider the underlying geometry:

At one extreme, if you fly from north to south (or vice versa) you'll experience little or no change in time zone no matter how long you fly for.

At the other extreme, if you fly from east to west the time zones will go in the wrong direction. Depending on how you define the question there might be an "answer" in this direction, where you fly west for n hours and go back by n time zones. But for the rest of this question, let's assume you want later time zones.

If you fly from west to east, you're chasing the time zones in the right direction... but unless you have a supersonic aircraft, you can't go fast enough - you will travel east slower than the rate at which the earth is spinning, so (as long as you go far enough for more than a 1 time zone change, i.e. to get out of rounding errors) the number of hours you fly for will always be more than the number of time zones you gain.

So to match the two, you must fly diagonally : north-east or south-east. The exact angle will depend on the speed of your plane and the latitude that you start from. And then, you have to find a city along that line!


If I build a concrete surface on a 1km x 1km area and painted it black, will I create convection (updraft from ground)? Providing there are low wind conditions and a clear sky, will the 1 km x 1 km black surface create an upward convention?

Let’s say I were to do this at 60$^\circ$ N longitude in march month, the elevation of the site to be at 2000 feet, by just building a surface of 1 km x 1 km and painted in black, how high would the convection go? What other side effects will it cause? My goes to have air going up to 7000 feet constantly.

In principle: yes. But i won't give exact numbers for a mere km², though i'd expect a skilled paraglider under optimal atmopsheric conditions to be able to make some use of it ;-)

Actually, gliders use industrialized areas exactly because of this, a low albedo created by asphalt, concrete, dense construction/building.

How high and strong the resulting thermal convection will be depends on the weather, atmospheric layering, solar irradiation, wind, cloud cover, temperature and moisture. The ability of the ground to store heat (e.g. street canyons, or heat strored in the air of a high grown cornfield) also influences the triggering of convection. In some cases a rising thermal is visualized by a dust devil.

Thermal convection can be calculated/estimated and predicted from atmospheric soundings, instrumented weather balloons to measure the state of the atmopshere from ground to tropopause. The results are visualized in Skew-T or Stüve diagrams.

Reading and interpreting them is an art. But they can tell when the trigger temperature for thermals will be reached, how high they go, if there will be clouds and at which level their base is, etc.

Side effects of this are called urban heat islands or generally heat islands due to low albedo ground. They can cause local low pressure due to the rising air over the hotter ground, which can result in showers or even thunderstorms in the evening/afternoon.

Quite obviously, these effects can also be triggered by naturally occurring differences in ground albedo and plant cover. They are a driving force for local weather phenomena.


If we burn all carbon in the Earth's crust, by how much would the atmospheric concentration of $CO_2$ increase? If we burn all carbon in the Earth's crust, by how much would the atmospheric concentration of CO2 increase?

Short answer: $\rm CO_2$ levels would increase by $209,460$ $\rm ppm$ by using all the available oxygen.

Long answer: The mass of carbon in Earth's crust is $9(10)^{22}$ $\rm gC$ (grams of Carbon). The mass of the atmosphere is $5.15(10)^{21}$ $\rm g$. If all the carbon was burned, or otherwise moved to the atmosphere, then the limiting factor would be the $\rm O_2$ (molecular oxygen) available for carbon to bond with to make $\rm CO_2$. Oxygen makes up $209,460$ $\rm ppm$ of the atmosphere. So, assuming the carbon bonded with all available $\rm O_2$ and the remaining carbon did not enter the atmosphere, then there would be an additional $209,460$ $\rm ppm$ of carbon dioxide in the atmosphere.


Historically, how did the total amount of salt / NaCl on Earth develop? After reading this question about the origin of NaCl on Earth, I'm wondering: How did the amount of NaCl develop historically? Is it a more or less linear growth until today? Was it such a linear growth and then reached a level that it maintains until now? Or was that more complex over the billions of years of Earth's existence?

Here is a graph describing modern estimates of ocean salinity throughout the Phanerozoic (past ~550 million years).
The original PDF of this study can be found here:

Hay, W.W., Migdisov, A., Balukhovsky, A.N., Wold, C.N., Flögel, S. and Söding, E., 2006. Evaporites and the salinity of the ocean during the Phanerozoic: Implications for climate, ocean circulation and life. Palaeogeography, Palaeoclimatology, Palaeoecology, 240(1-2), pp.3-46.

As this graph shows, it is not linear at all and actually decreasing since the end of the Precambrian. This is a result of many factors including changing volume of ocean water, and the global chloride cycle. The details are all described in that paper.


Is there a data source for annual global precipitation? Is there a data source for cumultative global precipitation by year? Units would presumably be mass, volume, or height of water equivalent. Also, what about cumulative global precipitation over land by year?

In case you want to learn something about precipitation first:

http://www.nasa.gov/mission_pages/GPM/main/index.html

There is the most important information about current missions with some videos, too.
If you click on the education link a little lower on the page you can learn a lot about how the data is retrieved and about earth's water cycle itself.. if that's not interesting to you and you just want the data this here is the right way:

http://giovanni.sci.gsfc.nasa.gov/giovanni/

The site does not just provide data about precipitation but about all measurements that are made from space such as amounts of aerosols, evaporation rates, heat flux, vegetation, wind speed and so on.. you'll see it all on the left site of the page.. just make a tick on the dataset you need and all measurements will pop up on the middle of the screen, make another tick on the information that suits your needs, choose the time span you wish and the area of the map you want to know more about under "show map".. if you're done with that just click "Plot data" on the right bottom corner and a map with the data you wish will show up.

I hope that helps!


Stratigraphy of limestone applied to the Great Pyramid of Giza The Great Pyramid of Giza is made from blocks of limestone quarried from the Giza plateau. The pyramid is composed up of about 200 layers, with individual  blocks varying in thickness from 20 to 60 inches.  How feasible would it be to identify which layers of limestone in the quarry were used in the various layers of the pyramid? For example, if a core sample were taken from the quarry could individual blocks from the pyramid be matched to a position along the core?

Theoretically? Maybe! Practically? No. Identifying the general quarry site by comparing lithology is often possible. It might be possible to identify a block as coming from a particular level in a quarry by detailed characterisation of micro-fossil assemblages. Geochemical analyses could assist. However the practical problems are almost certainly insurmountable:

The limestone rocks used in the pyramids were deposited in a marine reef environment, so deposition rates could be quite high. This means that with small age differences between the top and bottom there may be negligible variation in the micro-fossils.
The parts of the quarry from which rock was taken are, by definition, not there to be sampled. You would have to take your comparison core from the edge of the quarry and you would presumably be looking at the stratigraphy of the layer on the outside of the pyramid which may have come from the centre of the quarry. Reef limestones can change their characteristics significantly over short distances.
The exposed limestone blocks will have been altered by weathering - so would probably need destructive sampling by coring to gather material for comparison.
As an analogy, matching individual blocks to layers in a quarry would be like trying to match sawn wood to individual tree stumps in a forest. Conceptually possible, but would take an awful lot of work and even more luck.


What would the pH of rainwater be if carbon dioxide in the atmosphere doubled? Natural rain water is slightly acidic (pH = 5.6) since there is carbon dioxide in the atmosphere that dissolves to make carbonic acid.

What would be the pH of rainwater if carbon dioxide in the atmosphere was doubled to 800 ppm?

"Cutting the $CO_2$ in half would yield about 5.9."

(A valid approach because we expect reversibility in chemistry and physics.)

[pH of 5.9 for 200 ppm $CO_2$]

Atmospheric $CO_2$ of 200 would be about that of the last glacial maximum. So most of the difference between pure water at pH 7 and 5.9 would seem to be other gases dissolved in air.

Side note: A drop to much below 200 ppm $CO_2$ would extinguish all life on Earth except those that live at geothermal vents at the bottom of the ocean, because they do not use $CO_2$.

If a linear relationship is assumed, then if there were 800 ppm of $CO_2$, rain water would have a pH of 5.0.  Though, if we assume the log relationship for concentrations that userLTK mentions in the comments, it would be a pH of 5.3.

Bouttes, N., D. Paillard, D. M. Roche, V. Brovkin, and L. Bopp (2011), Last Glacial Maximum CO2 and δ13C successfully reconciled, Geophys. Res. Lett., 38, L02705, doi:10.1029/2010GL044499.


How accurately can explosions be triangulated from the IRIS seismogram data? There's a bunch of armed conflicts going on right now. What are the lower boundaries on accuracy of triangulating and timing man-made events from the IRIS datasets (accessible, for instance, with the ObsPy package)?

David Salzberg and Margret Marshall suggest that using green receiver functions (which might make more sense given how we determine the difference between explosions and earthquakes) can locate a detonation in North Korea with up to 0.5 km accuracy. I am assuming that IRIS data sets carry enough information to make receiver functions, as I think triangulation may require a much larger event than a single nuclear warhead (not sure about this point) to be very accurate.

For a little background on how we determine if an event is a explosion or an earthquake (I know this isn't part of the question, but still useful for others) we look at "beachball" diagrams(from csegrecorder.com):

As you can see, the "beachball" diagrams are visual representations of the three linearly independent stress vectors or moment tensors. I may be getting these colors reversed (so someone check on it) but "white" means dilation and "black" means compression. A fault rupture will have some compression and some dilation, meaning some "white" and some "black", but an explosion will appear to be fully compressed because an explosion pushes on every direction. Think about a slinky: when you push on it, there is compression, and when you pull on it dilates. So an Implosion, that pulls everything towards one point, would be extensional, and thus be fully "white".


How come the tide is low when the moon is high in the sky Happily spending my holidays in the Fiji Islands, I noticed a couple of days ago that when the moon was very high in the sky (my shadow was maybe  0.5-1m long) the tide was low. That really puzzled me. How does this happen?

This is based on the overly-simplified model of tides being the result of tidal bulges. As I explained in my answer to a related question on the physics.SE sister site, those tidal bulges do not and cannot exist.

Instead, the tides are dynamic responses to the tidal forcing functions from the Moon and the Sun, with the orbits of the three bodies about one another, the Earth's rotation rate, and the geometry of the oceanic basins all playing roles in the dynamic response. For any point on the surface of the Earth's oceans, the tides can be described in terms of a composition of frequency responses to those cyclic forcing functions. See the related question Definitions of tidal harmonic constituents…? for details.

At many places, Fiji included, the key component of the tides is the $M2$ tidal constituent, with a period of 12.42 hours. This is the principal response to the tides induced by the Moon. If Newton's equilibrium (tidal bulge) theory of tides was correct, every oceanic place on the Earth's surface would see high tide when the Moon is at zenith and nadir, and low tides when the Moon is at the horizon. Except for a few places, this is not what is observed. In Fiji, the situation is nearly reversed, as depicted by the tide chart below for Nadi on the west coast of Viti Levu, where high tides are more or less in sync with moonrise and moonset.

Source: https://www.tide-forecast.com/locations/Nandi-Waters-Viti-Levu-Fiji-Islands/tides/latest, accessed 2 November, 2017.

Laplace's dynamic theory of the tides says that the oceans' response to each of the tidal constituents comprises a set of amphidromic systems. Each such system is centered about an amphidromic point, a place with a null response to the component in question. A wave whose amplitude increases with distance from the the amphidromic point rotates about the amphidromic point at the constituent frequency. The $M2$ tidal response is depicted below. The $M2$ amphidromic points are at the centers of the dark blue areas. The white lines emanating from the amphidromic points are cotidal lines, curves along which high tide occurs at the same time. The different colors indicate height, which again increase with distance from the governing amphidromic point.

Source: http://en.wikipedia.org/wiki/File:M2_tidal_constituent.jpg.


The mean direction of waves in a directional distribution When modelling ocean waves, a directional distribution $D_f(\theta)$ is used together with a frequency spectrum $S(f)$ to describe the energy of waves at a particular frequency $f$ and angle $\theta$.

I understand that the directional distribution can be written as a Fourier series i.e.
$$D_f(\theta) = \frac{1}{2\pi}\left[ 1 + 2\sum_{n=1}^{\infty}\{a_n\cos(n\theta) + b_n\sin(n\theta) \} \right] $$
where $a_n = \int_0^{2\pi} D_f(\theta)\cos(\theta)\,d\theta$ and $b_n = \int_0^{2\pi} D_f(\theta)\sin(\theta)\,d\theta$.

In Kuik (1988), the mean wave direction, $\theta_0$, is found by calculating
$$\theta_0 = \arctan\left(\frac{b_1}{a_1}\right)$$
where $b_1$ and $a_1$ are the first order Fourier coefficients.

Alongside this definition, the author refers the reader to Borgman (1969) but I can't find this paper on the web.

My question is why is it only the first order Fourier coefficients $a_1$ and $b_1$ used in this calculation?

EDIT:
After giving this more thought, I think that the fact they are Fourier coefficients is somewhat of a coincidence.

If the directional distribution is seen as the PDF (as the integral of it is equal to 1) then $a_1$ and $b_1$ are more like the expected values that the cosine and sine of the angle $\theta$ take. The average values can then be used in the $atan2$ function to determine the mean angle.

It is helpful to understand the meaning of Fourier coefficients in the directional wave spectrum analysis.

A pitch-and-roll buoy measures time series of water elevation $\eta$, and slopes in both Cartesian directions, $\eta_x$ and $\eta_y$. The Fourier components $a_0$, $a_1$, $b_1$, $a_2$, $b_2$, are related to cross-spectra of elevation and slope time series (introduced by Longuet-Higgins 1963):

$$
a_0 = \dfrac{C_{11}}{\pi}
$$

$$
a_1 = \dfrac{Q_{12}}{\pi}
$$

$$
b_1 = \dfrac{Q_{13}}{\pi}
$$

$$
a_2 = \dfrac{C_{22}-C_{33}}{k^2\pi}
$$

$$
b_2 = \dfrac{2C_{23}}{k^2\pi}
$$

where the cross-spectra are:

$$
C_{11}(f) = C[\eta(f)\eta(f)] = \int_0^{2\pi} E(f,\theta)d\theta
$$

$$
Q_{12}(f) = -Q[\eta(f)\eta_x(f)]k^{-1} = -\int_0^{2\pi} E(f,\theta)\cos{\theta}d\theta
$$

$$
Q_{13}(f) = -Q[\eta(f)\eta_y(f)]k^{-1} = -\int_0^{2\pi} E(f,\theta)\sin{\theta}d\theta
$$

$$
C_{22}(f) = C[\eta_x(f)\eta_x(f)]k^{-2} = \int_0^{2\pi} E(f,\theta)\cos^2{\theta} d\theta
$$

$$
C_{23}(f) = C[\eta_x(f)\eta_y(f)]k^{-2} = \int_0^{2\pi} E(f,\theta)\sin{\theta}\cos{\theta} d\theta
$$

$$
C_{33}(f) = C[\eta_y(f)\eta_y(f)]k^{-2} = \int_0^{2\pi} E(f,\theta)\sin^2{\theta} d\theta
$$

From here you can see that the mean direction, as defined in the paper that you cite, is:

$$
\theta_0 = \arctan\left({\dfrac{b_1}{a_1}}\right) = \arctan\left({\dfrac{Q_{13}}{Q_{12}}}\right) = \arctan\left({\dfrac{\int_0^{2\pi}E(f,\theta)\sin{\theta}d\theta}{\int_0^{2\pi}E(f,\theta)\cos{\theta}d\theta}}\right)
$$

Because $Q_{12}$ and $Q_{13}$ are proportional to the integrated energy in the zonal and meridional directions, respectively, the arctangent of their ratio yields mean wave direction.

The authors could use higher moments to calculate direction, but this would have different physical meaning. In this specific case, using $\theta = \arctan\left({b_2/a_2}\right)$ would yield peak (dominant) wave direction in the context of pitch-and-roll buoys.

You see that the above analysis is inherently limited by the quantities that are measured, specifically $\eta$, $\eta_x$, and $\eta_y$. If higher order quantities, say $\eta_{xx}$, $\eta_{xy}$, $\eta_{yy}$ were available, the directional spectrum could be described at a higher accuracy even using just Fourier decomposition.

Further reading:

Oltman-Shay and Guza (1984)
Nondirectional and directional wave data analysis procedures at NDBC


What is this narrow band of clouds? I came across a narrow streak of clouds in Asia when playing with Google Earth.
Does anyone know what that is?

Given the weather patterns, it looks like Google Earth is generating those cloud overlays from some recent (i.e., the current day) satellite images.  That line is suspiciously close to the edge of disc line for Meteosat 8, which is located over the equator at 41.5 °E.  I suspect that it's just an artefact of how they're stitching together the various images from geostationary satellites.  The satellite zenith angle and data uncertainty get very large towards the edge of the disc, so most products for forecasting and research are clipped to exclude that region.

Image from EUMETSAT: https://eumetview.eumetsat.int/mapviewer/?product=EO:EUM:DAT:MSG:CLM-IODC


Why don't we feed the plastics to the termites? I may sound completely stupid but this idea hit me when I saw termites ate a plastic bag packed with papers today in my house. What are the challenges we're likely to face if we decide to do this?

Many animals can eat through plastic to get to what they really want to eat. Among them: termites, mice and myself when battling against a candy wrapper. However, as far as I understand, none of them can degrade the plastic. You can do the test yourself and I'll guarantee the plastic wrapper will come out the other end of your digestive system.

Nevertheless, several organisms have been identified to be able to actually eat and digest plastics, among them:

Bacteria: Like Ideonella sakaiensis discovered in japan.
Fungi: Like Aspergillus tubingensis discovered in the landfills of Pakistan, or Endophytic Fungi discovered in the Amazon, which seem to be edible, and there are some people offering small mushroom farms you can have at home to turn your polyester and polyurethane trash into food, they call it "FUNGI MUTARIUM".
Insects larvae: Like this plastic-eating waxworm.

That said, it could be possible that some species of termites eat and digest some kinds of plastic. But I don't know of any reports of it. In the comments, Jan Doggen pointed to a 2003 article of people working in that direction. But I guess given the lack of publications afterwards, it probably turned out that the termites were not digesting the plastics usefully. If you search the internet, you will find many references of termites chewing through plastics (like this one).


Was climate change inevitable anyway, or is there more to it? Climate change is now one of the greatest challenges facing humanity, but it seems that we have been here before; if we look at the world's climate over a much longer period than 150 years.

During the Paleocene-Eocene Thermal Maximum which occurred 50m years ago, the poles were ice-free tropical regions, with very high carbon levels, and temperature increases of as much as 5 degrees.

I am just wondering if what we are currently seeing was going to happen anyway? Has the discovery of oil, and the resultant human activity just accelerated an already inevitable outcome?

Earth's climate has changed in the past and will change in the future. It would have changed even if we didn't start spewing enormous amounts of CO2 to the atmosphere. The main difference here is the rate at which this is happening. Natural climate change unfolds over thousands or tens of thousands of years (btw - PETM was fairly rapid, but we don't know exactly how rapid). Current climate change happens much faster. You can see temperature records broken every single year.

Our modern civilisation developed in a fairly stable climate. We did not develop in a climate where we had ice-free poles, or higher sea levels. The geopolitical consequences of altering that too fast would be simply too much.


How to reduce the atmospheric temperature artificially? Is there any gas/substance which absorbs/reduces atmospheric temperature?

Ozone filters UV rays from the sun light and allows in infrared rays. In my knowledge these infrared rays are the cause of the heat. Do we have any provision to block these infrared rays at least for a targeted area? Does any gas or any other substance, natural or human-made, work in this way?

Ozone does not "filter" UV, it absorbs UV radiation (and undergoes some photo-chemistry reactions in the process).  By absorbing those wavelengths it prevents them from reaching us at the surface, but cause a temperature rise in the stratosphere.

We have plenty of gasses that behave this way for terrestrial radiation (however without the photo-chemical reactions).  Some of these gases are $\mathrm{CH_4}$, $\mathrm{CO_2}$ and $\mathrm{H_2 O}$, which we term greenhouse gases for their ability to readily absorb and radiate at terrestrial (infrared) wavelengths.  If concentrations are high enough, they would reduce some of the infrared spectrum from the sun reaching the surface, but they would also reduce terrestrial emissions into space.  Wherever these gases are doing the absorption one would also experience warming.

If you really want to reduce the global temperature solely through radiative forcing, the answer isn't to find and use a gas to absorb infrared radiation.  The answer is to reduce the concentrations of greenhouse gases in the region you are concerned with.  This will not greatly effect the incoming solar radiation, but it will allow greater radiative losses from the earth and atmosphere (lost to space), resulting in cooling.

You mention that "In my knowledge these infrared rays are the cause the heat".  While infrared radiation will make you hot (e.g. when the sun shines on your skin), the peak emission from the sun is in the visible spectrum and the primary energy source is in wavelengths shorter than infrared.  This shortwave radiation heats the surface of the Earth which in turn conducts heat to the atmosphere and radiates in the infrared.  Without the presence of greenhouse gases, our planet as is (current albedo, etc) would have a global mean temperature around 255 K, which is quite a bit colder than what we experience with greenhouse gases.  If you want to reduce temperatures, regionally or globally with a gas, your answer is to remove some of them, not add to them.


Using heat at depth to generate power I was thinking if there is any power plant in Earth that generates power by sending water at depths and getting steam in return?

Just some random idea:

This form of geothermal energy was investigated in Australia in the early 2000s.

A company, then called Geodynamics Australia but since renamed as ReNu Energy, constructed a geothermal power plant and during 2013 conducted a [160 day trial of a 1 MWe pilot plant] in the Hanbanero Field near the border between South Australia and Queensland(http://www.futuredirections.org.au/publication/geothermal-industry-future-australia/).

"The pilot plant used steam generated through an EGS (Enhanced/Engineering Geothermal System) that exploits thermal energy from hot rocks over four kilometres below ground level. Following the trial, it was concluded that it may be economically viable to supply heat energy to nearby gas production operations, but not the supply of electrical energy. No further work has been done on the project, and the company completed site environmental remediation works in 2016."

The pilot plant had many technical challenges:

"EGS (also described as a hot rock system) extracts heat from hot rocks deep in the earth and conveys it to surface where it is used to generate electricity (see Figure 1 above). This involves a heat exchange arrangement. In most EGS projects, water is pumped into an injection well where it takes on heat as it moves through fractures in hot rock, and is then extracted to surface from a production well. In the Habanero pilot plant trial, temperatures seen on surface from the production well were up to 215 degrees by the time the trial was finished – certainly demonstrating the capacity to extract heat. Another critical factor in how much electricity can be generated though is the rate at which the high temperature steam is produced. In the Habanero trial, up to 19 kg/s of flow could be sustained. This contrasts with (according to the 2014 report to ARENA) flow rates in the order of 80 to 100 kg/s that would be required for economic viability. Whereas the initial intention of the Habanero project was to create a network of man-made fractures to allow optimal rates of circulation, in actuality, fluid was circulated through a naturally occurring fault in the rock (a geological feature which was not anticipated). The challenge of establishing and optimising fluid circulation between wells might be addressed in future through fracturing techniques that have resulted from ongoing technology development in tight gas or oil wells."

One of the issues with developing geothermal energy in Australia, from deep hot rocks, is the remoteness of the potential generation sites from regions that could utilize the energy. This makes the use of such geothermal systems currently unviable.


The sun rises up at 7:30 in 25°E, What time will the sun rise up in 15°E? I've a question about time calculating.

  The sun rises up at 7:30 in 25°E, What time will the sun rise up in
  15°E?

What I've tried

$$25° - 15° = 10°$$

By the way

$$10° . 4 = 40 \space \text {minutes}$$

However, it seems wrong.

Your maths is not detailed, but you have the correct answer.

The Earth rotates at 15 degrees per hour (360 degrees/24 hours).

The time required to rotate 10 degrees is 10/15 = 0.666 hours, which is 40 minutes.


What is the meaning of -999 degrees Celsius? I was checking the weather forecast just now, and it is showing that it "feels like -999 °C". I never heard or saw -999 °C before. I searched other weather channels, and they were showing that it feels like 2 °C. What is the meaning of -999 °C, irrespective of the weather?

The value -999 is likely the "fill value" used in the dataset when data is missing or is not being properly filtered or handled when displayed.  In the specific case on the website you cite, it is likely a problem with the algorithm for wind chill (the "feels like" temperature this time of year).

It isn't a physical value and only means the value is missing. Furthermore, -999°C is not a possible value because absolute zero is –273.15°C, and it's not possible to be colder than this (at least not in any meaningful way, and certainly not because of wind chill). The coldest recorded temperature on earth is around –90°C.


How do we know that we are NOT in an "Ice Age?" Historically, the earth has had five "Ice Ages" Each of them lasted millions, tens of millions, or hundreds of millions of years.

The last ice age reportedly ended perhaps 10,000 years ago. That seems to represent the beginning of recorded history.

Except that a period 10,000 years is a "rounding error" in the span of millions of years. If it was true that "we haven't been in an ice age for at least 1 million years, then I would believe that we were "clear" of  the last one. But what is to say that the last 10,000 years isn't just an usually warm "spell" in what would otherwise be a cold period of say, a million years?

I think you're slightly confused by some of the terminology. (Caveat: I'm a geophysist, take anything I say with a grain of salt!)

We're currently in an interglacial during a prolonged period of icehouse climate (most of the Cenozoic).

During most of the Earth's history, the overall climate tends to be much warmer and more stable on the million-year scale. These periods are referred to as "greenhouse" climates.  During these times, there's relatively little evidence for extensive ice sheets (either on continents or over the ocean) at the poles.

However, during times of "icehouse" climate, the Earth becomes cooler.  Ice masses develop at the poles and due to regular preturbations in Earth's orbit (Milankovich cycles), continental ice sheets tend to advance and retreat.  (Assuming there are continents near the poles.)  This leads to regular glacial and interglacial periods.

Presumably, regular fluctuations in climate due to Milankovich cycles would occur during periods of greenhouse climate.  However, ice sheets cause positive feedback that amplifies the effect of the Milankovich cycles.  This is largely due to ice and snow's albedo: they reflect more sunlight back into space instead of adsorbing it.  Therefore greenhouse Milankovich cycles were probably smaller.  (There's probably a lot more to the story. I'm not an ideal person to ask.)


If water is a renewable resource, why is there water scarcity? Water is a renewable resource, renewed by the water cycle.

If we use water too quickly, water should get evaporated and come down again as precipitation.

So, Why is there water scarcity if water is a renewable resource?

What I miss in the other answers is water quality.

Even is there is sufficient water available it is often of insufficient quality  for the purpose you want it for:

We often don't drink straight out of rivers
We don't irrigate with salt water

etc.

So you need to spend effort to prepare it for use. This adds to the experience of shortage.


Winds in a northern hemisphere hurricane In the Northern Hemisphere, do surface winds around the center of a hurricane move counterclockwise and inward or counterclockwise and outward?

I'm guessing they move inward.

The winds in a hurricane move cyclonic and inward at the surface and anti-cyclonic and outward in the upper troposphere.

Cyclonic winds are counter-clockwise in the northern hemisphere and clockwise in the southern hemisphere.  Anti-cyclonic is the opposite of cyclonic.

See this answer for a more detailed discussion of the winds within a tropical cyclone.


Would a hurricane increase intensity due to increasing translation speed? For example, if a hurricane has 200km/h wind speed, and it increases the translation speed from 10km/h to 30km/h, would the intensity become 220km/h?

It is not the quite the numerical relationship you have described  in your question but YES the effect of translation(or forward speed) upon the intensity of tropical cyclones in the tropical oceans has been studied in this peer reviewed reference - Tropical cyclone translation speed and intensity correlation

The conclusions that the authors have reached  is that there is a positive correlation with translation speed and tropical cyclone intensity. The explanation for this correlation is that a tropical cyclone with a faster translation speed reduces the exposure time of the cooling to the TC. The sea surface cooling is induced by the effect of the surface winds on the ocean and and strong winds can bring cooler subsurface water to the ocean surface . A slower moving TC is exposed to the sea surface cooling for a longer period of time and hence contributes to a negative SST feedback and has lesser chance for  reaching higher intensity.

You  can read a popular science version of the peer reviewed reference in this website - Hurricane speed intensity


How does geothermal heating work? I have heard that geothermal heating is a way of generating energy from the temperature difference between the inner layers of the Earth and the Earth's crust.  How is it possible to extract this energy?  Must one be near a tectonic plate boundary to gain access to this temperature gradient?

There are currently two ways to extract geothermal energy: one mainstream, one still at the experimental / demonstration level.
Pollack et al (2010) estimate the global geoethermal heat loss at 44 TW (by comparison, civilisation's rate of energy use is about 20 TW).
As the USA Department of Energy notes, all methods rely on the combination of three factors:

heat underground,
fluid to carry the heat to the surface, and
permeability to allow the fluid to flow.

The mainstream geothermal method
This happens close to tectonic plate boundaries. Reservoirs of hot liquid are tapped, to bring the hot fluids to the surface. Typically the temperature of the fluids is well over 100 &degC;, and can be used for electricity generation, as well as for heating.
The experimental method is Hot Dry Rock
aka Enhanced Geothermal System or EGS.
As Pavel V. commented, this is being piloted in the Czech Republic, amongst other places, and allows geothermal heat extraction from places that are not close to tectonic plate boundaries. From that link:

the projected depth of the ... borehole was between 2100 and 2500 m. The drilling diameter ranges from 393,7 mm for the starting upper section to 152 mm at the borehole bottom

One tube is placed inside another, inside the borehole. Water can be injected down the inner tube, and returned up through the gap between tubes, or the other way around.
That paper reproduces the following chart from Šafanda, J., Dědeček, P., Krešl, M., Čermák, V. [2007] Report from geothermal research for PVGT-LT1 (in Czech):

As you can see, these temperatures are much lower, so are unsuitable for electricity generation, because one must accept very low efficiencies (Carnot's Law). They are used for space- and water-heating.


Do tornadoes usually avoid areas with many tall buildings? In my mind, tornadoes usually appear in flat areas with some wooden houses, but not places like cities with many tall buildings. It is especially rare to see a tornado destroying a series of skyscrapers. Is this true? If so, what is the reason?

It is especially rare to see a tornado destroying a series of skyscrapers.

Yes, it's rare, but not for the reasons you think. The reason that tornados striking downtown areas is rare is that cities are but a tiny fraction of the land area in places where tornados most often occur, and downtown areas are but a tiny fraction of the land area in those cities.

That said, downtown St. Louis, Missouri (the center of tornado alley) has been hit by tornados six times, and downtown Shrevesport, Louisiana (the center of Dixie alley), five times. The list of tornados that have hit the downtown areas of large cities goes on and on and on.


How does total pressure affect relative humidity and how do I correct my hygrometer for pressure? Obviously pressurized air can hold more water than regular air, so an increase in overall pressure would result in an inversely proportional decrease in relative humidity. However, I can't seem to find any equations which reflect this.

Basically my issue is that my thermohygrometer calculates the relative humidity assuming that atmospheric pressure is 101.3kPa but where I am measuring it the pressure is about 140kPa. This can be fixed by multiplying the reading by (101.3kPa/140kPa), but I am having a hard time expressing that scientifically and mathematically.

So, what is the mathematical relationship between pressure and relative humidity?

The maximum vapor pressure in air is a function of the temperature and not of the total pressure. So when you humidify air at 12°C you get a maximum vapor pressure of approximately 1.4 kPa. However, when the air is humidified at 140 kPa the vapor pressure is still only 1.4 kPa. Therefore, you have a water/air ratio of 1.4/140= 1%.

Therefore, for every mol of water, you will have 100 mol of air. So when the total pressure drops to 100 kPa, the mol/volume fractions will not change. This is also valid for the vapor pressure ratio. Therefore, in this case the vapor pressure will still be 1% of the total pressure. Thus, the vapor pressure will only be 1.0 kPa at a total pressure of 100 kPa.

Therefore, to "correct" an "under-pressure" reading you will need to multiply your "under pressure" reading by (101.3 kPa)/(P_actual), where P_actual is your actual total pressure in kPa.


Amount of Distortion at Continental Collisions Something that I've wondered for a while now - When continents collide, how much of the landmass gets distorted by the collision before it settles?

To perhaps give a clearer idea of what I'm talking about, I'll use an example - How much of the Indian Plate is still intact, and how much is now either driven under the Eurasian Plate or contorted into a different shape i.e. the Himalayas?

Any answers would be greatly appreciated

I think of the plates not as icebergs floating around in a sea but as pieces of porridge skin on the surface of gently simmering porridge, moved about by convection forces from below.

The Indian plate became fused with the Australian plate about 43 million years ago, so now there is only one plate, the Indo-Australian plate. Since then, the whole mass has been moving north at about 4 cm per year. A relatively small amount, perhaps a quarter of what was once the Indian plate, has crumpled against the Eurasian plate to form the Himalayas.

I don't know what the total mass of the Indo-Australian plate is, but it must be many trillions of tons. Imagine the momentum of many trillions of tons of rock moving at 3 mm per month, in addition to which there are convection forces from the mantle pushing it along! Bulldozing the Himalayas must have slowed it down a little, so 43 million years ago it was moving even faster.


Where did the continents come from? In animations that show the location / drift of the continents over the aeons, it seems that the continents (or cratons?) are just floating / drifting around, sometimes fusing or splitting, but having almost non-changing cumulative surface area over time.

This rises the question how that asymmetry between continents and oceans originally formed.  Of course continental crust is less dense than oceanic crust, and once formed the lighter one would just float around on the denser one.

Are there scientific theories how and when this originally started?  Like in liquid, hot, early Earth, the lighter stuff would just form a less-dense layer covering the whole planet.

And then?  Was it a smooth process of accumulating / clumping continental stuff over time, and after accumulation they just keep drifting around? Or was it some cataclysmic event like early Earth being hit by a Mars-sized object that formed the Moon, and that impact lead to a since-then persisting asymmetry in continents vs. oceans because some of the layers where ripped off locally?

note: Answers to What do continents  "lay" on? do not answer my question because they don't even mention continent genesis.

The answer is that we don't know. I could give you a list of all the considerations that go into evaluating that question, but it will probably be incomplete and outdated very soon. There were several important papers published in the recent years, including some very readable reviews that you might be interested in (1 2 3). But, there isn't a consensus. A friend just had a paper accepted in which she shows Yet-Another-Breakthrough™ in this field.

However, the general idea is that most of the continental lithosphere formed early on (Hadean to Archean) and since then the rate has slowed down.

Continental crust forms, as you said, by formation of "lighter stuff". Melting ultramafic mantle rocks leads to formation of basalt, and further melting of basalt leads to granites (and other related rocks) which are indeed lighter in colour, and also lighter in density so they end up at the top, and then "float" as the continental lithosphere on top of the mantle. The problem is that you can also remelt over and over again the continental lithosphere, recycling the already buoyant rocks and not increasing the net volume of the continents. It's this balance which is unknown. How much lithosphere is "juvenile" (i.e. formed by melting of mantle) and how much is recycled.

To make matters worse, continental lithosphere can also be destroyed. The most obvious case is erosion from the continents which then settles as a thin sedimentary cover on the oceanic crust and then lost to the mantle during subduction. Another process is delamination. Some rocks such as basalt are normally less dense than the mantle rocks. However, when metamorphosed at high pressure to become eclogite, they paradoxically become denser than the mantle, leading to the bottom part of the lithosphere to be "ripped off" and sink down into the mantle, thus destroying the continental lithosphere from the bottom. Here's a nice visualisation of this process:

Source for the image.


What are the differences in terms of appearance between sandstone and limestone? On the field, I find it very difficult to differentiate the two types of rocks just by looking at them.I know that you should do some analysis of the sample before any conclusions. But, for example in structural geology, I have to draw a geologic cross section and differentiate sedimentary layers.

By looking at pictures I find it very hard to tell whether the rock is limestone or sandstone.

The main difference is micro-texture. You cannot differentiate between both from a picture or a view, you need to go to the field and:

Break the rock with your hammer to observe
the rock in fresh-cut.
Add a bit of water to remove dust and correctly observe the colour and texture of the rock.
Finally use a magnifying glass to determine what the rock is made
of.

You should differentiate them as sandstones are always composed of grains. The texture is always grain-supported.

Limestones are composed of calcite. The texture use to be different, but it migth have also carbonated grains.

Rebooted Dunham classification of carbonates. Image from: commons.wikimedia.org

You can differentiate them in all cases unless you have a grainstone limestone with sand-grade grains (63μm-2mm). Then you need to identify the carbonated grains that can be fossils or oolites.

Oolites. Image from: commons.wikimedia.org

Fossils are clearly differenciated from sandstone grains with a magnifying glass, and oolites have a concentrical structure as shown in the picture.

Alternatively you can use dissolved hydrochloric acid or vinegar. As limestones are composed by calcite you will see a reaction, but you could also see the reaction on a sandstone if it has carbonated components.

Once you have determined the composition of a rock bed you can do the cartography at the lab with a stereoscope and aerial pictures or do it at the field again painting the limits on your map.


Can a braided fluvial create thick relatively continuous siltstone beds? I am attempting to interpret an outcrop I went to.  I suggested it was created by a braided fluvial.  The one thing I cannot explain are prominent alternating siltstone beds with the sandstone beds.  There are smaller lenses of silstone that I can suggest are abandoned channel fill, but I do have other siltstone beds about 0.2 - 0.5 m thick and seem pretty continuous (laterally continuous up to ~ 10-15 m as far as I can see).

Braided channels are usually found in estuaries. Estuaries are usually on the coast. That being the case, it seems possible that now and again a high tide brought in some sand which covered the silt-filled channels, which then brought more silt, thus creating the layered structure you describe. If the site is now a long way from the coast, it doesn't follow that it always was.


Did the surfaces of the oceans freeze over entirely during the snowball Earth periods? According to the snowball Earth theory, all of Earth froze over at least once (possibly twice) in the far past. However, Wikipedia says that there are opponents of the theory who state that tropical portions of the oceans were not entirely covered. These responses have been motivated both by sedimentary deposits (also used by proponents of the theory) and by computer simulations.

Is there any consensus as to whether or not the surface of Earth's oceans froze over entirely during either of the two main snowball Earth periods (~ 650 million years ago and ~2.1 billion years ago)?

To be clear, my question is about surface ice, not underwater freezing.

By looking at the recent literature on the question, I see neither a consensus or a definitive answer on the extent of the snowball earth.

Since the Neoproterozoic Snowball Earth event is more recent, there seem to be more evidences available and therefore more relevant research on this event than the 2.1 BY one - and will consider in the following only the latest event.

This paper suggest that the ice cover was likely global but some open water would have persisted in equatorial regions where complex life would have been able to survive during the glaciation.

This other paper suggest that glaciogenic sequences (sedimentary rocks containing  sediments deposited by glacier actions, for modern analog non-consolidated examples see till or moraine) in high latitudes (North) contains fragments of carbonates which was likely formed near the equatorial zone and somewhat transported by glacier action.

This yet another paper modeled how glaciation likely reach the equator, where this paper suggest that open sea must exists.

Other papers suggested that once a significant part of the Earth was covered by glaciers, albedo effect significantly cooled the planet until some kind of massive volcanic eruption interrupted the cycle.

The only consensus I figure is that it most likely happened (a late Proterozoic Snowball Earth Event) but it still remain to be seen if a thick cover of ice 'a la Encalade' covered the Earth at one time or another, or not.


If the earth's core is made up of copper/other metals other than iron-nickel alloy, how our magnetosphere will be? We know that our earth's core is made up of iron-nickel alloy and it is spinning to create magnetic flux all around our planet to create our life-saving Magnetosphere. And this also influences our 9.8 m/s gravitational force I think. So what if our core (both inner-solid and outer-liquid)is made up of copper or any other metals. How will it differ from our Magnetosphere and gravitational force?

On a quick approach:

Magnetism. The copper itself have a weak magnetism, so a copper core will not create a magnetosphere. Chek here or here.
Gravity. The Iron density is 7.874 g/cm³ and the nickel density is 8.908 g/cm³. Copper density is 8.96 g/cm³. So with those density data, the core will be heavier. (The actual core is supposed to have a 9.9-12 g/cm³ density) It is supposed that this additional mass will increase the gravity. What it is not well determine is the % of iron and nickel at the core, so, we can talk on probably a raising on between 5-15% the actual gravity force.

Regarding some other metals, it will depends on magnetism and density.

Hope it helps!


Why does the intensity of sunlight depend on your latitude? People at the equator get to bask in more sunlight than Santa Clause and other inhabitants of the arctic regions. Not quite as pronounced, but they get more than me too.

Why is the sunlight more intense closer to the equator and less intense farther away from it?

When I posted this question, I was not thinking about the possible ambiguities, such as "Are you talking about the exposure across a surface area with some non-perpendicular angle to the sun," or "Are you talking about the light gathered by an optic facing the sun?" There is a difference. Since "basking in sunlight" was the example use case, let us assume exposure across a surface area which is lying on the ground.

As noted in the comments, this answer applies to things like sun-bathing and solar panels, but it does not apply so much to a specific point-receptor like an eyeball. If all objects in question are pointing directly at the sun, then the angle of incidence is equal for all of them and this answer does not apply.

For an optic facing its target, the amount of atmosphere that the light passes through is a very large influencer. At higher latitudes, the sun is not directly overhead, and so the light is not coming straight down through the path of least atmosphere. Instead, it comes in at an angle, passing through more of the atmosphere before it gets to you.

For sun-bathers, solar panels, and the ground in general, the sunlight absorbed and reflected does depend very much on what is described in this answer. For that reason, more expensive solar panels are mounted on devices which alter their angle to face the sun for increased light exposure. And a sun-bather could likewise increase their exposure by mounting their platform at an angle. This is the direction the rest of the answer will take.

The answer is similar to the answer to some other questions, such as "Why does the solar power intensity change with the season?" and "Why does the solar intensity change with the height of the sun in the sky (ie: with the time of day)?"

The very short, non-technical version (tl;dr)

Each unit (think "beam of sunlight") is spread over a larger area.

That might not seem intuitive at first, but that is the answer in a nutshell. To see why, continue to the long version.

The mechanics of the situation

The sun is huge, but at great distances it can be thought of as a point-source of light. The sun emits light in every direction, so its energy, the electromagnetic radiation, can be thought of as beams of sunlight coming out from it in every direction. However, the farther away you get, the fewer of those sunlight beams cross your path, and if you get far away you can see, from the below illustration, that the lines which cross your path are becoming ever more parallel and directional.

However, the sun is very, very, very far away - many millions of miles (and kilometers) away. At this distance, the electromagnetic radiation is all arriving at roughly parallel trajectories. That is, the light reaching us can be viewed as coming from a directional light source; it's all going in the same direction. You will see my crude illustration of that coming up after the next picture, but first...

When this happens, the angle at which the light arrives relative to the surface of the earth that it is arriving at, known as the angle of incidence, is what makes all the difference here. In the image below, we have light arriving from the left, perhaps from a flashlight(s), and two objects we want to see. The surface of the object below will be illuminated more, but the surface of the object above will be illuminated less.

Notice the object above gets 2 units of light to illuminate its entire surface, whereas the object below gets 5 units of light for its surface.

Looked at another way, the object above has to spread 2 units of light across its entire surface, where the object below gets that same amount of light concentrated on only a smaller part of it.

Now back to Earth

We see the same thing happen. Notice in the below picture that I have labeled the diameter of the area covered by my arbitrary unit of light. The distance is literally measured in image pixels. Near the equator, 1 unit of light illuminates 27 pixels, but closer to the pole the same amount of light needs to be spread out over 61 pixels of Earth's surface, so it's illuminated less.

And that is the answer to the question, and the related questions mentioned in my introduction.

Some specific calculations

The specific ratio of illumination difference for the two areas in the above image would be (approximately) simply the division of the two numbers.

The area closer to the pole is getting 27/61 as much light per longitudinal unit cross-section (a north/south strip of land) as the area closer to the equator, or 0.4426 times as much light.

Or, in the other direction, the area at the equator is getting 61/27 as much light, or 2.26 times as much light. So the equator area is getting more than twice as much light.

We can easily see from this why the sun might appear very much brighter or dimmer depending on the angle, and therefore depending on the latitude. This also explains the temperature difference: the equator is receiving a lot more energy, so it gets hotter.

Answer applied to seasons

I said earlier that this also answers the question about the illumination change for the seasons. The only extra bit of information that we need to understand that is this: Earth's pole around which it revolves is not perfectly perpendicular to the direction that the light is arriving from. Because of this, the specific angle changes as Earth orbits around the sun. The orbit around the sun takes a year (That's what a year is, by definition), so this change of angle over time has a yearly cycle during which the angle of incidence (the fancy term from above again, which just means "the difference of angle of the light to the object it is hitting") changes. This change is large enough to further change the sunlight that we receive, in the exact same way as discussed above.

Further notes about temperature

Just like before, this also affects temperature. The temperature, however, cannot be calculated by a simple ratio of area coverage. The temperature is affected by multiple other complicated systems. One of the major temperature influencers is the weather. Another of which is the difference between energy absorbed and energy lost (a physics thing: Earth is not a "black body" radiator).

Weather affect on temperature is very complicated and would require its own post much longer than this one, but to sum it up super succinctly: stuff like the wind and rain and such change the temperature and also move it around.

The physics black-body thing likewise would require its own long post, but again, super succinctly: Earth is also putting out its own energy into space at the same time it is receiving energy from the sun, and the affect this has on temperature is a fine balancing act. Once that balancing act tips to one side, the temperature swings way over in that direction: during the spring when temps are rising the energy absorption is enough higher than the energy radiation that the temperature rises, and the opposite is true for autumn when temps are falling.

The above descriptions for both weather and block-body radiation are way oversimplified.

That's it

And now the answer applies fully to seasonal changes as well, and it applies not only to illumination but also to temperature. I will leave it as an exercise for the reader to think about the daily affect (Why is it so much easier to look at the sun when it is low on the horizon just before it sets?)

And, as stated before, there are other influencers for these things; this just happens to be the biggest one.


How much water do aquifers contain? How much does it take to deplete them? How much water does a typical aquifer contain, and is casual well use is likely to deplete them?

Your answer can take many forms, but an ideal question would cover these sort of areas- would a single well that someone took 50 liters from a day deplete one? Do you need industrial use of hundreds of meters cubed? Typically how vulnerable are they to depletion by people. Is there a lot of variability in size, or are they usually fairly large? Does it take months, years, decades?

Inspired by Mad Max, where there is a large mountain aquifer supplying a couple hundred people and I have no idea how long it would take to deplete.

Previous questions have covered this in vague terms but haven't given a general picture of what it's like for people in general, especially in more remote areas where they may not have access to the largest of aquifers, which apparently can last decades. I'm more curious what it's like for a random farmer.

It depends on the size of the aquifer and the amount of precipitation falling on the exposed areas of the aquifer. These have been used to provide water form many major cities.
The Trafalgar square fountains were initially powered by artesian pressure (as London is below the altitude of the North downs / Chilterns). Eventully too much water was extracted leading to salt water incursions in places. This aquifer is still in use but backed up by a number of surface reservoirs surrounding the London area.

"
The Report of the Royal Commission on Water Supply (1869) shows that, by 1869, a number of large conurbations were supplied by groundwater. Nottingham and parts of Liverpool and Birkenhead relied on wells in the Permo-Triassic sandstones, Sunderland and South Shields on the Permian Magnesian Limestone and Croydon and parts of south east London on the Chalk. Government offices around Westminster together with the fountains in Trafalgar Square were supplied by wells sunk to the Chalk in 1844 (Amos, 1860).
"

More from here http://www.groundwateruk.org/British-Hydrogeology-a-brief-history.aspx

More detail on the London aquifer:
 https://assets.publishing.service.gov.uk/government/uploads/system/uploads/attachment_data/file/735451/2018_Final.pdf


What is it called when water forms to look like a root structure? I've tried looking it up on google but I can't find anything. I'm talking about when a small amount of water forms to look like a root hair structure, thin and branching out.

The term you are looking for is Dendritic, it literally means "having a branched form resembling a tree" and is used to describe the shape of rivers but can also apply to anything with that pattern.