What does the land under Antarctica’s ice sheet look like?
ubique.americangeo.org
ubique.americangeo.org
The lack of ice weight would allow the land to rebound upwards:
https://polarjournal.ch/en/2022/07/13/what-greenland-and-ant...
The seismic tremors caused by the uplift are so minor that no layperson even knows about them.
The volume of the Antarctic Ice Sheet is 26.5 million cubic kilometres. If you use the Spherical Cow Theorem and put all that ice into an sphere, the radius is 185km https://www.wolframalpha.com/input?i=sphere+of+26.5+million+... . The gravity caused by that sphere at the surface is 0.047m/s^2 https://www.wolframalpha.com/input?i=gravity+of+24%2C380%2C0...
Compared to the usual acceleration of gravity that is almost 9.8m/s^2, it's only a 0.0048 = 0.48%. That is the slope of the real see compared to an ideal sea where there is no gravity from the ice. I'm using a spherical ice instead of a sheet of ice. With a sheet of ice the effect would be much smaller, but I'm too lazy to look up.
The gravity from the ice decrease as 1/r^2 when you go farther. We must integrate the slope to get the volume. I'm going to use a flat Earth, to simplify the calculation, but as the 1/r^2 reductions is quite fast, it's not a problem. With a curved Earth it will be more, but not too much.
Also, we must multiply by 2*pi*r to get the volume in the flat surface all around the sphere instead of the surface of a vertical cut of the sea.
The radios of the Earth is 6371km, so the "distance" to the north pole is 6371kmpi=20000km and we can cut the integral there https://www.wolframalpha.com/input?i=Integral+from+185+to+20... The result is 48,000 km^3 of water. But melting the Antarctic Ice sheet will release 24,300,000 km^3 So it's only a .2% more.
In other units, melting all the Antarctic Ice would increase the sea level like 58m (190ft). The additional effect of the gravity is less than .3m (1ft) And I expect that a calculation if a sheet shaped ice will give a much smaller result.
First, let's calculate the additional high. It's better to imagine the 0 is in the oposite pole, and the additional height is very similar to the old integral without the 2*pi*r https://www.wolframalpha.com/input?i=Integral+from+x+to+2000... i.e. 0.0048 * 185^2 * (1/r - 1/20000)
Now we must integrate it again, with the 2*pi*r Integral from 185 to 20000 of (0.0048 * 185^2 * (1/r-1/20000)), that is only 606km^3, so 1/80 of my previous calculation. So, fixing the last part of my comment with the new numbers:
But melting the Antarctic Ice sheet will release 24,300,000 km^3 So it's only a .0025% more. In other units, melting all the Antarctic Ice would increase the sea level like 58m (190ft). The additional effect of the gravity is less than .0004m (4mm, 1/8 inch)
The end of the 7-second animated video on your link shows a landmass that looks much more like the rest of the Earth.
Utterly fascinating to see a "hidden continent" like this -- I'm so used to unfamiliar maps being fantasy (like Westeros from Game of Thrones), so it's quite remarkable to see an unfamiliar map that is on our very planet.
http://www.seismolab.org/model/antarctica/lithosphere/index....
https://scitechdaily.com/high-precision-map-shows-what-the-l...
Which I assume is using data from here
Edit: without the underwater parts: https://i.imgur.com/tR67Nwq.jpg
One pattern that is curiously consistent is that broadly land tends to be antipodal to ocean (and not just because there’s more ocean in general - there’s a genuine statistical skew) - so having land at one pole and none at the other matches that overall pattern. (https://www.geodatos.net/en/antipodes)
Basically a surprisingly large proportion of antipodal pairs have land at one end and water at the other (although water-water is also common).
In other words it would certainly be more surprising to have land at both poles; and we should be unsurprised if we find land at neither or land at one and not the other.
So by that measure the poles are fairly typical of other antipodes.
Regardless, it's just an interesting snapshot in time considering continental drift.
It’s easy to fall into some counterintuitive fallacies reasoning about this though because we’re considering a spherical surface pairwise and you might be tempted to say ‘ah, but we’re counting the antipodal land twice so that probably accounts for the 15%/30% thing’ - but I don’t think that follows; that 15% with land antipodal to it can be thought of as two sets of 7.5% of the land, which are antipodal to one another - the ‘double counting’ doesn’t work that way to eliminate the anomaly.
We have a 30% chance of picking a random point on Earth and getting land. Because earth is so big and these theoretical points are so small I'm going to assume that one point of land existing doesn't effect the odds of another piece of land existing. So that's two antipodal points, each with a 30% chance of being land.
Putting that into a probability calculator (again, long time since statistics) that gives us odds of both point A & B being land as 9%. If we flip that around and go with the 30% land and 70% water odds we end up with a chance of 21%. It seems like there's actually MORE land that's antipodal to other land than we'd expect statistically? I must have some bad math here.
Not sure where 21% comes from - if 3/10 of points are on land, 9/100 antipode-pairs should be land-land, and that means 9/100 / 30/100 = 9/30 = 30% of land should have land opposite it, and conversely 70% should have water?
But the true numbers are 15%/85% - on the actual earth, land is more likely to have ocean opposite it than random chance would suggest.
Analyzing whether items sufficiently far from the ‘expected’ random distribution as to be statistically significant is a rather different exercise though.
This is the issue.
A large contributor of the "point antipodal to land is water" anomaly is the typical size of land masses. If, for example, the average size of land masses was 30% of earth's area, then we'd have a single large land mass (Pangaea style), and all antipodal points would be water.
The approximation of independence here doesn't reflect reality. Eurasia alone is more than 10% of the land area of Earth.
EDIT: also, do we know if there were no other continents that submerged?
If we distribute chunks of land randomly on a sphere we would expect roughly the same amount of land on any randomly-chosen hemisphere. But what can we say about the average amount of land in the hemisphere with the most land?
My intuition says that in an Earth-like sphere, with 29% land and 71% water, there's a high chance of there existing a hemisphere with almost all of the land.
I have no idea how to prove this empirically, but it seems like a really nice maths problem.
Suspect there’s something to do with the fractal dimension of the coastlines as well. Agreed - sounds like a rich vein of math.
The locations of the continents spend the vast majority of the time either approaching a clustered state, or leaving it.
Because the thick continents with land are a minority of the surface [2], They only briefly pass through a period of even dispersal when passing from one clustered state to another.
>My intuition says that in an Earth-like sphere, with 29% land and 71% water, there's a high chance of there existing a hemisphere with almost all of the land
Can you explain more? If you have 7 or so plates with land, why would you expect them to have a high chance of being in the same hemisphere? The math is the same as flipping a coin 7 times. Flip the first and see which side it lands on, then flip the remaining 6 and see how many match the same hemisphere. What am I missing?
https://www.youtube.com/watch?v=gQqQhZp4uG8
https://www.imperial.ac.uk/news/199334/scientists-find-link-...
I'm not saying that a particular hemisphere has a high change of having most of the land; I'm saying that there will be a hemisphere that has this chance.
29% of Earth is land and 80% of this land is in the land hemisphere, so this problem would be the equivalent of rolling a D48 14 times and placing the dice on a table so that any 11 of the rolled numbers are visible if I look at it from the top.
I think that is also fairly solvable. let me think on it. I cant understand your example, but is this the same?:
Roll a d100 once per continent (e.g. 7 times) and write the numbers down. What percent of the time can you pick a range 50 numbers that captures all 7 results. the range can wrap around from 100 to 0.
The problem would be easy to monte carlo, but I would have to think about how to solve it in closed form.
Will update if I have a solution.
My question is the probability of being able to choose 11 of the 14 continents so that there exists a hemisphere that contains all of them, like the land hemisphere contains 80% of the Earth's land.
1)When you say you are using the canonical definition of hemisphere, do you mean North and South? Or are you talking about East and West as well?
2) Out of curiosity, why 14 continents?
If you are using North and South as pre-defined hemispheres, ignoring East and West, I do think it can be reduced down to a coin flip if you want to prove that plates are unlikely to almost all be in the same hemisphere. This is because allowing them to overlap allows more clustering, not less.
12/14 continents is >80% and 11/14 is <80%.
If you drop a continent randomly, it has a 50/50 chance of being in the north or south.
The probability of 12 or more being in the northern hemisphere is 0.647%. The probability of 12 or more being in the southern hemisphere is 0.647%. Together, the probability of 12 or more being in the same N/S hemisphere is 1.294%
You can do the same for E and W, getting another 1.294%, for a total of 2.588% chance. This is allowing the continents to overlap, so the real chance would be lower.
By "canonical definition of hemisphere" I'm talking about the surface of any half-sphere that covers half of the Earth; the "joke" in my proof is that you can choose any one.
For any two continents there's always a hemisphere that contains both of them (assuming you don't worry about the semicircle in the border of the hemisphere). If you put three equidistant continents with the maximum possible distance between them (so that they form an equilateral triangle whose centre is the centre of the sphere) then you can't do anything about it.
Think it like shining a light from some point in space that lights 80% of the land in Earth.
That is what I initially meant by having "freedom to define hemispheres however you want", eg set arbitrary centers.
That certainly makes it a more interesting problem and increases the chances substantially.
"Almost all remaining life on Earth has adapted to live in Antarctica. Elsewhere life is extinct except for some evidence of biological activity in the Himalayan mountains."
It is interesting how much more archipelago-like Antarctica is. I wonder if that would mean fewer interior deserts than a large continent like Australia has
I recommend we don't adopt this program of rotating the earth. It's almost certainly worse than just inventing fission power and pumping desalinated sea water to wherever we want to de-desertify. We'd probably also be better off if we wasted quintillions of dollaeuroyuans on a program to terraform Mars that ends in complete failure and the deaths of millions of workers over a century, but that's also an outcome I recommend we avoid.
There are lots of countries closer to the equator - they are wetter.. large body masses have continental clima, which enocourages the growth of dessert, but they usually do so only when the war of forrest vs savannah is lost by the forrest. Which can cling to areas quiet defiantly, for example the last big forrests in syria, that once covered the middle east
Does not have to mean close to the equator.
Moist tropical air barely reaches the very top of Australia - instead it hits places like Java, where the abundant rainfall and numerous active volcanoes provide ridiculously fertile land that currently supports 150 million people - 5x that of Australia - despite being 50 times smaller.
Well, Australia claims about half of Antarctica so as long as we can enforce that claim maybe it's not the end of the country. Just have to declare part of it "New Southern New South Wales" and take it from there.
> I wonder if that would mean fewer interior deserts than a large continent like Australia has.
There Arabian pensinsula are the Sahara are surrounded on three sides by sea, but they're still very dry. The Sahara and Arabian deserts go right up to the coast. As mentioned in my reply to your other reply, as far as I know it's not a lack of water but what happens to the water once it's evaporated that causes the dryness in those areas - it goes up into the high atmosphere and comes down in rain over the tropics. Wish I remembered the name though.
I have always wondered how this would turn out. Currently not many other countries recognise our claims, but so far there's been no reason to enforce them. Given out proximity we are in a good position to enforce them, and the likes of BHP and Rio Tinto would love access to all the possible resources.
It's the other way around - the subtropical ridges are high pressure high altitude masses of dry air that fall to the surface around 30 N & S. This dry air is then sucked back to the equator as the Trade Winds, which further dry out the sub tropics.
> Wish I remembered the name though.
Hadley cell?
Scientific implications would be IceCube would no longer work, because the premise is based on the detector(s) being drilled 2 km into the glacier. You can't just bury it in rock, it specifically needs a transparent medium and ice is surprisingly good for neutrino propagation. The other telescopes would probably be fine, provided it was still sufficiently dry. SPT and BICEP are mostly taking advantage of the location - e.g. high altitude, extremely low humidity and extremely radio quiet, to a lesser extent the ability to stare at the same patch of sky all year. I guess you'd lose a lot of the environmental advantages, but maybe it would still be a desert.
There isn't a particularly good reason why everything is sited there beyond geopolitics. You could pick pretty much any place on the polar plateau and it would be fine. We just happened to luck out at how suitable the ice was for certain experiments.
The station is on a moving/shearing ice sheet though, which has some minor implications (like the relative position of the pole changes, so we have to move the marker each year by about 10 m).
Is there any benefit to having in a place that rotates in situ rather than going around in circles in a given day?
And yet I'm saddened by the huge waste of it all, as this can be brushed away by one "charismatic" speaker that enough masses will follow. Voters will not want to do anything. People burn their gas stoves out of spite.
Scraped.