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Comment by jaspax | Hacker News Reader

Wait. My intuition says that the mass of ice cannot possibly have a gravitational influence on nearby water big enough to create the effects described. Back of envelope calculation: surface area of Antarctica is on the order of 10M km^2, with an average thickness of 1 km (a high estimate), means 10M km^3 of ice; some further calculations from volume to mass multiply out to 10^19 kg of ice. (Note that we're assuming that 1kg of ice is 1 cubic deciliter, as with liquid water; this is definitely wrong but let's go with it since this is just an order-of-magnitude estimate). The mass of the Earth is on the order of 10^24 kg, so this is 10^-5 of the mass of the earth.

Can you get really get 50m of local displacement from 10^-5 Earth masses?

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At a distance of 100 km, the gravitational acceleration caused by the greenland ice sheet is .019 m/s^2 or ~0.2% g. Doesn't sound like a lot but it's still 524 times higher than the effect of the moon's gravity which creates the tides. The ice sheet has the same gravitational effect as the moon at a distance of approximately 2300 km.

This is the answer which makes the most sense to me.

Hmm, but a water molecule 100 km from an Arctic ice sheet is also experiencing the pull of an "equal" weight of water 100km the other direction, albeit ever so slightly offset angularly.

The lunar system is seemingly much less symmetrical but we still get some amphidromic points (points of zero tidal range) and neep tides (lowest "high" tide) can be very small.

That pull would be there regardless though.

The interviewee is talking about ice on land (Greenland and Antarctica), not about ice in the ocean. If the ice on land melts, then those water molecules migrate away from Greenland because they drop to sea level. That loss of mass would happen regardless of the gravitational pull of water 200 km away and would change the gravitational gradient of the water 100 km away.

Another way to think of it: there's not an "equal" weight of water above Greenland because it's raised above sea level by the land. The weight of rock underneath the ice sheet needs to be included as well.

> Doesn't sound like a lot but it's still 524 times higher than the effect of the moon's gravity which creates the tides.

Contrary to common belief, tides are not caused by the direct influence of the moon's gravity (it's far too weak to have any effect)[1]. The tidal forces are caused by the gravitational gradient from the moon (and the "centrifugal" forces from our path around the earth-moon barycenter), and I don't believe you'd get the same effects from a gravity source on the surface of the earth.

Even a lot of very respectable scientists and textbooks get this wrong.

[1] See https://www.youtube.com/watch?v=pwChk4S99i4 for a pretty good explanation

Just to add to this answer and perhaps save you the click, what he is referring to is that the force pulling the water upward at high tide is not the direct gravitational pull of the moon. It is that the gravitational force from the moon at the edge of the earth is greater than the force at the center (because it is closer to the moon). Similarly the force on the opposite side of the earth is less (because that side is further from the moon than the earth's center). So the water molecules are drawn away from the center of the earth (near side because they are being pulled slightly harder than the center of the earth, and far side because they are being pulled slightly less hard than the center of the earth). Hence the high tides on both sides of the earth, not just the side closer to moon.

Or, to put it another way, the surfaces of the Earth closest to and farthest away from the Moon are traveling at the same orbital velocity of the Earth. However, they should be in different orbits; the point closest to the Moon is too slow for the orbit it is in and the point farthest is too fast. The former wants to into a lower orbit while the latter wants to go into a higher orbit.

In a further translation from gibberish:

Or, to put it another way, the surfaces of the Earth closest to and farthest away from the Moon are traveling at the same orbital velocity around the center of the Earth/Moon system. However, they should be in different orbits; the point closest to the Moon is too slow for the orbit it is in and the point farthest away is too fast. The former wants to into a lower orbit while the latter wants to go into a higher orbit.

If I correctly understood the video linked above, the primary cause is actually due to the tidal acceleration (e.g. moon's gravity) of objects on sides 1 and 3 (see diagram below), relative towards the earth's surface, being mostly radially inward [1]. The majority of the ocean water along the sides of Earth is being pulled in very slightly, and in aggregate across the massive surface of the ocean, this results in enough pressure to push up the water at sides 2 and 4. Tides are pushed up due to pressure, not pulled up.

The analogy they used is that tides are more like a pimple being squeezed than taffy being stretched.

[1] See timestamp 4:45 in the video: https://www.youtube.com/watch?v=pwChk4S99i4&feature=youtu.be...

---

Diagram:

   1
 4 E 2      M
   3

E = Earth

M = Moon

Numbers = 4 "sides" of the Earth, relative to the Earth-Moon line

It seems this would made the effect from the icebergs bigger rather than smaller, because the gradient decays by r^3 instead of r^2, so the distance is more important?

Like, the gravitational acceleration is a = GM/r^2 while the gradient is da/dr = -2GM/r^3

So for moon vs glacier at 1000km you'd get

- 2 * (Gravitational constant) * (mass of moon) / (391184 km)^3 = - 1.638×10^-13 reciprocal seconds squared

- 2 * (Gravitational constant) * (1e19 kg) / (1000 km)^3= -1.335×10^-9 reciprocal seconds squared

It's the gradient applied over the whole ocean that causes the tides. In other words, the forces are not enough to create any kind of local change in the depth of the ocean. It's not a gravitational "pulling" as we're accustomed to think about, but more of a global squeezing of water from that gradient applied to the entire ocean. I'll have to give some more thought to the idea of a gravity source on the surface of the earth, but I doubt it would work the same way (we're comparing a relatively small mass in a concentrated location on the earth with that same mass distributed mostly evenly over the earth).

Nothing of this disproves the effect of the polar ice. The forces still apply, they're just not shifting daily.

> The forces still apply, they're just not shifting daily.

No, the forces are completely different. If we have an object on the surface of the earth that has enough mass to roughly produce the same nearby gravitational acceleration as that felt by the moon (which is minuscule and undetectable by most instruments), that object would not produce changes in ocean levels as we see with the moon. Again, the oceans are not rising/falling due to the moon's gravity pulling on them. It only happens because the moon is far enough away that its tiny gravitational acceleration on the earth is (1) felt everywhere on earth, and (2) felt everywhere on earth in slightly different amounts.

For a smaller, closer object (even with similar nearby gravitational acceleration), the tidal forces will not be the same because that gravitational acceleration will fall off to near zero in a very short distance.

You are mixing tidal forces with local attraction forces due to ice. Even though the forces due to ice are not felt everywhere in the planet, they do have an effect that can be calculated, and it is stronger in the body of water closer to the poles. The result of this force causes water attraction in the polar regions. So the effect that is felt in other parts of the planet is not due to the gravity force, but due to the displacement of water happening in polar regions - after all the water needs to go somewhere.

I was originally responding to the claim that "it's still 524 times higher than the effect of the moon's gravity which creates the tides" - that's why I mentioned tidal forces. If the ice creates that much direct, local gravitational attraction[1], then my point is we shouldn't be comparing it to the moon at all (because the forces involved are not the same as those involved with the moon and tides).

[1] Even the claim about the ice sheet (and its melting) contributing significantly (via gravity) to global sea level change seems dubious since, as noted elsewhere in this discussion, the Earth Gravitational Model appears to be affected much more by factors other than ice sheet thickness or surface features.

As you said, it's the gradient which matters. The gradient determines how much the sea level changes over a distance. But, the sea level change is distinct from the tides. The difference in water level between two connected locations at the same moment in time is the same for the ocean and the lake.

The video you linked to compares lakes and oceans because the lunar tides vary with time. The lake level difference between Cleveland and Buffalo at 6 will be the same as the sea level difference between New York and Providence at ~5:30. You need to compare your sea level to the sea level a quarter of the way around the world to understand why your local sea level changes from 6:00 to 12:00.

1) the center of the ice sheet can be much closer to the water than the center of the earth. Some water around greenland is at 600 km from the center of the ice sheet, and 6000 km from the center of the earth. This factor 10 difference is squared because gravity's strength drops off with the square of distance.

2) Both forces are orthogonal, so you take the ratio to get an idea of how much the ice sheet attraction is slanting the water surface. This might be a very small angle, but if you have a small angle sustained over hundreds of kilometers then you can arrive at a height difference of meters. E.g. if the ratio is 10^5, then you have a 1 meter height difference at 100km distance (ignoring that the ratio actually changes over that distance to simplify).

So... why wouldn't this show up consistently in the EGM96 data?

EGM96 is a definition of Earth's equipotential height constructed by measuring it with satellites like GRACE. Its as close to a definition of true sea level as you're going to get. But it doesn't have this kind of consistent uplift near tall masses throughout the model. The Southern Ocean's height is sinusoidal about Antarctica. We do see an increase in equipotential height in the Andes, and in the eastern Pacific nearby. But near the Himalayas the equipotential height is lower.

https://en.wikipedia.org/wiki/File:Earth_Gravitational_Model...

I guess because plainly visible "tall masses" are not the only things contributing to the geoid? E.g. earth crust thickness might dominate.

In the context of this discussion, we're not so much interested in the absolute shape of the geoid, as in how much it might change because of melting ice sheets

You’re missing one of the effects gravity and ice have. From the article:

> The second thing that happens is that this gravitational attraction that the ice sheet exerts on the surrounding water diminishes. As a consequence, water migrates away from the ice sheet. The third thing is, as the ice sheet melts, the land underneath the ice sheet pops up; it rebounds.

The land underneath the glacier or ice sheet (and it has to be on land, because ice displaces it’s melted volume when floating) pops up and increases in altitude (from the centre of earth) due to the drop in weight.

This popping up effect will of course affect surround land not under the ice because rock isn’t that flexible.

I'm not a physicist. I did get curious and went to Wikipedia. You're mostly spot on:

>It covers an area of almost 14 million square kilometres (5.4 million square miles) and contains 26.5 million cubic kilometres (6,400,000 cubic miles) of ice.[2] A cubic kilometer of ice weighs approximately one metric gigaton, meaning that the ice sheet weighs 26,500,000 gigatons.

26,500,000 gigatons is 2.65e+18 kg in scientific notation.

I compared this to the moon, which is 7.35 x 10^22 kg, or about 30,000 times as heavy. The moon does create quite some tidal effects, while at 384,400 km distance.

Since gravity is inversely proportional to the square of the distance Both the 50m of local displacement as well as the 2000km distance until it sufficiently cancelled out sound believable to me.

I think you are off by an order of magnitude, it should be 2.65 10^19 kg. one gigaton is 10^12 kg , 26,500,000=2.65*10^7

You're right, thanks!

the mass of the iceshield / mass of the moon = 5 10^-4. But at a distance of 2000km compared to 4 10^5 km for the moon, the iceshield should have similar gravitational pull. the inverse square of 2000km / the inverse square of 4 10^5 km = 4 10^4. (4e4=(1/2000000^2)/(1/(4e8)^2))

Distance and mass seem to cancel out almost perfectly

I found this Minute Physics video about "sea level" very enlightening [1], and I think it can help here (just replace mountains by "huge masses of ice").

[1] https://www.youtube.com/watch?v=q65O3qA0-n4

If the Earth were perfectly round the center of mass would be in the exact center, but if it had lopsided bulge on one side the center of gravity would shift in that direction. The water, being liquid, will want to form a sphere around that new center of mass. That deviation from center means a corresponding increase in water around the surface in that direction. So intuitively, if you think about it this way, 30 something meters deviation from center doesn’t seem too counterintuitive.

Hard to imagine a floating ice cube having an effect on a sphere with mountains, though.

It floats, so it bulges out from the earth, shifting that mass away from perfect center. Given that it’s not liquid, it bulges out non uniformly in one direction and biases center of mass in one specific direction. It also doesn’t matter too much that it’s not particularly high, as it it covers a large area in the general direction. Anything on the northern hemisphere would bias center of mass in that direction.

EDIT: I’m not totally sure about the bulging out part of the floating ice is true, but there are also glaciers on land such as over Greenland that would shift the center of mass.

Is an effect like this similar to the difference between the water level of the Atlantic and Pacific? It never made intuitive sense to me that the Panama Canal for example had to deal with a significant water level difference between the two sides, as the two bodies of water are contiguous at other points on the Earth’s surface.

The Panama Canal has locks mainly to raise (by 26m) the water-level over the land. It's less work to flood the area than to dig a giant channel between the oceans.

https://armyengineer.com/history/panama/engineers/How_Canal_...

The average sea level difference on each side is minor in comparison (20cm). Plus, the average sea level isn't constant through the whole ocean. It's variable depending on location due to different salt concentrations (salty water is less dense).

https://oceanservice.noaa.gov/facts/globalsl.html

Salty water is more dense.

Whoops, of course. It's easier to float in the ocean than in a lake.

Oh man, thanks for this! I did not realize that it was essentially to hoist the boats over uneven terrain, always assumed the canal was a straight dug channel. Never enough to go search out the answer, but it always tickled the back of my mind when it came up. Thank you!

The force vector from a polar ice sheet doesn't get cancelled out by the rest of the earth though (because they aren't pointing in opposite directions). Plus, you have to account for the force of the ice sheet, plus the water it attracts, plus the water that those combined attract, etc.

I'm no good with calculus so can't run that back-of-the-envelope for you, but it doesn't seem all that surprising to me.

Even if we ignore all that, and treat everything as a point mass, it's still believable, I think.

The earths radius is very approximately 6.4km. So 1km out from the ice sheet, we have 1(1/6.4^2) (earth) vs 10^-5(1/1^2) (ice), which is very roughly 1/2500th of the effect of earths gravity. But the earths gravity is strong! The forces it exerts on the ocean are titanic - it doesn't seem outside the realms of belief that even this small percentage of the force could produce an observable effect, when the forces are so huge, and the differences are only measured in meters.

I think part of the reason we find this difficult to grasp intuitively is we don't really have a good mental model of just how titanic many of these forces are - huge numbers are just not something we, as a species, are good at understanding.

Earth’s radius is 6371 km, not ~6 km.

Radius of 6 km is comparable to the size of Halley's comet :-)

Let's see,.. Mitrovica's robust computer modelling vs. your back of the envelope calculations...

Wonder which one I'd put money on.

Do you have a link to a paper, I couldn't find one on google

The one linked in the article? https://advances.sciencemag.org/content/1/11/e1500679

If your fancy computer model doesn’t approximately equal your back of the envelope calculation, that’s generally a knock against the computer model rather than the napkin math. Exceptions exist, but for a system as linear as this you should be able to separate out gravity and look at it on its own no problem.

Check out the Canvendish experiment! Watch gravity work in a classroom.

https://youtu.be/MbucRPiL92Q

G is 6.67e-11, so for 1kg of matter 1000km away from a center of that mass the force is (e-11 x e19 / e(6x2)), around e-4 newtons or e-5 kg of weight equivalent. Idk what’s that worth ocean-wise.

This isn't intuitive to me either, given that the ice sheet is less dense than the surrounding water. I mean, it floats. Never mind the effect of the surrounding landmass, which is WAY more dense. I'm not ready to call his theory "fact" like Nautilus is.

That said, I'm not a geophysicist.

I don't think the comparison with the surrounding land mass is relevant here. The landmass might have a much larger effect than the ice, but the landmass will not melt. While my mental image considers the water to envelop the world as a sphere, I actually have no idea how large the differences in the distance from the core to the water surface is in different parts of the world.

> I actually have no idea how large the differences in the distance from the core to the water surface is in different parts of the world

Taking the core to be a point, the distance to the sea's surface would depend on factors such as

* The earth’s variation from a pure sphere (6,378.137 km (3,963.191 mi) at the Equator and 6,356.752 km (3,949.903 mi) at the poles) [0, 1]

* The local depth of the sea (up to 10,984 metres for the Mariana Trench [2])

* Tidal effects [3]

[0] https://en.wikipedia.org/wiki/Spheroid

[1] https://en.wikipedia.org/wiki/Figure_of_the_Earth

[2] https://en.wikipedia.org/wiki/Mariana_Trench

[3] https://en.wikipedia.org/wiki/Tide

And, I suppose, the centripetal force caused by the rotation. But maybe that is the whole reason for earth being a spheroid?

And correct me if I'm wrong, but the local depth of the sea should not matter, as water would fill the depth before the levels stabilized. But perhaps the local volume of the water affects the size of the effects of tidal forces and local land mass.

Mostly related to your line of inquiry, the point on land that is furthest from the center of the Earth is Mount Chimborazo in Ecuador.[1] It is located one degree south of the equator and is very tall to begin with.

[1] https://oceanservice.noaa.gov/facts/highestpoint.html

Parent is writing about Antarctica, the ice sheet is not floating but would be replaced by air which is a lot less dense.

And the land rising effect is certainly there, its well known from our norther hemisphere where satellite measurements have tracked for quite a while now how i.e northern Europe still rises in comparison to the southern parts due to the not being covered by the ice age glaciers anymore.

The arctic ice sheet floats. Greenland is an island and Antarctica a continent, both of them surrounded by a little floating ice.

Melting that ice shifts a km-deep layer of ice from a fixed position above nearby sea level to being part of the liquid ocean. This means that it doesn't exert any gravitational pull on the ocean nearby, so the water-covered part of the globe becomes a little bit more spherical. Not much more spherical: The article says 30-50m on the coast of Greenland, which is a very small fraction of the earth's radius.

The 30-50m column of water is distributed elsewhere.

Replying to myself, this implies that near Greenland, the gravitational effect of the ice cap should be ~30 times as big as the moon's (which causes a tide of a little over a meter). Which doesn't seem horribly implausible.

The moon causes a rise of about 54cm, and the sun around 28cm.

from the wikipedia entry for tides.

https://en.wikipedia.org/wiki/Tide#Amplitude_and_cycle_time

An amplitude of 54cm is about a meter peak-to-peak.

Not only that, but I think the whole idea here is just plain wrong. From the article:

> this gravitational attraction that the ice sheet exerts on the surrounding water diminishes. As a consequence, water migrates away from the ice sheet.

When an ice sheet melts, it doesn't create a void in its place. What was once ice becomes liquid water. That liquid water actually has a slightly higher density than it had when in ice form, but it will occupy slightly less volume in the ocean. Mass is conserved, and the net effect in terms of gravity is essentially zero.

If anything, you'd have slightly higher local gravity in that part of the ocean (due to a higher concentration of liquid water vs ice), but again, zero net change looking at the entire ocean.

If the ice was above sea level to start, then it does create a void (OK, a space filled with air). The mountain of ice turns into liquid water and flows all over the earth.

Gotcha. I thought the article was referring to polar ice in general, not just specifically ice sheets (over land areas). Still, the gravity produced by that mass doesn't seem to be enough to make a real difference (see my other comment questioning the comparison with the moon/tides).

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