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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

by jaspax 6y ago

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?

_petronius 6y ago
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.
- jfindley 6y ago
  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.
  - silvestrov 6y ago
    Earth’s radius is 6371 km, not ~6 km.
    - ljcn 6y ago
      Radius of 6 km is comparable to the size of Halley's comet :-)
chrischen 6y ago
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.
- larrik 6y ago
  Hard to imagine a floating ice cube having an effect on a sphere with mountains, though.
  - chrischen 6y ago
    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.
    - chrischen 6y ago
      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.
- Infernal 6y ago
  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.
  - 6y ago
larrik 6y ago
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.
- hansbo 6y ago
  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.
  - KineticLensman 6y ago
    > 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 https://en.wikipedia.org/wiki/Spheroid [1] https://en.wikipedia.org/wiki/Figure_of_the_Earth https://en.wikipedia.org/wiki/Figure_of_the_Earth [2] https://en.wikipedia.org/wiki/Mariana_Trench https://en.wikipedia.org/wiki/Mariana_Trench [3] https://en.wikipedia.org/wiki/Tide https://en.wikipedia.org/wiki/Tide
    - hansbo 6y ago
      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.
wruza 6y ago
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.
YetAnotherMatt 6y ago
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.
- rshnotsecure 6y ago
  FYI this is a paid advertorial. The first clue is that there is no TLS support. The second, much larger clue, is the note at the bottom that "financial support is provided by http://fondation-bertarelli.org" http://fondation-bertarelli.org".
- DalasNoin 6y ago
  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
  - YetAnotherMatt 6y ago
    You're right, thanks!
- DalasNoin 6y ago
  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
mikro2nd 6y ago
Let's see,.. Mitrovica's robust computer modelling vs. your back of the envelope calculations... Wonder which one I'd put money on.
- mhh__ 6y ago
  Do you have a link to a paper, I couldn't find one on google
  - mehrdadn 6y ago
    The one linked in the article? https://advances.sciencemag.org/content/1/11/e1500679 https://advances.sciencemag.org/content/1/11/e1500679
- themeiguoren 6y ago
  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.
jjk166 6y ago
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.
- jaspax 6y ago
  This is the answer which makes the most sense to me.
- pbhjpbhj 6y ago
  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.
  - mehrdadn 6y ago
    That pull would be there regardless though.
  - rovolo 6y ago
    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.
- curiousgeorgio 6y ago
  > 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 https://www.youtube.com/watch?v=pwChk4S99i4 for a pretty good explanation
avianlyric 6y ago
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.
wcoenen 6y ago
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).
- brandmeyer 6y ago
  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_1996.png https://en.wikipedia.org/wiki/File:Earth_Gravitational_Model...
  - wcoenen 6y ago
    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
kilburn 6y ago
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 https://www.youtube.com/watch?v=q65O3qA0-n4
PopeDotNinja 6y ago
Check out the Canvendish experiment! Watch gravity work in a classroom. https://youtu.be/MbucRPiL92Q https://youtu.be/MbucRPiL92Q
curiousgeorgio 6y ago
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.
- dmurray 6y ago
  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.
  - curiousgeorgio 6y ago
    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).

by jaspax 6y ago

_petronius 6y ago

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.

jfindley 6y ago
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.
- silvestrov 6y ago
  Earth’s radius is 6371 km, not ~6 km.
  - ljcn 6y ago
    Radius of 6 km is comparable to the size of Halley's comet :-)

chrischen 6y ago

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.

larrik 6y ago
Hard to imagine a floating ice cube having an effect on a sphere with mountains, though.
- chrischen 6y ago
  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.
  - chrischen 6y ago
    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.
Infernal 6y ago
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.
- 6y ago

larrik 6y ago

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.

hansbo 6y ago
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.
- KineticLensman 6y ago
  > 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 https://en.wikipedia.org/wiki/Spheroid [1] https://en.wikipedia.org/wiki/Figure_of_the_Earth https://en.wikipedia.org/wiki/Figure_of_the_Earth [2] https://en.wikipedia.org/wiki/Mariana_Trench https://en.wikipedia.org/wiki/Mariana_Trench [3] https://en.wikipedia.org/wiki/Tide https://en.wikipedia.org/wiki/Tide
  - hansbo 6y ago
    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.

wruza 6y ago

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.

YetAnotherMatt 6y ago

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.

rshnotsecure 6y ago
FYI this is a paid advertorial. The first clue is that there is no TLS support. The second, much larger clue, is the note at the bottom that "financial support is provided by http://fondation-bertarelli.org" http://fondation-bertarelli.org".
DalasNoin 6y ago
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
- YetAnotherMatt 6y ago
  You're right, thanks!
DalasNoin 6y ago
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

mikro2nd 6y ago

Let's see,.. Mitrovica's robust computer modelling vs. your back of the envelope calculations... Wonder which one I'd put money on.

mhh__ 6y ago
Do you have a link to a paper, I couldn't find one on google
- mehrdadn 6y ago
  The one linked in the article? https://advances.sciencemag.org/content/1/11/e1500679 https://advances.sciencemag.org/content/1/11/e1500679
themeiguoren 6y ago
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.

jjk166 6y ago

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.

jaspax 6y ago
This is the answer which makes the most sense to me.
pbhjpbhj 6y ago
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.
- mehrdadn 6y ago
  That pull would be there regardless though.
- rovolo 6y ago
  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.
curiousgeorgio 6y ago
> 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 https://www.youtube.com/watch?v=pwChk4S99i4 for a pretty good explanation

avianlyric 6y ago

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.

wcoenen 6y ago

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).

brandmeyer 6y ago
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_1996.png https://en.wikipedia.org/wiki/File:Earth_Gravitational_Model...
- wcoenen 6y ago
  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

kilburn 6y ago

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

PopeDotNinja 6y ago

Check out the Canvendish experiment! Watch gravity work in a classroom. https://youtu.be/MbucRPiL92Q https://youtu.be/MbucRPiL92Q

curiousgeorgio 6y ago

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.

dmurray 6y ago
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.
- curiousgeorgio 6y ago
  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).