*A cute town in Greenland*

I just made a full-fledged calculation in Mathematica and I can confirm the basic statements by Eduardo. They're kind of funny. Here is the summary:

Indeed, if the average sea level rise induced by the melted ice of Greenland is 7 meters, the rise of the sea level at the antipodal point would be around 8.3 meters.

However, near the center of Greenland, the sea levels would actually drop. Imagining the Greenland ice (today) as a mass point, the sea level rise would be zero about 1,700 kilometers from the center of this ice.

Below 1,700 kilometers, the elevation change would be negative, and it would behave as -1/D for a very small distance D. The simple formula implies about 25 meters of sea level drop at the distance of a few hundreds of meters.

Of course, the point mass approximation can only be trusted if D is kind of greater than the actual radius of the ice - around 1000 kilometers. Later, I may make a more accurate calculation that takes the actual pancake-like shape of the ice into account.

However, it's still clear that Iceland, which is only 1,000 kilometers from the center of Greenland, would see its sea level decrease. The Western European beaches which are about 3,000-3,500 km from the center of Greenland would experience about 4 meters sea level rise - about 1/2 of what the people on the other side of the globe would experience.

**Calculation**

Here is a sketch of the basic steps you need to know in order to understand or reproduce my calculation in Mathematica.

First, you define various constants - Newton's constant, the radius of the Earth, the total mass of the Greenland ice (density times sea level rise of 7 meters times 2/3 of the Earth's surface), and others.

Second, you imagine that after the Greenland ice melts, the surface of the ocean will be spherical because it will be the surface where the gravitational potential from a spherically symmetric mass is equal to an appropriate constant.

The actual Earth is not quite spherical - but the additional non-uniformities from the rotation and gravity induced by mountains etc. may be additively added to the effects of the Greenland ice. In the first approximation, the effects may be linearized and added. The second-order effects are negligible.

Third, you must realize that before the ice melts, the Earth's gravitational field is not spherical. Instead, you may write the gravitational potential, as a function of position on Earth's surface and altitude, as the sum of the gravitational potential from the spherical Earth plus the gravitational potential from the body of Greenland ice.

The average elevation of the ice is 2,132 meters so the average height of its center-of-mass is one half of it. But this won't play too much role. One kilometer is negligible relatively to the radius of the Earth - and even relatively to the horizontal extent of the ice. What will matter is the distance of places on globe from the center of Greenland - and the elevation may change it at most by 1 kilometer.

**If I haven't made it clear yet, the new effect comes from the gravity induced by the ice itself.**

So how do we calculate this effect? Well, qualitatively, if you add the ice above Greenland, its gravity will be attracting the nearby liters of seawater more strongly than the distant ones. That's why in the presence of the Greenland ice, the sea level will actually be higher near Greenland.

This argument is most obvious "right below" the mass of the ice. Today, the seawater is being attracted in the direction "up".

How do you calculate the actual shape of the sea level? Well, you need to realize that the surface of the oceans sits at a constant value of the gravitational potential. This potential is equal to -"GM/R" where "G" is Newton's constant, "M" is the mass of the source of the gravitational field, and "R" is the distance.

Whenever you get a "wrong" gravitational potential on some point of the surface, you may compensate it to the right value simply by changing the elevation by "h": the gravitational potential changes by "gh".

Consequently, the only thing I need is to calculate the gravitational potential from the body of ice at various points of the globe. I parameterize the Earth in modified spherical coordinates that have "theta=0" in the center of Greenland - a shifted "North Pole".

The gravitational potential goes like "1/R" where "R" is the distance from the center of Greenland. You know the mass, the distance, Newton's constant, everything you need. If you divide this gravitational potential as a function of the spherical coordinate "theta" by "g", the gravitational acceleration 9.8 m/s^2, you get the required correction of the sea level to the elevation.

So we have found this correction to the elevation and subtract its average over the globe - so that the average of the new, corrected sea level correction over the globe equals zero.

Here, I am cheating a bit. The continents are not uniformly distributed while I assumed that they are. But it's not too big a mistake because a vast majority of the water will be shifted at distant enough places where the sea level rise will be around 7 meters, anyway. And the land-ocean distribution in this bulk is pretty uniform.

We have obtained the required sea-level correction (that averages to zero), as a function of the distance from the center of Greenland. We may simply add 7 meters to get

*The x-axis is the distance from the center of Greenland measured along the surface; the y-axis is the total sea level rise from the melted ice.*

You see that below 1,700 km from the center of Greenland, the sea level rise is negative. In the mass point approximation, the graph would continue as "-1/D" for very short distances. I believe that up to -10 meters could be realistic.

About 3,000 km from the center of Greenland, e.g. in London, the sea level rise would already be positive but only 1/2 of the average rise of 7 meters. At the opposite point of Greenland, the sea level rise would converge to something like 8.3 meters.

The counting may be applied to smaller sea level rise; the sea level rise at various places will be the same percentage of the average sea level rise.

Clearly, the melting ice is not a problem for the immediate vicinity of the island whose ice melts. In fact, some very close points could experience dropping sea level. For example, for the Europeans and the Americans, it's much more important what happens in the Antarctica. And be sure it won't melt anytime soon.

Thanks for the link to Climate Audit

**Source**

Sorry, I don't want to post the source as files because it is too simple and unrefined. So you may just copy-and-paste it if you need it:

gNewton = 6.67428*^-11 meter^3/kg/sec^2;

massEarth = 5.9742*^24 kg;

radiusEarth = 6378000 meter;

acceleration = gNewton*massEarth/radiusEarth^2;

(* Greenland ice is approximated by a point mass *)

massGreenlandIce = 510*^12 meter^2*2/3*7 meter*1000 kg/meter^3;

heightGreenlandIce =

2132 meter/2; (* average elevation in the middle, therefore 1/2 *)

(* Elevation won't play role for the vast majority of the surface *)

(* After Greenland ice melting, the Earth is sphere of radiusEarth *)

\

(* The average increase of sea level is 7 meters over the ocean which \

is 2/3 of the surface *)

(* The nontrivial task is to compute the deformation of the shape \

before the melting *)

(* theta goes from 0 - Greenland - to pi - the opposite point on \

Earth *)

distanceSeaGreenlandIce[

theta_] := (2*radiusEarth + 1*heightGreenlandIce)*Sin[theta/2];

potentialFromIce[theta_] := -gNewton*

massGreenlandIce/distanceSeaGreenlandIce[theta];

compensatingElevation[theta_] := potentialFromIce[theta]/acceleration;

Plot[compensatingElevation[theta]/meter, {theta, 0, Pi}]

averageCompElevation =

Integrate[compensatingElevation[theta]*Sin[theta], {theta, 0, Pi}]/

Integrate[Sin[theta], {theta, 0, Pi}]

(* the denominator is equal to two - written just for clarity *)

centeredElevation[theta_] :=

compensatingElevation[theta] - averageCompElevation;

totalElevation[theta_] := 7 meter + centeredElevation[theta];

(* the average of centeredElevation over surface equals zero *)

Plot[centeredElevation[theta]/meter, {theta, 0.1, Pi},

PlotRange -> Full]

(* The total rise of sea level in meters *)

Plot[totalElevation[theta]/meter, {theta, 0.1, Pi}, PlotRange -> Full]

(* the same thing as a function of the distance from Greenland along \

surface, trustworthy from 500 km to 2 times radiusEarth *)

Plot[totalElevation[thetaa/6378.]/meter, {thetaa, 1000, Pi*6378.},

PlotRange -> Full]

massEarth = 5.9742*^24 kg;

radiusEarth = 6378000 meter;

acceleration = gNewton*massEarth/radiusEarth^2;

(* Greenland ice is approximated by a point mass *)

massGreenlandIce = 510*^12 meter^2*2/3*7 meter*1000 kg/meter^3;

heightGreenlandIce =

2132 meter/2; (* average elevation in the middle, therefore 1/2 *)

(* Elevation won't play role for the vast majority of the surface *)

(* After Greenland ice melting, the Earth is sphere of radiusEarth *)

\

(* The average increase of sea level is 7 meters over the ocean which \

is 2/3 of the surface *)

(* The nontrivial task is to compute the deformation of the shape \

before the melting *)

(* theta goes from 0 - Greenland - to pi - the opposite point on \

Earth *)

distanceSeaGreenlandIce[

theta_] := (2*radiusEarth + 1*heightGreenlandIce)*Sin[theta/2];

potentialFromIce[theta_] := -gNewton*

massGreenlandIce/distanceSeaGreenlandIce[theta];

compensatingElevation[theta_] := potentialFromIce[theta]/acceleration;

Plot[compensatingElevation[theta]/meter, {theta, 0, Pi}]

averageCompElevation =

Integrate[compensatingElevation[theta]*Sin[theta], {theta, 0, Pi}]/

Integrate[Sin[theta], {theta, 0, Pi}]

(* the denominator is equal to two - written just for clarity *)

centeredElevation[theta_] :=

compensatingElevation[theta] - averageCompElevation;

totalElevation[theta_] := 7 meter + centeredElevation[theta];

(* the average of centeredElevation over surface equals zero *)

Plot[centeredElevation[theta]/meter, {theta, 0.1, Pi},

PlotRange -> Full]

(* The total rise of sea level in meters *)

Plot[totalElevation[theta]/meter, {theta, 0.1, Pi}, PlotRange -> Full]

(* the same thing as a function of the distance from Greenland along \

surface, trustworthy from 500 km to 2 times radiusEarth *)

Plot[totalElevation[thetaa/6378.]/meter, {thetaa, 1000, Pi*6378.},

PlotRange -> Full]

So, when the ice of Antarctica melts, the earth axis will flip/wobble due to the change in gravity?

ReplyDeleteCalculating the consequences of the gig (Greenland ice gravity) and aig (Antarctica ica gravity) should be even more fun.

Dear datarimlens, "flip" and "wobble" are probably too strong words. But of course, such melting would probably change the moment of inertia, which would change the length of the day - a tiny bit - and it would also move the axis of rotation (i.e. the position of the North and South pole) to a slightly different place on the globe/crust. I am afraid that the quantitative changes will be so unspectacular that you don't even want to calculate it unless you are responsible for inserting the leap seconds etc. Cheers, LM

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