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The answers to this question say that the sea level will rise 66m if all the polar ice, etc, melts.

How long will this take?

Transporting incredible amounts of heat energy to the poles and injecting it into the ice, a good insulator, so that it melts must be very, very slow. The Netherlands, London, and all the rest will be long gone for other reasons, I imagine, long before it could be completed.

I am looking for an order of magnitude minimum under which it is provably impossible for the ice to all melt any faster. Wild and crazy assumptions are welcome, so long as they err on the side of making the ice melt faster than it would in reality - that way the estimate will always be a minimum.

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    $\begingroup$ I hope everyone has seen this picture: grist.files.wordpress.com/2009/11/humble-oil.jpg $\endgroup$
    – tobias47n9e
    Commented May 1, 2014 at 15:43
  • $\begingroup$ Need to take into account albedo; and for anything other than an extremely rough-and-ready calculation, need to take into account contact with seawater, seawater temperatures, etc. $\endgroup$
    – kaberett
    Commented May 1, 2014 at 15:49
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    $\begingroup$ Do you mean you're looking for a realistic lower boundary on time estimates for everything to melt? Also, what does it mean to melt all polar ice? No glaciers left in the mountains, or "just" the ice sheets? I think this could be a great question if you specify a little better exactly what kind of answer you're looking for. $\endgroup$
    – gerrit ♦
    Commented May 1, 2014 at 17:20
  • $\begingroup$ I think it is correct to assume that, whatever happens, the antartic ice sheet will be the last to go. So, therefore, I do not care how ling it takes to melth the glaciers in the himalyas, etc. The lower boundary does not have to be realistic, just a provable minimum. $\endgroup$
    – ravenspoint
    Commented May 1, 2014 at 17:22
  • $\begingroup$ Note for anyone who saw my earlier estimate of 63 years. I mistook kJ for J, so had three orders of magnitude error. Please double check for other mistakes. $\endgroup$
    – ravenspoint
    Commented May 1, 2014 at 17:42

3 Answers 3

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The question is requesting an answer that has no practical application. So rather than improving on some hypothetical calculation, I will describe the problem and hopefully make the difficulty providing a real answer to a time frame clear.

First: "polar ice" Antarctica is located in a polar position and consists of a multi-km thick ice sheet. The Arctic is covered by sea ice except for smaller glaciers and Greenland, which is not located in a polar position equal to Antarctica. One has to remember that Antarctica is circulation-wise isolated from the rest of the planet by circumpolar circulation, both in th esea and atmosphere, a situation not met in the Arctic. Hence the term polar ice is very fuzzy and should be avoided. It is better to speak about the objects directly.

Second: an ice sheet such as Greenland and Antarctica is not a stagnant sheet of ice, it has its own dynamics and which can change over time depending on forcing and change in forcing.. Greenland is largely land based but with outlet glaciers that terminate in the sea. Although mass loss through these have seen an increase in loss, much of the Greenland mass gain and loss is through surface mass balance (snow accumulation minus ablation, melt). Antarctica is a different beast since East Antarctica is largely land based while West Antarctica is to a large extent sitting on ground below (in places several km) below sea level. Te latter situation is more unstable since the ice could disintegrate by calving (loss through ice berg formation), a process which is potentially much quicker than surface melt (of which there is currently very little in Antarctica). So one part of Antarctica is much more unstable than the other.

Now, several other processes are at play. Ocean currents are sen to produce massive amounts of submarine melt on both Greenland outlet glaciers as well as on the sea-terminating edges of Antarctica. In places, circulation models estimate up to equal loses by calving an submarine melt. This occurs at the same time as there is hardly any surface melt to speak of. So calculating mass loss by surface melt is clearly a lost cause when calving and submarine melt dominate.

Finally, in this still simplified picture, the dynamic response of the ice sheets to changes in forcing is not well understood. This means we do not know how unstable East Antarctica really is. That Antarctica will totally disappear seems very unlikely any time soon and time frames of 10^3-10^4 years is a minimum. Greenland, not being a polar ice sheet is more sensitive in that as it loses mass it is also on average lower in altitude, experiencing more melt. There appears to be a point of no return for Greenland, under which elevation the ice sheet would not be able to recuperate unless the climate cooled significantly. Again the time to melt off the ice is long.

despite the seemingly slow increase in sea level, much infra structure in the coastal environments is threatened by only dm to m changes in ea level so the 66.1 m quoted from IPCC is not something that will happen in a life time but that does not mean the rates are not already problematic.

As a PS, I will add some relevant references once I am back home in case someone misses them at the moment.

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  • $\begingroup$ The practical application of the answer I am asking for is, of course, propaganda. $\endgroup$
    – ravenspoint
    Commented May 1, 2014 at 18:52
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Don't rest so easy, ravenspoint.

The Laurentide ice sheet at its maximum extent was larger than the Antarctic ice sheet is now. The bulk of that ice sheet melted in two pulses of 2000 years each, separated by the ~1000 year long Younger Dryas. During the first pulse, the Laurentide lost 5400 km3 of ice pear year. During the second pulse, it lost ice at an even faster clip, 5600 km3 of ice pear year.

Source: Dyke, A. S. (2004), "An outline of North American Deglaciation with emphasis on central and northern Canada" in Quaternary Glaciations- Extent and Chronology, Part II, p. 373-424, J. Ehlers and P. L. Gibbard, eds, Elsevier.

You want a SWAG, so I'll use 5000 km3 per year. That's just 5000 years for the Antarctic ice sheets to melt. The Laurentide ice sheet was able to melt quickly because a good chunk of it was outside of the polar circle. That's not the case for the Antarctic ice sheet, so this 5000 years most likely is an overly aggressive estimate.

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  • $\begingroup$ How thick was the Laurentide ice sheet? $\endgroup$
    – ravenspoint
    Commented May 1, 2014 at 18:29
  • $\begingroup$ It was over three kilometers thick in places. $\endgroup$
    – David Hammen
    Commented May 1, 2014 at 18:33
  • $\begingroup$ I understand that Antartic ice is 2km thick for most of its extent. $\endgroup$
    – ravenspoint
    Commented May 1, 2014 at 18:36
  • $\begingroup$ To pick a very minor point about the last paragraph: Most/all of the Antarctic ice sheet is south of the polar circles. (I understand what you meant of course.) $\endgroup$
    – Torbjørn T.
    Commented May 1, 2014 at 19:02
  • $\begingroup$ @TorbjørnT. - I changed "south of the polar circle" to "outside of the polar circle". Hope that helps, and thanks for the critique. $\endgroup$
    – David Hammen
    Commented May 1, 2014 at 19:03
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We need some numbers.

Antarctic ice is about $2\,\mathrm{km}$ thick.

enter image description here

Let's replace this with a $1\,\mathrm{km}$ blanket of ice equivalent insulator. In this case, if the mean air temperature ( currently -57C ) rises to $10\,^{\circ}\mathrm{C}$, then energy arrives at the bottom of the blanket ( where I assume the temperature has risen to 0 celsius and the pressure has fallen to 1 atmosphere ) at the rate of ... $0.02\,\mathrm{W/m^2}$.

Antarctica is $14 \times 10 ^{6} \,\mathrm{km}^2$ so we are able to deliver energy at $3\times 10 ^7\,\mathrm{W}$

$$\frac{6 \times 10 ^ {24}}{3 \times 10 ^ 7} = 2 \times 10^{17}\,\mathrm{s} \approx 7 \times 10^{9} \mathrm{years}$$

I think we can all relax!

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    $\begingroup$ This is completely wrong. Let's start with your mass. 26.5 million kilometers at 0.9167 grams per cubic centimeter is 2.4*10^19 kg, not 2.4*10^14. You are off by a factor of one hundred thousand. Multiplying by 333 kJ/kg (heat of fusion of water) yields 8*10^24 joules, not 6*10^19. Using that new value in lieu of the old in your final calculation yields a value of 8.5 billion years. That's obviously wrong. The Laurentide ice sheet was bigger than the Antarctica ice sheet is now, and most of that larger ice sheet melted in 5000 years. $\endgroup$
    – David Hammen
    Commented May 1, 2014 at 18:00
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    $\begingroup$ You're still off by a factor of ten. (6e24 joules) / (3e7 watts) is 6.3 billion years. WolframAlpha calculation. Think about it for a second or two: Does this answer make any sense? Keep in mind how quickly the ice sheets melted at the start of the Holocene, and also keep in mind that Earth has gone through periods where almost the entire planet was frozen. $\endgroup$
    – David Hammen
    Commented May 1, 2014 at 18:41
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    $\begingroup$ Does your new result of 7 billion years make one bit of sense? That's a rhetorical question. The answer is no. $\endgroup$
    – David Hammen
    Commented May 1, 2014 at 18:55
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    $\begingroup$ @ravenspoint - Think about it like this: For reductionism to work you need to simplify your system enough so it can be described by arithmetic formulas. But Earth Science problems are very complicated systems in which you have to consider feedback loops and extreme non-linearity. Even adding more variables to the equation won't help because you either have more unknowns or have to assume things. $\endgroup$
    – tobias47n9e
    Commented May 1, 2014 at 19:34
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    $\begingroup$ @ravenspoint - if it was a unique, unprecedented situation, then starting from assumptions/first principles would be fine. However, we have some historical events to compare with - if some assumptions prove that timeframe of X isn't possible, but we have observed X, then it proves that those particular assumptions are somehow wrong. $\endgroup$
    – Peteris
    Commented May 1, 2014 at 20:30

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