Steven Dutch, Natural and Applied Sciences, University
of Wisconsin - Green Bay

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- Mass of the earth 5.9736 E+24 kg
- Surface Area - Land: 148,000,000 sq. km
- Surface Area - Water: 362,000,000 sq. km
- Surface Area - Total: 510,000,000 sq. km
- Mean Radius 6371.01 km
- Volume of the Earth: 1.0832 E+12 km
^{3} - Mass of the Oceans: 1.3370 E+21 kg
- Volume of the Oceans: 1.377 E+09 km
^{3} - Mass of the Cryosphere (Ice) 2.6 E+19 kg
- Volume of the Cryosphere 29,000,000 km
^{3}

The mass of the world's ice is 1/200,000 of the total mass of the earth. If we take that mass from the pole and put it along the equator, we'd expect it to slow the earth's rotation by about 1/200,000. Since there are 60 x 60 x 24 = 86,400 seconds in a day, the slowdown would be 0.37 seconds.

This is simple and intuitive, but not very accurate. Nevertheless, it suggests that the effect of melting the ice caps wouldn't be very great. The reason it's not very accurate is that rotation depends greatly on how mass is distributed around the rotation axis.

Rotating objects possess a quantity called *moment of inertia* that
plays much the same role in rotational motion that mass does in linear motion.
For example, the linear momentum of a moving object is p = mv, where m is the
mass, v is velocity and p is linear momentum. For a rotating object, the formula
is J = Iω, where I is moment of inertia,
ω is angular velocity in radians per second (there
are 2 pi radians in 360 degrees) and J is angular momentum. Similarly, for an
object moving in a straight line, kinetic energy is given by K = 1/2 mv^{2}. For a rotating object, K = 1/2 I
ω^{2}.

Obviously I is going to depend on the mass of the object and how it's
spinning. It takes a lot less energy to get a rod spinning at 100 rpm around its
axis than perpendicular to it. It takes a lot less effort to get a playground
carousel spinning if the passengers are all in the center than if they're on the
outside circumference. For almost all objects, I = kmr^{2}, where m is the mass, r is the radius and k is some constant.
Some values of k we will find useful:

- For a uniform sphere, k = 2/5 = 0.4. This is a bit counterintuitive, since we'd expect a fundamental shape like a sphere to have some nice neat value like 1 or 1/2, but there it is.
- The earth's moment of inertia enters into formulas for precession, so we can determine the value of k for the real earth. Since the earth is denser near the center, k is smaller than 0.4. It's 0.33, or a hair less than 1/3.
- For a thin spherical shell, k = 2/3
- For a flat disk, k=1/2

We're going to want the moment of inertia of the earth and its rotational velocity. We can calculate them, but the figures are already tabulated.

- Equatorial moment of inertia: 8.0095 E+37 kg m
^{2} - Rotational velocity 7.27220 E-5 radians/sec

Since J = Iω, we have J = 5.624 E+33 kg m^{2}
radians/sec. Don't worry about the weird looking units.

Now, we melt the cryosphere and add it to the oceans. We add 26,000,000 cubic kilometers of water (because ice has a density of 0.92, remember) and add it to 362,000,000 sq. km of ocean. That comes out to 72 meters of water. It will actually be less, because as sea level rises, the oceans will cover larger areas. Also, any ice below sea level (a pretty large amount in Antarctica) won't contribute to sea level rise at all. Exact modeling of sea level change will have to include the isostatic depression of oceanic crust and the thermal expansion or contraction of sea water. It gets very hairy.

For our purposes, we won't care about the exact rise of sea level, for
reasons that will become clear. We care
that we're taking ice out of the polar regions and creating a thin global shell
of water. Since the only thing that matters to the earth's rotation is how far
the ice is from the pole, we can lump Greenland and Antarctica together and
approximate them as a disk of ice 2500 km in radius. So the moment of inertia of
this spinning disk of ice is 1/2 x (2,500,000 m)^{2} x 2.6 E+19 kg =
8.125 E+31 kg m^{2}. That's about a millionth of the total earth's
angular momentum.

We take that ice, melt it and create a thin spherical shell of water (the
gaps created by the continents don't affect the result very much). The shell has
a radius of 6,371,010 meters and a mass of 2.6 E+19 kg, so its moment of inertia
is 2/3 x (6,371,010 m)^{2} x 2.6 E+19 kg = 7.03 E+32 kg m^{2} or over eight times
that of the polar ice. Talk about bang for the buck. The increase comes from
taking all that mass and redistributing a lot of it at low latitudes. The net
increase in the earth's moment of inertia is 7.03 E+32 kg m^{2} gained - 8.125 E+31 kg
m^{2} lost or 6.22 E+32 kg m^{2} gained.

From here on, it's simple proportion. Since J = Iω,
and the earth's angular momentum stays constant as long as there is no outside
disturbance, any increase in I will be offset by the exact same decrease in
ω. The change in I, 6.22 E+32 kg m^{2},
compared to the total moment of inertia of the earth, 8.0095 E+37 kg m^{2},
amounts to an increase of 7.77 x 10^{-6}. So to conserve angular
momentum, we need to slow the earth down by the same amount. There are 60 x 60 x
24 = 86,400 seconds in a day, so a decrease of 7.77 x 10^{-6} amounts to
0.67 seconds or about 2/3 of a second. The day will become about 2/3 of a second
longer. We'll have more time to get stuff done. Like pile sandbags along the
coasts. Yay. The actual figure will probably be less because ice below sea level
doesn't contribute any effect - it just melts and is replaced by sea water.

If the ice caps melt, changes in the earth's rotation will be very far down on our list of concerns.

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*Created 12 March 2007; Last Update
02 June, 2010
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