MadSci Network: Earth Sciences

Re: Will earth rotation be affected by glaciers melting?

Date: Wed Sep 15 13:56:55 1999
Posted By: Jason Goodman, Graduate Student, Massachusetts Institute of Technology
Area of science: Earth Sciences
ID: 936097893.Es

You're absolutely right that melting ice sheets will cause a movement of mass from the equator to the poles. This will, by conservation of angular momentum, cause the Earth's rotation to slow down.

In fact, we can estimate the magnitude of the effect mathematically. Let's consider the most extreme case, in which the entire Antarctic ice sheet (which makes up the vast majority of the ice on the planet) were to melt. The mass of this ice sheet is about 2e19 kg (I'm using engineering notation: 2e19 means 2 * 10^19). Its current distance, on average, from the Earth's axis of rotation is very roughly 1500 km. Thus, its moment of inertia is roughly I_ice=M*L^2 = 4.5e31 kg-m^2. (You could get a better estimate by integrating rho*l^2 over the volume of the ice cap, but this is close enough.)

If this ice melts, this mass will spread out evenly over the surface of the Earth's oceans. While you could do another integral to calculate the new moment of inertia precisely, a rough estimate will do: on average, since the Earth is 6300 km in radius, the mass is now about 5000 km from the Earth's axis of rotation, so the new moment of inertia is I_water=5e32 kg-m^2.

The moment of inertia of the Earth right now is I0 = 8.1e37 kg-m^2. If the ice caps are melted, the new moment will be I_new = I0 - I_ice + I_water. The law of conservation of angular momentum states that

I0*f0 = I_new*f_new
where omega0 is the current rotation rate of the Earth (1/86400 secs), and f_new is the rotation rate after we melt the ice. We define df = f_new - f0 and dI = I_water-I_ice = 4.5e32, and solve for df to get
I0*f0 = (I0 + dI)*(f0+df)
0 = dI*f0 + I0*df + dI*df
df = - f0 * dI/(dI+I0)
The fractional change in Earth's rotation rate, df/F is the most useful number. Note also that since dI is much smaller than I0, dI + I0 is nearly equal to I0. Making that approximation, df/f0 = dI/I0 From the description above, dI/I0 is 5.5e-6, or .00055%. So the length of a day will change by .00055%, or about .5 seconds. But any day length change will be the least of our problems if all of Antarctica melts: this would cause a rise of global sea level of 55 meters!

Current models predict a sea level change of maybe 20 centimeters over the next century: 1/300th of the change given above. Thus, the redistribution of mass is 1/300th as much as for a fully-melted ice-cap, with a proportional change in dI and thus df. This would lead to a change of length-of-day of perhaps 1.4 milliseconds.

You may be interested to learn that the length of the day can be measured to small fractions of a millisecond by carefully measuring the radio signals from quasars and pulsars, whose position is known very precisely. Scientists doing this experiment do notice changes in the length of day due to mass distributions on earth: in particular, atmospheric storm systems redistribute air around the globe, and cause pressure variations which exchange angular momentum between atmosphere and Earth. Air motions associated with the seasonal cycle cause the day to be about 1 millisecond shorter during the summertime than in the winter. On longer time-scales, changes in ocean currents also cause length-of-day variations.

For more information on how the length of day relates to weather and climate (though ice sheets are unfortunately not covered), see Chapter 11 of "Physics of Climate", by Peixoto and Oort.

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