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August 27, 2014

August 27, 2014

Posted by **Jordan** on Friday, April 9, 2010 at 12:07am.

- physics -
**Count Iblis**, Friday, April 9, 2010 at 1:23pmMake an approximating assumption by modeling the deposited water as an effective extra uniform surface density of negligible thickness added to the surface. The surface densisty is then:

sigma = M_ice/(4 pi R^2)

where M_ice = 2.76E+19 kg and R is the radius of the Earth.

You can then evaluate the contribution to the moment of intertia from such surface density.

The surface element on the sphere in spherical coordinates is given by:

R^2 sin(theta) dtheta dphi

The mass element is thus:

dm = sigma R^2 sin(theta) dtheta dphi

The distance to the rotation axis is R sin(theta), so we have:

I_ice = sigma R^4 sin^3(theta) dtheta dphi

If you evaluate this, you find:

I_ice = 2/3 M_ice R^2

The total moment of inertia of the Earth is:

I = 2/5 M R^2

where M is the mass of the Earth

(in case of a uniform volume density, you need to integrate over the radius, which would have yielded an 1/5 R^5 term compared to the above derivation and we would have divided by 4/3 pi R^3 instead of 4 pi R^2, i.e. we would have ended up with an overall extra factor of 3/5 and 3/5 * 2/3 = 2/5)

In the mass redistrubution process the total angular momentum is conserved, so you have:

(I + I_ice) omega_new = I omega_old

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