Length of the day change if polar ice caps melt.

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SUMMARY

The discussion focuses on calculating the change in the length of a day if the polar ice caps melt, assuming the Earth is a sphere and the ice is uniformly distributed at the axis of rotation. The mass of the ice is given as 2.3 x 1019 kg, while the mass of the Earth is 6.4 x 1024 kg. The moment of inertia for both a solid sphere and a spherical shell is utilized in the calculations. The final result indicates a change in the length of the day of approximately 0.207 seconds after correcting an initial miscalculation.

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Jomaho
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Homework Statement


Question requires assumption that the Earth is spherical, all the ice is located at the axis of rotation.
Basically if all the ice was to melt uniformally over the surface of the earth, what would be the change in the length of the day?
Mass of ice: m= 2.3*10^19 kg
Mass of earth: M= 6.4*10^24 kg (doesn't say whether this takes into account the ice, I assumed it does)

Homework Equations


Moment of inertia of a solid sphere: I=2m(r^2)/5
Moment of inertia of a spherical shell I=2m(r^2)/3
m=mass
r=radius

The Attempt at a Solution


I went about it as follows:

Initially: I1=2M(r^2)/5
After: I2=2(M-m)(r^2)/5 + 2m(r^2)/3

I then used conservation of momentum:
L= I1w1 = I2w2

Finally using:
Time diff = 2*pi*((1/w2)-(1/w1))
Where 2*pi/w1 = 24hours

This eventually canceled down to:
Time diff = (4*pi*m)/(3*w1*M)

I got the answer to be 0.621s. Would someone mind checking my method and answer?
 
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Scratch that end answer, I've got 0.207s. Must not have divided by 3.
 

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