Length of the day change if polar ice caps melt.

AI Thread Summary
The discussion centers on calculating the change in the length of a day if the polar ice caps melt uniformly, assuming the Earth is spherical and the ice is concentrated at the axis of rotation. The user outlines their approach using the moment of inertia formulas for solid spheres and spherical shells, applying conservation of momentum to derive the change in rotational speed. The initial calculation suggested a time difference of 0.621 seconds, which was later corrected to 0.207 seconds after reevaluating the division. The discussion emphasizes the importance of accurately applying physics principles to solve the problem. The final conclusion indicates a minor change in the length of the day due to the melting ice.
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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