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[SOLVED] Change in earth's day length due to water displacement

 
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Nov12-07, 08:59 PM   #1
 

[SOLVED] Change in earth's day length due to water displacement


1. The problem statement, all variables and given/known data

About 10000km3 of water is held behind dams in reservoirs around the world. Most reservoirs are at mid-latitudes, whilst the bulk of the world's oceans are concentrated near the equator. By using conservation of angular momentum, estimate by how much the overall movement of water into reservoirs has changed the length of the day.
[The earth has mean density 5.5 x10^3 kgm-3, radius R = 6400km and radius of gyration
0.58R. The density of water is 1gcm-3, and, for the purposes of this question, the density
of sea water is not significantly different .

2. Relevant equations

i have found that radius of gyration, k has the relation k^2 = I/m


3. The attempt at a solution

i have drawn the earth and showing the reservoirs as bulges at 45 degrees from the equators just to get an idea of what is going on. I guess i need to calculate the change in moment of inertia from no reserviors to having reservoirs at these latitudes. Then seeing how this changes the radius of gyration? Any ideas?
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Nov12-07, 10:12 PM   #2
 
Okay im confident that ive calculated the different moments of inertia of the before reserviors scenario and after reservoirs. So how can i calculate the difference in rotation period between these two different mometns of inertia?
Nov13-07, 12:43 AM   #3
D H
 
Mentor
Use conservation of angular momentum: [itex]I_-\omega_- = I_+\omega_+[/itex], where the - and + subscripts denote the before reservoirs scenario and after reservoirs scenario. The rotation period is related to the angular velocity via [itex]T=2\pi/\omega[/itex].
Nov13-07, 09:16 AM   #4
 

[SOLVED] Change in earth's day length due to water displacement


Quote by D H View Post
Use conservation of angular momentum: [itex]I_-\omega_- = I_+\omega_+[/itex], where the - and + subscripts denote the before reservoirs scenario and after reservoirs scenario. The rotation period is related to the angular velocity via [itex]T=2\pi/\omega[/itex].
Brilliant, thanks for all your help guys. Out of interest, i calculated the change in the day length equals 2.07 x 10^-4 seconds shorter.
Nov13-07, 09:34 AM   #5
D H
 
Mentor
You're welcome. Thread marked as 'solved'.
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