Posted by **Qay** on Saturday, December 4, 2010 at 4:21pm.

Find the rotational kinetic energy and angular momentum about its axis of the earth due to its daily rotation. Let ME = 6 x 1024 kg, RE = 6.4 x 106 m and ùE = 1.157 x 10-5 rev/sec.

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**bark**, Saturday, December 4, 2010 at 4:45pm
angular momentum of earth = Ie*ùe

Putting ùe = 2ð/T = 2ð/(24*60*60) = 7.27*10^-5

If we assume average density of earth = 5500 kg/m³, and R = 6400 km,

Ie = 8ðñeR^5 / 15 (formula for moment of inertia (MOI) of sphere with density ñe and radius R)

Ie=9,9 * 10^22

angular momentum of earth = Ie*ùe

= 9,9 * 10^22* 7.27*10^-5= 7,2*10^18

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