Question
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.
Answers
bark
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
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