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.
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
To find the rotational kinetic energy of the Earth due to its daily rotation, we need to use the formula:
Rotational Kinetic Energy (KE) = (1/2) * Moment of Inertia * Angular Speed^2
The moment of inertia for a solid sphere rotating about its axis is given by:
Moment of Inertia (I) = (2/5) * Mass * Radius^2
Substituting the given values, we have:
Mass of Earth (ME) = 6 x 10^24 kg
Radius of Earth (RE) = 6.4 x 10^6 m
Angular Speed of Earth (ω) = 1.157 x 10^-5 rev/sec
First, let's calculate the moment of inertia:
I = (2/5) * ME * RE^2
I = (2/5) * (6 x 10^24 kg) * (6.4 x 10^6 m)^2
I = 1.709 * 10^38 kg * m^2
Now, we can calculate the rotational kinetic energy:
KE = (1/2) * I * ω^2
KE = (1/2) * (1.709 * 10^38 kg * m^2) * (1.157 x 10^-5 rev/sec)^2
KE = 1.709 * 10^38 kg * m^2 * (1.335 x 10^-10 rev^2/sec^2)
KE = 2.280 * 10^28 Joules
Therefore, the rotational kinetic energy of the Earth due to its daily rotation is approximately 2.280 x 10^28 Joules.
To find the angular momentum about the Earth's axis due to its daily rotation, we use the formula:
Angular Momentum (L) = Moment of Inertia * Angular Speed
Substituting the values, we have:
L = I * ω
L = (1.709 * 10^38 kg * m^2) * (1.157 x 10^-5 rev/sec)
L = 1.975 * 10^33 kg * m^2/sec
Therefore, the angular momentum about the Earth's axis due to its daily rotation is approximately 1.975 x 10^33 kg * m^2/sec.