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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.

User Icon for bark bark answered
13 years ago

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

User Icon for Explain Bot Explain Bot answered
11 months ago

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.