# Thread: What is the magnitude of the angular momentum of the Earth due to rotation about its axis?

1. ## What is the magnitude of the angular momentum of the Earth due to rotation about its axis? var addthis_config = {"data_track_clickback":false};

Assume the Earth is a uniform solid sphere with radius of 6.32 x 10^6 m and mass of 5.98 x 10^24 kg. Find the magnitude of the angular momentum of the Earth due to rotation about its axis.

2. Moment of inertia of a sphere = (2/5)MR^2. Plug in the values of M and R and calculate.

Angular velocity of earth = (2*pi radians)/(number of seconds in a day). Plug in the values and calculate.

Angular momentum = (moment of inertia) x (angular velocity). Plug in the values and calculate.

3. L = Iw = kmr^2 w; where w = 2pi/24 hours is the angular velocity, r = R cos(lat) is the rotation radius of the sphere we call Earth, k = 2/5 is a shape density dependent value, and m = 5.98E24 kg is the mass of Earth.

We assume the angular momentum at latitude = 0 degrees, the equation; so that r = R = 6.32E6 meters. Change the angular velocity w to radian per second for unit consistency, and you have everything in L = kmr^2 w; so you can do the math.

4. angular momentum, L
moment of inertia, I
angular velocity = w

L = I w

now in the case of a spherical object the moment of inertia about the line passing through the centre of mass is I = (2/5) * M * R^2
where M = 5.98 * 10^24 kg, is the mass of the body and R = 6.32 * 10^6 m is the radius of the body.

we can work out the angular velocity of the earth since we know the earth take 24 hours to complete a full 2 pi radian rotation

so w = 2 * pi / (24 * 60 * 60 ) radian per second
So from the first equation

L =(( 2/5) * 5.98 * 10^24* (6.32 * 10^6 )^2 * 2 * pi / (24 * 60 * 60 ) )kg meter square per second

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