Calculating Earth's Angular Momentum: A Model of a Uniform Sphere

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The discussion focuses on calculating the angular momentum of the Earth modeled as a uniform sphere. The Earth’s rotation period is established as 86400 seconds, leading to an angular velocity of approximately 7.27 x 10^-5 radians per second. The initial calculation for rotational inertia uses the formula I = 2/3MR^2, yielding an angular momentum of 1.17 x 10^34 kg·m^2/s. However, this result is identified as incorrect due to the use of the wrong moment of inertia for a solid sphere. The conversation concludes with a clarification that the Earth is not hollow, implying the need for a different moment of inertia calculation.
Bashyboy
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The problem is:

"Model the Earth as a uniform sphere. Calculate the angular momentum of the Earth due to its spinning motion about its axis."

So, I first pointed out the the Earth rotates about its axis in 24 hrs x 3600 s / 1 hrs = 86400 s

Then I noted that a fixed point on the Earth rotates through a angle of 2pi, tracing out the circumference of the earth, in 86400s: \omega=\frac{2\pi}{86400s}=7.27\cdot 10^{-5}

The rotational inertia of the sphere being I=2/3MR^2

Angular momentum being L=(2/3)(5.97\cdot 10^{24})(6.37\cdot 10^6)^2(7.27\cdot 10^{-5}=1.17\cdot 10^{34} kg\cdot m^2/s However, this answer is wrong. What did do wrong?
 
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You've used the wrong moment of inertia. (Hint: is the Earth solid or hollow?)
 
Hmm, it certainly isn't hollow. Thank you very much for the help.
 
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