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

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SUMMARY

The discussion focuses on calculating the angular momentum of the Earth modeled as a uniform sphere. The Earth completes one rotation every 86400 seconds, leading to an angular velocity of ω = 7.27 × 10-5 rad/s. The initial calculation for angular momentum used the formula L = (2/3)MR2 with M = 5.97 × 1024 kg and R = 6.37 × 106 m, resulting in L = 1.17 × 1034 kg·m2/s. However, the moment of inertia used was incorrect, as the Earth is not a hollow sphere but a solid one.

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  • Understanding of angular momentum and its calculation
  • Familiarity with the concepts of rotational inertia
  • Basic knowledge of spherical geometry
  • Proficiency in using physical constants such as mass and radius of the Earth
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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: [itex]\omega=\frac{2\pi}{86400s}=7.27\cdot 10^{-5}[/itex]

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

Angular momentum being [itex]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[/itex] 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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