How Is the Angular Momentum of Earth Calculated?

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SUMMARY

The angular momentum of the Earth about its rotation axis is calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. The moment of inertia for a uniform sphere is given by I = (2/5)Mr², with M being the mass of the Earth (6.0 x 1024 kg) and r being the radius (6.4 x 106 m). The angular velocity ω is determined using the Earth's rotation period, yielding a value of approximately 1.992 x 10-7 rad/sec. The calculated angular momentum was found to be 1.958 x 1031 kg·m²/s, which is significantly lower than the expected value of 7.1 x 1033 kg·m²/s, indicating an error in the calculation process.

PREREQUISITES
  • Understanding of angular momentum and its formula (L = Iω)
  • Knowledge of moment of inertia for a uniform sphere (I = (2/5)Mr²)
  • Familiarity with angular velocity and its calculation (ω = 2π/T)
  • Basic physics concepts related to rotational motion
NEXT STEPS
  • Review the calculation of moment of inertia for different shapes
  • Learn about the Earth's rotation period and its impact on angular velocity
  • Explore the implications of angular momentum in astrophysics
  • Investigate common errors in angular momentum calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators seeking to clarify concepts related to angular momentum and its applications.

sammythesp3rmy
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Homework Statement


Determine the angular momentum of the Earth about its rotation axis (assume the Earth is a uniform sphere).


Homework Equations


L = Iω
I = (2/5)Mr^2
ω = 2[itex]\pi[/itex]/T


The Attempt at a Solution


I = (2/5)(6.0 X 10^24 kg)(6.4 X 10^6m)^2
ω = 1.992 X 10^-7 rad/sec

L = (2/5)(6.0 X 10^24 kg)(6.4 X 10^6m)^2 * (1.992 X 10^-7 rad/sec)
What I get from this is 1.958 X 10^31 and the answer is supposed to be 7.1 X 10^33. where did I go wrong? Thank you
 
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