Calculating Angular Momentum of the Earth

No information is given in the statement of the problem; so we must look up pertinent data before we can calculate $L=\mathrm{I\omega }$. First, according to Figure 10.12, the formula for the moment of inertia of a sphere is

so that

Earth’s mass $M$ is $5\text{.}\text{979}\times {\text{10}}^{\text{24}}\text{kg}$ and its radius $R$ is $6\text{.}\text{376}\times {\text{10}}^6\text{m}$. The Earth’s angular velocity $\omega$ is, of course, exactly one revolution per day, but we must covert $\omega$ to radians per second to do the calculation in SI units.

Substituting known information into the expression for $L$ and converting $\omega$ to radians per second gives

Substituting $\mathrm{2\pi }$ rad for $1$ rev and $8\text{.}\text{64}\times {\text{10}}^4\text{s}$ for 1 day gives

This number is large, demonstrating that Earth, as expected, has a tremendous angular momentum. The answer is approximate, because we have assumed a constant density for Earth in order to estimate its moment of inertia.