Determine the angular momentum of the Earth (a) about its rotation axis (assume the Earth is a uniform sphere), and (b) in its orbit around the Sun (treat the Earth as a particle orbiting the Sun).

The rotational inertia of an object is called its moment of inertia. A rotating object consists of many particles located at different distances from the axis of the rotation.

The sum of the mass of each particle (m) multiplied by the square of their distances (r) from the axis of rotation gives the moment of inertia (I) of that object, i.e.,

\(I = \sum {m{r^2}} \)

In this problem,the moment of inertia of the Earth rotating about its own axis is the same as that of a uniform solid sphere of mass m and radius R rotating about its own axis, i.e.,\(I = \frac{2}{5}m{R^2}\).The moment of inertia of the Earth rotating about the Sun at a distance \(R'\)from the center of the Sun, if it is treated as a particle, is:\(I' = m{R'^2}\)

The mass of the Earth is\(m = 5.98 \times {10^{24}}\;{\rm{kg}}\).

The radius of the Earth is\(R = 6.38 \times {10^6}\;{\rm{m}}\).

The distance between the center of the Earth and the center of the Sun is\(R' = 1.496 \times {10^{11}}\;{\rm{m}}\).

Since the Earth completes one rotation about its axis in one day, the angular velocity of the Earth about its axis is:

\(\begin{aligned}{c}\omega = 2\pi \;{\rm{rad/day}}\\ = \left( {\frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{day}}}}} \right) \times \left( {\frac{{1\;{\rm{day}}}}{{24\;{\rm{h}}}}} \right) \times \left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)\\ = \frac{\pi }{{43,200}}{\rm{rad/s}}\end{aligned}\)

Since the Earth completes one revolution about the Sun in one year, the angular velocity of the Earth about the Sun is:

\(\begin{aligned}{c}\omega ' = 2\pi \;{\rm{rad/y}}\\ = \left( {\frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{y}}}}} \right) \times \left( {\frac{{1\;{\rm{y}}}}{{365\;{\rm{d}}}}} \right) \times \left( {\frac{{1\;{\rm{d}}}}{{24\;{\rm{h}}}}} \right) \times \left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)\\ = \frac{\pi }{{15,768,000}}{\rm{rad/s}}\end{aligned}\)

Consider the Earth to be a uniform solid sphere rotating about its own axis withangular velocity\(\omega \). Thus, the angular momentum of the Earth will be:

\(\begin{aligned}{c}L = I\omega \\ = \left( {\frac{2}{5}m{R^2}} \right)\omega \\ = \frac{2}{5} \times {\left( {5.98 \times {{10}^{24}}\;{\rm{kg}}} \right)^2} \times {\left( {6.38 \times {{10}^6}\;{\rm{m}}} \right)^2} \times \frac{\pi }{{43,200}}{\rm{rad/s}}\\ = 7.08 \times {10^{33}}{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

So, the angular momentum of the Earth about its rotation axis is \(7.08 \times {10^{33}}{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\).

Consider the Earth to be a particle that revolves around the Sunwith angular velocity\(\omega '\). Thus, the angular momentum of the Earth will be:

\(\begin{aligned}{c}L = I'\omega '\\ = \left( {m{{R'}^2}} \right)\omega '\\ = {\left( {5.98 \times {{10}^{24}}\;{\rm{kg}}} \right)^2} \times {\left( {1.496 \times {{10}^{11}}\;{\rm{m}}} \right)^2} \times \frac{\pi }{{15,768,000}}{\rm{rad/s}}\\ = 2.67 \times {10^{40}}{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

So, the angular momentum of the Earth in its orbit around the Sun is \(2.67 \times {10^{40}}{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\).