Calculate the angular momentum of the Earth: (a) Calculate the magnitude of the translational angular momentum of the Earth relative to the center of the sun. See the data on inside back cover. (b) calculate the magnitude of the rotational angular momentum of the Earth. How does this compare to your result in part (a)?

(The angular momentum of the Earth relative to the center of the sun is the sum of the translational and rotational angular momenta. The rotational axis of the Earth is tipped $23.5°$away from a perpendicular to the plane of its orbit.

(a) Period of rotation of Earth about the Sun, $T=1$year

$=1year\left(\frac{365day}{1year}\right)\left(\frac{24hr}{1day}\right)\left(\frac{3600s}{1hr}\right)$

$=3156000s$

The angular speed of the Earth rotation about the Sun is calculated as

$\omega =\frac{2\pi rad}{T}$

$$\begin{gathered} =\frac{2\pi rad}{31536000s} \\ =1.99\times {10}^{-7}rad/s \end{gathered}$$

We can treat the Earth as a solid sphere, so the moment of inertia of the Earth about its axis of rotation is

$$\begin{gathered} I_{rot}=\frac{2}{5}{M}_{Earth}{R}_{Earth}^2 \\ =\frac{2}{5}\left(6\times {10}^{24}kg\right){\left(6.4\times {10}^{6}kg\right)}^2 \\ =9.83\times {10}^{37}kg·m^2 \end{gathered}$$

The moment of inertia of the Earth relative the center of the Sun is

$$\begin{gathered} I=M_{Earth}{d}^2 \\ =\left(6\times {10}^{24}kg\right){\left(1.5\times {10}^{11}kg\right)}^2 \\ =1.35\times {10}^{47}kg·m^2 \end{gathered}$$

The rotational angular momentum of an object rotating with angular speed $\left(\omega \right)$in terms of its moment of inertia is

$L_{rot}=I\omega ......\left(1\right)$

The magnitude of the Earth's translational angular momentum in relation to the Sun's center is provided by the sum of the translational and rotational angular momenta.

$$\begin{gathered} L_{rot}=I_{rot}\omega +I\omega \\ =\left(I_{rot}+I\right)\omega \\ =\left(9.83\times {10}^{37}kg·m^2+1.35\times {10}^{47}kg.m^2\right)\left(1.99\times {10}^{-7}rad/s\right) \end{gathered}$$

$=2.686\times {10}^{40}kg·m^2/s$