(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sum. Is it reasonable to model it as a particle? (b)Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix E and the astronomical data in Appendix E.

According to the question, for a particle of mass m moving in a circular path at a distance r from the axis, $I=m{r}^2$and $v=r\omega$.

For a uniform sphere of mass M and radius R and an axis through its center, $I=\frac{2}{3}M{R}^2$. The earth has mass $m_E=5.97\times {10}^{24}kg$, radius$R_E=6.38\times {10}^{6}m$and orbit radius $r=1.50\times {10}^{11}m$. The earth completes one rotation on its axis in $24\text{h}=86,400\text{s}$and one orbit in$1\text{year}=3.156\times {10}^7\text{s}$.

Use the formula for angular momentum,

$$\begin{gathered} L_z=I{\omega }_z \\ =m{r}^2{\omega }_z \\ =\left(5.97\times {10}^{24}kg\right){\left(1.50\times {10}^{11}m\right)}^2\left(\frac{2\pi rad}{3.156\times {10}^7\text{s}}\right) \\ =2.67\times {10}^{40}{\text{kgm}}^{\text{2}}\text{/s} \end{gathered}$$

Thus, the angular momentum is $2.67\times {10}^{40}{\text{kgm}}^{\text{2}}\text{/s}$.

Apply formula of angular momentum and solve as follows:

$$\begin{gathered} L_z=I{\omega }_z \\ =\left(\frac{2}{3}M{R}^2\right){\omega }_z \\ =\frac{2}{3}\left(5.97\times {10}^{24}kg\right){\left(6.38\times {10}^{6}m\right)}^2\left(\frac{2\pi rad}{86,400\text{s}}\right) \\ =7.07\times {10}^{33}{\text{kgm}}^{\text{2}}\text{/s} \end{gathered}$$

Thus, the magnitude of the angular momentum of the earth rotate around axis through the north and south is$7.07\times {10}^{33}{\text{kgm}}^{\text{2}}\text{/s}$.