91影视

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 $v=r蝇$.

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 $m_E=5.97脳{10}^{24}kg$, radius$R_E=6.38脳{10}^{6}m$and orbit radius $r=1.50脳{10}^{11}m$. The earth completes one rotation on its axis in $24\text{h}=86,400\text{s}$and one orbit in$1\text{year}=3.156脳{10}^7\text{s}$.

Use the formula for angular momentum,

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

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

Apply formula of angular momentum and solve as follows:

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

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