Determine the ratio of the earth's angular momentum about its axis...

The angular momentum of the earth about its own axis due to its spinning motion is given by

$L_s = I_s \omega_s = \dfrac{2}{5} M R^2 \left(\dfrac{2\pi}{T_s}\right)<br><br>where$M is the mass of the earth, $R$ is its radius, and $T_s$ is the time period of its rotation ($1 \text{ day}$).

The angular momentum of the earth about the sun due to its orbital motion is given by

$L_o = I_o \omega_o = M r^2 \left(\dfrac{2\pi}{T_o}\right)$

where $r$ is the radius of the earth's orbit and $T_o$ is the time period of its revolution ($365 \text{ days}$).

The ratio of the two angular momenta is

\dfrac{L_s}{L_o} = \dfrac{\dfrac{2}{5} M R^2 \left(\dfrac{2\pi}{T_s}\right)}{M r^2\left(\dfrac{2\pi}{T_o}\right)} = $\dfrac{2}{5} \left(\dfrac{R}{r}\right)^2 \left(\dfrac{T_o}{T_s}\right)<br><br>Substituting the given values,$R = 6400 \text{ km}$,$r = 1.5 \times 10^8 \text{ km}$,$T_o = 365 \text{ days}$, and$T_s = 1 \text{ day}$:<br><br>\dfrac{L_s}{L_o} = \dfrac{2}{5} \left(\dfrac{6400}{1.5 \times 10^8}\right)^2 \times \left(\dfrac{365}{1}\right)$

$\dfrac{L_s}{L_o} = \dfrac{2}{5} \times \left(\dfrac{6.4}{1.5} \times 10^{-5}\right)^2 \times 365$

$\dfrac{L_s}{L_o} = 146 \times \left(4.267 \times 10^{-5}\right)^2$

$\dfrac{L_s}{L_o} = 146 \times 18.2 \times 10^{-10} \approx 2.66 \times 10^{-7}$