Respuesta :
Givens.
• Mass of the Moon = 7.35x10^22 kg.
,• The radius of the orbit = 3.84x10^8 m.
First, find the tangential velocity.
[tex]v=\frac{2\pi r}{T}=\frac{2\pi\cdot3.84\times10^8m}{2.36\times10^6s}=1022.35\cdot\frac{m}{s}[/tex]Observe that we use the period in seconds, not in days. The transformation we performed is
[tex]T=27.3\text{days}\cdot\frac{86400\sec}{1day}=2.36\times10^6\sec [/tex]Once we have the tangential velocity, we can find the angular momentum of the Moon around the Earth.
[tex]\begin{gathered} L=mvr=7.35\times10^{22}\operatorname{kg}\cdot1022.35\cdot\frac{m}{s}\cdot3.84\times10^8m \\ L=2.9\times10^{34}\operatorname{kg}\cdot\frac{m^2}{s} \end{gathered}[/tex]Therefore, the angular momentum around the Earth is 2.9x10^34.
The angular momentum of the moon can be found using the following formula.
[tex]L=I\cdot w[/tex]Where I is the moment of inertia of the moon and w is the angular velocity.
The moment of inertia of a spherical solid is
[tex]I=\frac{2}{5}mr^2[/tex]The mass of the moon is 7.35 × 10^22 kg and its radius is 1.74 × 10^6 m. So, the inertia is
[tex]\begin{gathered} I=\frac{2}{5}\cdot7.35\times10^{22}\operatorname{kg}\cdot(1.74\times10^6m)^2 \\ I=8.9\times10^{34}\operatorname{kg}\cdot m^2 \end{gathered}[/tex]Using its angular velocity of 2.7x10^-6 rad/s, we find its angular momentum.
[tex]\begin{gathered} L=2.7\times10^{-6}rad/s\cdot8.9\times10^{34}\operatorname{kg}\cdot m^2 \\ L=2.4\times10^{29}\operatorname{kg}\cdot\frac{m^2}{s^{}} \end{gathered}[/tex]The angular momentum of the moon onto itself is 2.4x10^29.
At last, divide both momentums to compare them.
[tex]\frac{L_{\text{earth}}}{L_{\text{moon}}}=\frac{2.9\times10^{34}\operatorname{kg}\cdot\frac{m^2}{s}}{2.4\times10^{29}\operatorname{kg}\cdot\frac{m^2}{s^{}}}=1.2\times10^5[/tex]Therefore, the angular momentum around the Earth is 1.2x10^5 times greater than the angular momentum of the Moon onto itself.
