Chapter 13, Problem 55PQ

To determine

The total angular momentum of Earth - Moon system.

Answer to Problem 55PQ

The total angular momentum of the system is $\underset{\_}{3.60\times {10}^{34}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}$.

Explanation of Solution

The rotating system considered here is the Earth – Moon system. The total angular momentum of the system is equal to the sum of angular momentum of Moon orbiting the earth, the Moon’s rotational angular momentum, and the Earth’s rotational angular momentum. All these are acting in the same direction.

Write the equation to find the orbital angular momentum of Moon.

  $L_o=r{M}_{m}v$                                                                                                          (I)

Here, $L_o$ is the orbital angular momentum of Moon, $r$ is the distance of Moon from Earth, $M_m$ is the mass of Moon, and $v$ is the orbital speed of Moon.

Write the equation to find the orbital speed of the Moon.

  $v=\frac{d}{T_m}$                                                                                                                (II)

Here, $d$ is the orbital distance travelled by Moon and, $T_m$ is the time period of rotation of moon.

The orbit of Moon around the earth is circular. Therefore the distance travelled by Moon is equal to the perimeter of the circular orbit which is $2\pi r$.

Rewrite equation (II).

  $v=\frac{2\pi r}{T_m}$                                                                                                          (III)

Substitute equation (III) in (I) to get $L_o$.

  $\begin{array}{c}{L}_o=M_{m}r\left(\frac{2\pi r}{T_m}\right)\\ =M_m\left(\frac{2\pi r^2}{T_m}\right)\end{array}$                                                                                             (IV)

Now the equation for Moon’s rotational angular momentum is to be found.

Write the equation to find Moon’s rotational angular momentum.

  $L_{\text{moon}}=I_m{\omega }_m$                                                                                                     (V)

Here, $I_{\text{moon}}$ is the rotational angular momentum of Moon, $I_m$ is the rotational inertia of Moon, and ${\omega }_m$ is the angular velocity.

Here Moon can be treated as a solid sphere to find the moment of inertia.

Write the equation to find the momentum of inertia of moon.

  $I_m=\frac{2}{5}{M}_m{R}_m{}^2$

Here, $I_m$ is the moment of inertia of moon, $M_m$ is the mass of moon, and $R_m$ is the radios of moon.

Substitute above equation in (V).

  $L_{\text{moon}}=\frac{2}{5}{M}_m{R}_m{}^2{\omega }_m$                                                                                         (VI)

Write the equation to find the angular velocity of moon.

  ${\omega }_m=\frac{2\pi }{T}$

Here, $T$ is the orbital period of moon.

Since the same side of moon always faces the earth, its orbital period is equal to the rotational period of moon. Thus replace $T$ by $T_m$ in above equation and substitute back to equation (VI).

  $\begin{array}{c}{L}_{\text{moon}}=\frac{2}{5}{M}_m{R}_m{}^2\left(\frac{2\pi }{T_m}\right)\\ =\frac{4\pi }{5}\frac{M_m{R}_m{}^2}{T_m}\end{array}$                                                                                 (VII)

Same procedure used to find the rotational angular momentum of Earth also.

Write the equation to find Earth’s rotational angular momentum.

  $L_{\text{earth}}=I_e{\omega }_e$                                                                                                (VIII)

Here, $I_{\text{earth}}$ is the rotational angular momentum of earth, $I_e$ is the rotational inertia of earth, and ${\omega }_e$ is the angular velocity.

Write the equation to find the momentum of inertia of Earth.

  $I_e=\frac{2}{5}{M}_e{R}_e{}^2$

Here, $M_e$ is the mass of earth and $R_e$ is the radius of earth.

Substitute above equation in (VIII).

  $L_{\text{earth}}=\frac{2}{5}{M}_e{R}_e{}^2{\omega }_e$                                                                                         (IX)

Write the equation to find the angular velocity of earth.

  ${\omega }_e=\frac{2\pi }{T_e}$

Here, $T_e$ is the orbital period of earth.

Substitute above equation in equation (IX).

  $\begin{array}{c}{L}_{\text{earth}}=\frac{2}{5}{M}_e{R}_e{}^2\left(\frac{2\pi }{T_e}\right)\\ =\frac{4\pi }{5}\frac{M_e{R}_e{}^2}{T_e}\end{array}$                                                                                 (X)

The total angular momentum of the system is equal to the sum of angular momentum of Moon orbiting the earth, the Moon’s rotational angular momentum, and the Earth’s rotational angular momentum.

Write the equation to find the total angular momentum of the system.

  $L=L_o+L_{\text{moon}}+L_{\text{earth}}$                                                                                       (XI)

Here, $L$ is the total angular momentum of the system.

Conclusion:

Substitute $7.35\times {10}^{22}\text{kg}$ for $M_m$, $3.84\times {10}^8\text{m}$ for $r$, and $2.36\times {10}^6\text{s}$ for $T_m$ in equation (IV) to get $L_o$.

  $\begin{array}{c}{L}_o=\left(7.35\times {10}^{22}\text{kg}\right)\left(\frac{2\pi {\left(3.84\times {10}^8\text{m}\right)}^2}{2.36\times {10}^6\text{s}}\right)\\ =2.89\times {10}^{34}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\end{array}$

Substitute $7.35\times {10}^{22}\text{kg}$ for $M_m$, $1.738\times {10}^6\text{m}$ for $R_m$, and $2.36\times {10}^6\text{s}$ for $T_m$ in equation (VII) to get $L_{\text{moon}}$.

  $\begin{array}{c}{L}_{\text{moon}}=\frac{4\pi }{5}\frac{\left(7.35\times {10}^{22}\text{kg}\right){\left(1.738\times {10}^6\text{m}\right)}^2}{\left(2.36\times {10}^6\text{s}\right)}\\ =2.36\times {10}^{29}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\end{array}$

Substitute $5.98\times {10}^{24}\text{kg}$ for $M_e$, $6.3871\times {10}^6\text{m}$ for $R_e$, and $86400\text{s}$ for $T_e$ in equation (X) to get $L_e$.

  $\begin{array}{c}{L}_{\text{earth}}=\frac{4\pi }{5}\frac{\left(5.98\times {10}^{24}\text{kg}\right){\left(6.3871\times {10}^6\text{m}\right)}^2}{86400\text{s}}\\ =7.10\times {10}^{33}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\end{array}$

Substitute $2.89\times {10}^{34}{\text{kgm}}^{\text{2}}/\text{s}$ for $L_o$, $2.36\times {10}^{29}{\text{kgm}}^{\text{2}}/\text{s}$ for $L_{\text{moon}}$, and $7.10\times {10}^{33}{\text{kgm}}^{\text{2}}/\text{s}$ for $L_{\text{earth}}$ in equation (XI) to get $L$.

  $\begin{array}{c}L=\left(2.89\times {10}^{34}+2.36\times {10}^{29}+7.10\times {10}^{33}\right){\text{kgm}}^{\text{2}}/\text{s}\\ =3.60\times {10}^{34}{\text{kgm}}^{\text{2}}/\text{s}\end{array}$

Therefore, the total angular momentum of the system is $\underset{\_}{3.60\times {10}^{34}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}$.