Chapter 13, Problem 60PQ

To determine

The rotational inertia of the Earth-Moon system around its center of mass.

Answer to Problem 60PQ

The rotational inertia of the Earth-Moon system around its center of mass is $1.08\times {10}^{40}\text{kg}\cdot {\text{m}}^2$ .

Explanation of Solution

Assume that origin is at the Earth’s center.

Write the expression for the center of mass of a system.

  $x_{\text{CM}}=\frac{1}{M}{\displaystyle \sum mx}$                                                                                                         (I)

Here, $x_{\text{CM}}$ is the position of the center of mass of the system, $M$ is the total mass of the system, $m$ represents the individual masses and $x$ is the distance of $m$ from the origin.

The system consists of the Earth and the Moon.

Since the origin is taken at the center of the Earth, value of $x$ for the Earth will be zero.

Expand equation (I) for the Earth-Moon system.

  $\begin{array}{c}{x}_{\text{CM}}=\frac{M_{\oplus }\left(0\right)+M_{\text{Moon}}{x}_{\text{Moon}}}{M_{\oplus }+M_{\text{Moon}}}\\ =\frac{M_{\text{Moon}}{x}_{\text{Moon}}}{M_{\oplus }+M_{\text{Moon}}}\end{array}$

Here, $M_{\oplus }$ is the mass of the Earth, $M_{\text{Moon}}$ is the mass of the Moon and $x_{\text{Moon}}$ is the distance from the center of the Moon to the origin.

The mass of the Earth is $5.98\times {10}^{24}\text{ kg}$, the mass of the Moon is $7.36\times {10}^{22}\text{ kg}$, the distance from the center of the Moon to the center of the Earth is $3.84\times {10}^8\text{ m}$ and the radius of the Earth is $6.37\times {10}^6\text{ m}$ .

Substitute $7.36\times {10}^{22}\text{ kg}$ for $M_{\text{Moon}}$, $3.84\times {10}^8\text{ m}$ for $x_{\text{Moon}}$ and $5.98\times {10}^{24}\text{ kg}$ for $M_{\oplus }$ in the above equation to find the value of $x_{\text{CM}}$ .

  $\begin{array}{c}{x}_{\text{CM}}=\frac{\left(7.36\times {10}^{22}\text{ kg}\right)\left(3.84\times {10}^8\text{ m}\right)}{5.98\times {10}^{24}\text{ kg}+7.36\times {10}^{22}\text{ kg}}\\ =4.67\times {10}^6\text{ m}\end{array}$

The value of $x_{\text{CM}}$ is smaller than the radius of the Earth so that system’s center of mass is inside the Earth. The Moon is far from the center of mass, so that it can be modelled as a particle orbiting the system’s center of mass.

Write the equation for the rotational inertia of the Moon around the center of mass of the system.

  $I_{\text{Moon}}=M_{\text{Moon}}{r}^2$                                                                                                        (II)

Here, $I_{\text{Moon}}$ is the rotational inertia of the Moon around the center of mass of the system and $r$ is the perpendicular distance from the Moon to the center of mass of the system.

Write the equation for $r$ .

  $r=x_{\text{Moon}}-x_{\text{CM}}$

Put the above equation in equation (II).

  $I_{\text{Moon}}=M_{\text{Moon}}{\left(x_{\text{Moon}}-x_{\text{CM}}\right)}^2$                                                                                    (III)

The Earth can be modelled as a solid sphere rotating around a point somewhat offset from its own center of mass so that parallel axis theorem can be used to find the rotational inertia of the Earth.

  $I_{\oplus }=I_{\text{CM}}+M_{\oplus }{h}^2$                                                                                                    (IV)

Here, $I_{\oplus }$ is the rotational inertia of the Earth around the center of mass of the system, $I_{\text{CM}}$ is the rotational inertia of the Earth around its own center of mass and $h$ is the perpendicular distance between the center of mass of the system and the center of the Earth.

Write the equation for $I_{\text{CM}}$ .

  $I_{\text{CM}}=\frac{2}{5}{M}_{\oplus }{R}_{\oplus }^2$

Here, $R_{\oplus }$ is the radius of the Earth.

Since the center of the Earth is the origin, the perpendicular distance between the center of mass of the system and the center of the Earth will be equal to the position of the center of mass of the system.

  $h=x_{\text{CM}}$

Put the above two equations in equation (IV).

  $\begin{array}{c}{I}_{\oplus }=\frac{2}{5}{M}_{\oplus }{R}_{\oplus }^2+M_{\oplus }{x}_{\text{CM}}^2\\ =M_{\oplus }\left(\frac{2}{5}{R}_{\oplus }^2+x_{\text{CM}}^2\right)\end{array}$                                                                                            (V)

Write the equation for the rotational inertia of the Earth-Moon system around the center of mass of the system.

  $I=I_{\text{Moon}}+I_{\oplus }$                                                                                                           (VI)

Here, $I$ is the rotational inertia of the Earth-Moon system around the center of mass of the system.

Conclusion:

Substitute $7.36\times {10}^{22}\text{ kg}$ for $M_{\text{Moon}}$, $3.84\times {10}^8\text{ m}$ for $x_{\text{Moon}}$ and $4.66\times {10}^6\text{ m}$ for $x_{\text{CM}}$ in equation (III) to find $I_{\text{Moon}}$ .

  $\begin{array}{c}{I}_{\text{Moon}}=\left(7.36\times {10}^{22}\text{ kg}\right){\left(3.84\times {10}^8\text{ m}-4.66\times {10}^6\text{ m}\right)}^2\\ =1.06\times {10}^{40}\text{ kg}\cdot {\text{m}}^2\end{array}$

Substitute $5.98\times {10}^{24}\text{ kg}$ for $M_{\oplus }$, $6.37\times {10}^6\text{ m}$ for $R_{\oplus }$ and $4.66\times {10}^6\text{ m}$ for $x_{\text{CM}}$ in equation (V) to find $I_{\oplus }$ .

  $\begin{array}{c}{I}_{\oplus }=\left(5.98\times {10}^{24}\text{ kg}\right)\left(\frac{2}{5}{\left(6.37\times {10}^6\text{ m}\right)}^2+{\left(4.66\times {10}^6\text{ m}\right)}^2\right)\\ =2.27\times {10}^{38}\text{ kg}\cdot {\text{m}}^2\end{array}$

Substitute $1.06\times {10}^{40}\text{ kg}\cdot {\text{m}}^2$ for $I_{\text{Moon}}$ and $2.27\times {10}^{38}\text{ kg}\cdot {\text{m}}^2$ for $I_{\oplus }$ in equation (VI) to find $I$ .

  $\begin{array}{c}I=1.06\times {10}^{40}\text{ kg}\cdot {\text{m}}^2+2.27\times {10}^{38}\text{ kg}\cdot {\text{m}}^2\\ =1.08\times {10}^{40}\text{kg}\cdot {\text{m}}^2\end{array}$

Therefore, the rotational inertia of the Earth-Moon system around its center of mass is $1.08\times {10}^{40}\text{kg}\cdot {\text{m}}^2$ .