Relative to an origin at the center of the Earth, where is the center of mass of the Earth-Moon system? The mass of the Earth is $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{kg}$, the mass of the Moon is$\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{kg}$ , and the distance from the center of the Earth to the center of the Moon is$\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{m}$ . The radius of the Earth is 6400 kmOna can show that the Earth and Moon orbit each other around this center of mass.

• The mass of the Earth is$m_E$ =$6\times {10}^{24}\mathrm{kg}$
• The mass of the Moon is $m_M$ = $7\times {10}^{22}\mathrm{kg}$
• The distance from the center of the Earth to the center of the Moon is${\mathrm{r}}_{\mathrm{ME}}=4\times {10}^8\mathrm{m}$
• The radius of the Earth is 6400 km

The Earth’s radius is 6400 kilometers. So the center of the mass of Earth and Moon comes inside the planet that is 1700 kilometers below the surface.

The center of mass of the Earth-Moon system,

$\frac{{\mathbf{r}}_{\mathbf{com}}}{{\mathbf{R}}_{\mathbf{E}}}$

Where,

$r_{com}$is the center of mass of the Earth-Moon system

${\mathrm{R}}_{\mathrm{E}}$= Radius of Earth

And the center of mass of the Earth-Moon system is calculated by the formula,

${\mathbf{r}}_{\mathbf{com}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{M}}{\mathbf{r}}_{\mathbf{ME}}}{{\mathbf{m}}_{\mathbf{M}}\mathbf{+}{\mathbf{m}}_{\mathbf{E}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here,

${\mathrm{m}}_{\mathrm{M}}$= mass of Moon

${\mathrm{m}}_{\mathrm{E}}$= mass of Earth

${\mathrm{r}}_{\mathrm{ME}}$= the distance from the center of the Earth to the center of the Moon

$$\begin{gathered} {\mathrm{m}}_{\mathrm{M}}=7\times {10}^{22}\mathrm{kg} \\ {\mathrm{m}}_{\mathrm{E}}=6\times {10}^{24}\mathrm{kg} \\ {\mathrm{r}}_{\mathrm{ME}}=4\times {10}^8\mathrm{m} \end{gathered}$$

Substitute these values in Equation (2),

role="math" localid="1657793643155" $$\begin{gathered} {\mathrm{r}}_{\mathrm{com}}=\frac{{\mathrm{m}}_{\mathrm{M}}{\mathrm{r}}_{\mathrm{ME}}}{{\mathrm{m}}_{\mathrm{M}}+{\mathrm{m}}_{\mathrm{E}}} \\ =\frac{7\times {10}^{22}\times 4\times {10}^8}{\left(7\times {10}^{22}\right)+\left(6\times {10}^{24}\right)} \\ =\frac{2.8\times {10}^{31}}{6.07\times {10}^{24}} \\ =4.61\times {10}^6\mathrm{m} \end{gathered}$$

Hence, the center of mass of the Earth-Moon system is $\mathbf{4}\mathbf{.}\mathbf{61}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{m}$