At a particular moment in time the Moon is located $\langle 1.9\times10^8, 0, -1.9\times10^8 \rangle$ m in a coordinate system in which the origin is located at the center of the Earth.

Determine the gravitational force exerted by the Earth on the Moon.

• The relative position vector from the Earth to the Moon is $\langle 1.9\times10^8, 0, -1.9\times10^8 \rangle$ m
• Earth is origin of coordinate system $\langle 0,0,0 \rangle$ m
• G, the gravitational constant = $6.7\times10^{-11} Nm^2/kg^2$

• The mass of the Earth is not given but can be found online ($5.9\times10^{24} kg$)
• The mass of the Moon is not given but can be found online ($7.3\times10^{22} kg$)

• Assume no other forces acting on the moon.

• Gravitational Force: $\vec{F}_{grav} = -G\dfrac{{M}_E\,M_{m}}{|\vec{r}|^2}\hat{r}$

In order to find the gravitational force we must first calculate the distance between the moon and the earth

$$|\vec{r}_{M-E}| = \sqrt{(r_{M-E_x})^2+(r_{M-E_y})^2+(r_{M-E_z})^2}$$

$$= \sqrt{(1.9 \times 10^8 m)^2 + (0m)^2 + (-1.9 \times 10^8 m)^2}$$

$$= 2.7 x 10^8 m$$

We also must find the direction of this force. The direction of the force will be in the same direction as the radius vector. We can find the direction of a vector by computing the unit vector of $\vec{r}_{M-E}$

$$\hat{r}_{M-E} = \dfrac{\vec{r}_{M-E}}{|\vec{r}_{M-E}|}$$

$$= \dfrac{\langle 1.9 \times 10^8, 0, -1.9 \times 10^8 \rangle m}{2.7 \times 10^8 m}$$

$$= \langle 0.7,0,-0.7 \rangle$$

You now have everything needed to calculate the gravitational force exerted by the Earth on the Moon:

$$\vec{F}_{M-E} = \vec{F}_{grav\,on\,M\,by\,E}$$

$$= -G\dfrac{{m}_M{m}_E}{|\vec{r}_{M-E}|^2}\hat{r}_{M-E}$$

$$= (6.7 \times 10^{-11} Nm^2/kg^2)\dfrac{(7.3 \times 10^{22} kg)(5.9 \times 10^{24} kg)}{(2.7 \times 10^8 m)}\langle 0.7,0,-0.7 \rangle$$

$$= 1.0 \times 10^{29}N \langle 0.7,0,-0.7 \rangle$$

$$= \langle 7.0 \times 10^{28}, 0, -7.0 \times 10^{28} \rangle N$$