Question

1. the moon and earth rotate about their common center of mass, which is located about 4700 km from the center of earth. (this is 1690 km below the surface.) a) calculate the magnitude of the acceleration due to the moon’s gravity at that point. b) calculate the magnitude of the centripetal acceleration of the center of earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). comment on whether or not they are equal and why they should or should not be.

Explanation:

Step1: Recall gravitational - force formula

The gravitational acceleration $g$ due to a mass $M$ at a distance $r$ is given by $g = G\frac{M}{r^{2}}$, where $G=6.67\times 10^{- 11}\ m^{3}\ kg^{-1}\ s^{-2}$, the mass of the Moon $M_{m}=7.348\times 10^{22}\ kg$, and the distance $r$ from the Moon to the common - center of mass. The distance from the Earth's center to the common - center of mass is $r_{1}=4700\ km = 4.7\times 10^{6}\ m$, and the average distance from the Earth to the Moon is $d = 3.844\times 10^{8}\ m$. So the distance from the Moon to the common - center of mass is $r=d - r_{1}=3.844\times 10^{8}-4.7\times 10^{6}=3.797\times 10^{8}\ m$.

Step2: Calculate the acceleration due to the Moon's gravity

Substitute the values into the formula:
\[

$$\begin{align*} g&=G\frac{M_{m}}{r^{2}}\\ &=(6.67\times 10^{-11}\ m^{3}\ kg^{-1}\ s^{-2})\frac{7.348\times 10^{22}\ kg}{(3.797\times 10^{8}\ m)^{2}}\\ &=(6.67\times 10^{-11})\frac{7.348\times 10^{22}}{1.442\times 10^{17}}\\ &=\frac{6.67\times7.348\times 10^{-11 + 22}}{1.442\times 10^{17}}\\ &=\frac{49.011\times 10^{11}}{1.442\times 10^{17}}\\ &=3.4\times 10^{-5}\ m/s^{2} \end{align*}$$

\]

Step3: Calculate the centripetal - acceleration formula

The centripetal acceleration $a_{c}=\omega^{2}r$, where the angular velocity $\omega=\frac{2\pi}{T}$, and $T = 27.3\ d$. First, convert $T$ to seconds: $T=27.3\ d\times24\ h/d\times3600\ s/h = 2.359\times 10^{6}\ s$. Then $\omega=\frac{2\pi}{2.359\times 10^{6}\ s}=2.66\times 10^{-6}\ rad/s$. The radius of rotation of the Earth's center about the common - center of mass is $r = 4.7\times 10^{6}\ m$.

Step4: Calculate the centripetal acceleration

\[

$$\begin{align*} a_{c}&=\omega^{2}r\\ &=(2.66\times 10^{-6}\ rad/s)^{2}\times4.7\times 10^{6}\ m\\ &=6.976\times 10^{-12}\times4.7\times 10^{6}\\ &=3.28\times 10^{-5}\ m/s^{2} \end{align*}$$

\]

Step5: Compare the two accelerations

The acceleration due to the Moon's gravity $g = 3.4\times 10^{-5}\ m/s^{2}$ and the centripetal acceleration $a_{c}=3.28\times 10^{-5}\ m/s^{2}$. They are approximately equal. This is because the gravitational force exerted by the Moon on the Earth provides the centripetal force that causes the Earth to rotate about the common - center of mass. According to Newton's second law for circular motion $F = ma_{c}$, and the gravitational force $F = mg$, so $a_{c}=g$ when the gravitational force is the only force providing the centripetal force for the circular motion of the Earth about the common - center of mass.

Answer:

a) The magnitude of the acceleration due to the Moon's gravity at that point is $3.4\times 10^{-5}\ m/s^{2}$.
b) The magnitude of the centripetal acceleration of the center of Earth is $3.28\times 10^{-5}\ m/s^{2}$. They are approximately equal because the gravitational force exerted by the Moon on the Earth provides the centripetal force for the Earth's rotation about the common - center of mass.