What is the centripetal acceleration of the moon, given its orbital radius and period?

Given period $T = 12.1 \text{ days}$.
There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute.

$$T = 12.1 \text{ days} \times \frac{24 \text{ hours}}{1 \text{ day}} \times \frac{60 \text{ minutes}}{1 \text{ hour}} \times \frac{60 \text{ seconds}}{1 \text{ minute}}$$ $$T = 12.1 \times 24 \times 3600 \text{ s}$$ $$T = 1044240 \text{ s}$$

The formula for angular velocity is $\omega = \frac{2\pi}{T}$.

$$\omega = \frac{2\pi}{1044240 \text{ s}}$$ $$\omega \approx \frac{6.283185}{1044240} \text{ rad/s}$$ $$\omega \approx 6.0169 \times 10^{-6} \text{ rad/s}$$

The formula for centripetal acceleration is $a_c = r\omega^2$.
Given radius $r = 3.85 \times 10^8 \text{ m}$.

$$a_c = (3.85 \times 10^8 \text{ m}) \times (6.0169 \times 10^{-6} \text{ rad/s})^2$$ $$a_c = (3.85 \times 10^8) \times (3.6203 \times 10^{-11}) \text{ m/s}^2$$ $$a_c \approx 0.013938 \text{ m/s}^2$$

Rounding to three significant figures:

$$a_c \approx 0.0139 \text{ m/s}^2$$

The centripetal acceleration of the moon towards the earth is:

$$\text{0.0139 m/s}^2$$