Question

The moon completes one (circular) orbit of the earth in 27.3 days. The distance from the earth to the moon is 3.84×108 m. What is the moon’s centripetal acceleration?

Answer 1

Answer:

$2.72\cdot 10^{-3} m/s^2$

Explanation:

Let's start by calculating the angular velocity of the Moon. We know that the period is:

$T=27.3 d \cdot 24 \cdot 60 \cdot 60 =2.36\cdot 10^6 s$

So now we can calculate its angular velocity:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{(2.36\cdot 10^6)}=2.66 \cdot 10^{-6} rad/s$

The centripetal acceleration is given by

$a=\omega^2 r$

where

$\omega=2.66\cdot 10^{-6}rad/s$

$r=3.84\cdot 10^8 m$ is the radius of the orbit

Substituting,

$a=(2.66\cdot 10^{-6})^2(3.84\cdot 10^8)=2.72\cdot 10^{-3} m/s^2$