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

`2.72\cdot 10^(-3) m/s^2`

Step-by-step 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=(2\pi)/(T)=(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`