Question

Suppose the moon is moving in a circular path of radius 3.8*10^5km about the earth. The period of revolution is 27.5 days. Calculate the moon’s acceleration

Answer 1

Answer:

a ≈ $2.657360411*10^{-3}$ m/s^2

Explanation:

Let's convert the values into SI unit.

r = 3.8*10^5 km = 3.8*10^8 m

T = 27.5 days =  2376000 seconds

Now, apply the equation $T = \frac{2\pi r}{v}$

$v = \frac{2\pi(3.8e8) }{2376000}$

v ≈ 1004.886539 m/s

Now, apply the equation $a = \frac{v^{2} }{r}$

$a = \frac{1004.886539^{2} }{3.8e8}$

a ≈ $2.657360411*10^{-3}$ m/s^2