carolinedonalds99 carolinedonalds99
  • 16-03-2022
  • Physics
contestada

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

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samuellai03
samuellai03 samuellai03
  • 16-03-2022

Answer:

a ≈ [tex]2.657360411*10^{-3}[/tex] 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 [tex]T = \frac{2\pi r}{v}[/tex]

[tex]v = \frac{2\pi(3.8e8) }{2376000}[/tex]

v ≈ 1004.886539 m/s

Now, apply the equation [tex]a = \frac{v^{2} }{r}[/tex]

[tex]a = \frac{1004.886539^{2} }{3.8e8}[/tex]

a ≈ [tex]2.657360411*10^{-3}[/tex] m/s^2

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