Calculation the orbital period of Moon.

  • Thread starter avito009
  • Start date
  • #1
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Newton showed that if gravity at a distance R was proportional to 1/R2, then indeed the acceleration g measured at the Earth's surface would correctly predict the orbital period T of the Moon. (Remember Earths gravity causes the moon to orbit the Earth.) We can find the answer using MKS system.

Let’s calculate the gravitational field due to the Earth at a distance equal to the moon’s distance from the Earth, the radius of the moon’s orbit.

g = GM/r^2

Substituting r = 60 RE

g = GM/(60 RE)^ 2

Then squaring everything in the denominator

g = GM/(3600 RE^2)

Regrouping to get the term “(GM/RE^2)” alone

g = (GME/RE^2) (1/3600)

Then, using g = 9.8 m/s2 = GM/RE^ 2

g = (9.8 m/s^2) / 3600
g = 0.0027 m/s^2
g = 2.7 x 10^-­3m/s^2 towards the center of the Earth

But orbital period of the moon is 27.3 days. So how do I proceed from here to get the answer as 27.3?
 

Answers and Replies

  • #2
Bandersnatch
Science Advisor
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Find the formula for centripetal acceleration.
 
  • #3
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In stable orbit, gravitational acceleration ( what you calculated ) = centripetal acceleration ( v ² / r )
 

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