- #1
avito009
- 184
- 4
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?
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?