Calculating earth gravity by using centripetal acceleration

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To calculate gravitational acceleration at Earth's surface using the Moon's orbital data, the formula g=(4*π²R)/T² is applied, where R is the orbital radius and T is the period. The correct period must be in seconds for accurate results, as converting 27 days to minutes yields incorrect values. The discussion suggests that the focus should be on using the Moon's motion to determine Earth's mass first, then applying the gravitational formula to find the surface acceleration. The initial approach mistakenly calculates the Moon's acceleration rather than Earth's. Understanding the relationship between the Moon's orbit and Earth's gravity is crucial for accurate calculations.
ankhmor
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Homework Statement


I am asked to calculate the value of gravitational acceleration at Earth's surface given the following
Moon's period around the Earth - 27 days 8 hours
Moons orbit - 60.1 times the radius of Earth (6.38 x 10^6 m)

Homework Equations



Fc=(4*3.142mR)/T2
which works out to be
g=(4*3.142R)/T2

The Attempt at a Solution


I am able to get the right answer (9.8 m/s2) IF i convert 27 days to MINUTES instead of proper SECONDS.
so where am i going wrong?

ive tried another method where i didvided the moons speed squared by the orbital radius and i get the same wrong answer
2.76e-3
 
Last edited:
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If you just use g=(4*3.142R)/T2, with the circular motion of the moon, you will find the acceleration of the moon due to Earth's gravity. It will be MUCH smaller than 9.8.

I rather think you are supposed to use the given information about the moon's motion to calculate the mass of the Earth, then use the gravitational formula (with big G) to find the acceleration at the surface of the Earth (little g).
 
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