Calculating earth gravity by using centripetal acceleration

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SUMMARY

The discussion focuses on calculating gravitational acceleration at Earth's surface using centripetal acceleration derived from the Moon's orbital parameters. The formula used is g=(4*πR)/T², where R is the orbital radius and T is the orbital period. A critical error identified is the conversion of the Moon's period from days to minutes instead of seconds, leading to incorrect results. The correct approach involves calculating the mass of the Earth using the Moon's motion and then applying the gravitational formula with the universal gravitational constant (G) to determine the surface acceleration.

PREREQUISITES
  • Understanding of centripetal acceleration and its relation to gravitational forces
  • Familiarity with the formula for gravitational acceleration, g = GM/R²
  • Knowledge of orbital mechanics, specifically regarding the Moon's motion
  • Ability to convert time units accurately (days to seconds)
NEXT STEPS
  • Study the derivation of the gravitational force formula, F = GMm/R²
  • Learn about the calculation of orbital periods and their implications in celestial mechanics
  • Explore the concept of centripetal acceleration in detail, particularly in relation to planetary bodies
  • Investigate the universal gravitational constant (G) and its significance in gravitational calculations
USEFUL FOR

Students in physics, particularly those studying mechanics and gravitation, as well as educators and anyone interested in understanding the calculations related to gravitational forces and celestial motion.

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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