Question

What is the magnitude of the centripetal acceleration of an object on Earth's equator owing to the rotation of Earth?

Answer 1

The centripetal acceleration at the equator is r*ω^2 where ω = angular velocity of the Earth which = 2π/(24h*3600s/h) = 7.27x10^-5 rad/s

So a = 6.38x10^6m*(7.27x10^-5)^2 = 3.37x10^-2m/s^2

b) If a = 9.8 then a = r*(2π/T)^2...so T = 2π*sqrt(r/a) =