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I'm not quite sure of my ground here, but here are my thoughts: Data: equatorial radius of the earth (r) = 6378137 m. equatorial gravitational field strength (g) = 9.78 Nkg-1 In order for the rotation of the earth to perfectly cancel weight, the gravitational field strength g should equal the centripetal acceleration, a. a = w2r where w is the Earth's angular velocity. If T is the duration of a day in seconds, then we can substitute w = 2 x pi / T giving a = 4pi2r / T2 Equating a to g: 4pi2r / T2 = 9.78 Rearranging to find T: T = sqrt(4pi2r / 9.78) Substituting pi = 3.1415... and r = 6378137 T = 5074 seconds (nearest second) which is 1 hour, 24 minutes and 34 seconds, or about 17 times shorter than an ordinary day.

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Q: If the earth were to spin so fast that a person at the equator were weightless what would the length of a day be?
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What the time periods of a simple pendulum at equator of the earth?

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