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At what rate should the earth rotate so ...

At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?

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At earth's euator effective value of gravity is
`g_(eq) = g_(s) - omega^(2) R_(e )`
If `g_(eff)` at equator be zero, we have
`g_(x)-omega^(2)R_(e ) = 0`
or `omega = sqrt((g_(e ))/(R_(e )))`
Thus length of the day will be
`T = (2 pi)/(omega) = 2pi sqrt((R_(e ))/(g_(s)))`
`= 2xx3.14 sqrt((6.4xx10^(6))/(9.8))=5074.77 s`
`~=84. 57` min.
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