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Calculate that imaginary angular velocit...

Calculate that imaginary angular velocity of the Earth for which effective acceleration due to gravity at the equator becomes zero. In this condition, find the length (in hours) of a day? Radius of Earth `= 6400 km. g = 10 ms^(-2)`.

Text Solution

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At equator, `lambda = 0^(@)`, so acceleration due to gravity is
`g_(e) = g - R omega^(2)`
Given, `g_(e) = 0`
`:. R omega^(2) = g`
`rArr omega = sqrt((g)/(R )) = sqrt((10)/(6,400 xx 10^(3))) = 1.25 xx 10^(-3)` rad/s
New length of the day will be
`T = (2pi)/(omega) = (2 xx 3.14)/(1.25 xx 10^(-3)) = 5,0.24 s = (5,024)/(3,600)`
= 1.395 h
`~= 1.4h`
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