Suppose Earth is perfect sphere of radius `6.4 xx 10^(6) m`. It is rotating about its polar axis with a period of `1` day. What is the difference in the value of acceleration due to gravity on pole and at a place of latitude `60^(@)`?

Explanation

Here, `R = 6.4 xx 10^(6) m`,
`T = 1 day = 24 xx 60 xx 60 s = 8.64 xx 10^(4) s`
`omega = (2pi)/(T) = (2xx 3.142)/(8.64 xx 10^(4)) rad//s`
We know that, `g' = g - R omega^(2) cos ^(2) lambda`
At pole, `lambda = 90^(@), g_(P)^(@) = g - R omega^(2) cos^(2) 90^(@) = g`
At latitude `60^(@)`, `lambda = 60^(@), g_(L) = g - R omega^(2) cos^(2) 60^(@)`
`= g - R omega^(2) xx (1)/(4)`
`:. g_(P) - g_(L) = g - (g - R omega^(2)//4) = (R omega^(2))/(4)`
`= (6.4 xx 10^(6)) xx ((2 xx 3.142)/(8.64 xx 10^(4)))^(2) xx (1)/(4)`
`= 8.4 xx 10^(-3) ms^(-2)`