If the Earth were a perfect sphere of ra...

If the Earth were a perfect sphere of radius `6.37 xx 10^(6) m`, rotating about its polar exis with a period of `1` day `(= 8.64 xx 10^(4) s)` how much would the acceleration due to gravity differ from the poles to equator?

Text Solution

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Here, `R = 6.37 xx 10^(6) m`,
`T = 8.64 xx 10^(4) s`.
`omega = 2 pi //T = 2 xx 3.14//(8.64 xx 10^(4)) rad//s`
We know, `g' = g - R omega^(2) cos^(2) lambda`
At poles, `lambda = 90^(@)` and `cos lambda = 0`.
Let `g' = g_(p)`
`:. g_(p) = g - R omega^(2) (0) = g` ..(i)
At equator, `lambda = 0^(@)` and `cos lambda = 1`.
Let `g' = g_(e)`
`:. g_(e) = g - R omega^(2) (1) = g - R omega^(2)` ..(ii)
`:. g_(p) - g_(e) = R omega^(2) = 6.37 xx 10^(6) xx((2 xx 3.14)/(8.64 xx 10^(4)))^(2)`
`= 3.37 xx 10^(-2) ms^(-2)`

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