`implies` The angle made by the line joining a given place on the earth.s surface to the centre of the earth with the equatorial line is called the latitude `(lamda)` of that place.
`:.` For the equator latitude `lamda = 0^@`
and for the poles latitude `lamda = 90^@`
`implies` As shown in figure the latitude of the place P on the earth.s surface is `lamda = anglePOE.` At this position consider a particle of mass m. Two forces acting on it.
(1) Earth.s gravitational force mg is in `vec(PQ)` direction `" ".... (1)`
(2) Earth has an acceleration due to rotational motion. So, this particle is in the accelerated frame of reference
`implies` At this point the acceleration of the frame of reference is `((v^2)/r)` in `vec(PM) ` direction. Hence, friction acceleration of particle is `((v^2)/r)` . It is in `vec(PQ )` direction. Hence, frictional centripetal force
`(mv^2)/r`
`=mromega^2 " " [:. v = r omega]`
This force is in `vec(PQ)` direction .
`implies` The component frictional centripetal force in the direction of `vec(PR) = mr omega^(2) cos lamda " "...(2) `
`:.` From equation (1) and (2) effective force on P, F `= mg - mromega^(2) cos lamda`
If g is the effective gravitational acceleration of this particle then,
`mg. = mg = mromega^(2) cos lamda `
`:. B. = g - r omega^(2) cos lamda " "...(3)`
`implies` But from figure `MP=r=R_(E)cos lamda` ( `:. " From " DeltaOMP` )
`:. g. = g - R_(E) omega^(2) cos^(2) lamda`
OR
`implies` From equation (4), we get information about the variation in g with latitude due to the earth.s rotation.
Special Cases :
(1) At equator `lamda =90^(@) , cos lamda = 0`
`:. g. =g [1-(R_eomega^2cos ^(2) 90^@)/g]`
`:. g. =g [1-0]`
`:. g. = g` which is the maximum value of the effective gravitation acceleration.
