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Obtain an expression for the variation ineffective gravitational acceleration g' with latitude due to earth's rotation: |
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Answer» Solution :`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. |
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