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Obtain an expression for the variation ineffective gravitational acceleration g' with latitude due to earth's rotation:

Answer» <html><body><p></p>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 <a href="https://interviewquestions.tuteehub.com/tag/equatorial-2617989" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATORIAL">EQUATORIAL</a> line is called the latitude `(<a href="https://interviewquestions.tuteehub.com/tag/lamda-536483" style="font-weight:bold;" target="_blank" title="Click to know more about LAMDA">LAMDA</a>)`of that place.<br/> `:.`For the equator latitude `lamda = 0^@` <br/> and for the poles latitude `lamda = 90^@` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C08_E01_018_S01.png" width="80%"/> <br/> `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. <br/>(1) Earth.s gravitational force mg is in `vec(PQ)`direction `"".... (1)` <br/>(2) Earth has an acceleration <a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to rotational <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a>. So, this particle is in the accelerated frame of reference <br/> `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 <a href="https://interviewquestions.tuteehub.com/tag/centripetal-415325" style="font-weight:bold;" target="_blank" title="Click to know more about CENTRIPETAL">CENTRIPETAL</a> force <br/> `(mv^2)/r` <br/> `=mromega^2 "" [:. v = r omega]` <br/> This force is in `vec(PQ)` direction . <br/> `implies`The component frictional centripetal force in the direction of `vec(PR) = mr omega^(2) cos lamda ""...(2) ` <br/> `:.`From equation (1) and (2) effective force on P, F `= mg - mromega^(2) cos lamda` <br/>If g is the effective gravitational acceleration of this particle then,<br/> `mg. = mg = mromega^(2) cos lamda ` <br/> `:. B. = g - r omega^(2) cos lamda ""...(3)` <br/> `implies`But from figure `MP=r=R_(E)cos lamda` ( `:. "From " DeltaOMP` )<br/> `:. g. = g - R_(E) omega^(2) cos^(2) lamda` <br/> OR <br/> `implies` From equation (4), we get information about the variation in g with latitude due to the earth.s rotation. <br/>Special Cases :<br/> (1) At equator `lamda =90^(@) , cos lamda = 0` <br/> `:. g. =g [1-(R_eomega^2cos ^(2) 90^@)/g]` <br/> `:. g. =g [1-0]` <br/> `:. g. = g`which is the maximum value of the effective gravitation acceleration.</body></html>


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