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Explain the variation of 'g' with latitude.

Answer» <html><body><p></p>Solution :Variation of g with latitude: <br/> Wherever we analyze the motion of object in <a href="https://interviewquestions.tuteehub.com/tag/rotating-1191156" style="font-weight:bold;" target="_blank" title="Click to know more about ROTATING">ROTATING</a> frames we must take into account the centrifugal force. Even though we treat the Earth as an inertial <a href="https://interviewquestions.tuteehub.com/tag/frame-998974" style="font-weight:bold;" target="_blank" title="Click to know more about FRAME">FRAME</a>, it is not exactly correct because the Earth spins about its own axis. So when and object is on the surface of the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been 'mg'. <a href="https://interviewquestions.tuteehub.com/tag/however-1032379" style="font-weight:bold;" target="_blank" title="Click to know more about HOWEVER">HOWEVER</a>, the object experiences an additional centrifugal force due to spinning of the Earth. <br/> This centrifugal force is given by `momega^(2) R'`. <br/> where `lambda` is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is <br/> `a_(c) = <a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>^(2) R' cos lambda = omega^(2) R cos^(2) lambda` <br/> since `R' = R cos lambda` <br/> Therefore, `g' = g - omega^(2) R cos^(2) lambda""......(2)` <br/> From <a href="https://interviewquestions.tuteehub.com/tag/hte-2718409" style="font-weight:bold;" target="_blank" title="Click to know more about HTE">HTE</a> expression (2), we can infer that at equator ,`lambda = 0, g ' = g - omega^(2)R`. The acceleratoin due to gravity is minimum. At poles `lambda = 90, g' = g`, it is maximum. At the equator, g' is minimum. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_PHY_XI_V02_C06_E03_008_S01.png" width="80%"/></body></html>


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