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Explain the variation of 'g' with latitude. |
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Answer» Solution :Variation of g with latitude: Wherever we analyze the motion of object in ROTATING frames we must take into account the centrifugal force. Even though we treat the Earth as an inertial FRAME, 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'. HOWEVER, the object experiences an additional centrifugal force due to spinning of the Earth. This centrifugal force is given by `momega^(2) R'`. where `lambda` is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is `a_(c) = OMEGA^(2) R' cos lambda = omega^(2) R cos^(2) lambda` since `R' = R cos lambda` Therefore, `g' = g - omega^(2) R cos^(2) lambda""......(2)` From HTE 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. 
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