15. Centripetal Forces at the Equator The radius of the earth is RE-6.38-106 m.

Question

15. Centripetal Forces at the Equator The radius of the earth is RE-6.38-106 m. (a) Find the speed of a person standing at the equator. (This is the speed due to the rotation of the earth.) (b) Since the person is in a state of uniform circular motion, there is a centripetal force. Find Feent/W, the ratio of the centripetal force to the weight of the person. (c) If the rotation of the earth was faster, at some point the weight would not be large enough to supply the centripetal force. In this case loose objects on the equator would float away from the earth's surface. What would the length of a day be in this case? (In other words, find the period of the earth's rotation in this case.)

Explanation / Answer

(A) w = 2 pi / T = 2 pi / (24 x 3600 s)

w = 7.272 x 10^-5 rad/s

v = w r = (7.272 x 10^-5) (6.38 x 10^6)

v = 464 m/s

(B) Fcent = m v^2 / R = 0.0337 m

W = m g = 9.8 m

Fcent / W = 3.44 x 10^-3

(C) g = w^2 r

w = sqrt(9.8 x 6.38 x 10^6) = 7907 rad/s

T = 2pi /w = 7.95 x 10^-6 sec .......Ans

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