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Question
Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth’s rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? what is it at latitude θ? How does these accelerations compare with g = 9.8 m/s2?
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Solution
Radius of the earth (R) = 6400 km = 6.4 × 106 m
Time period (T) = 1 day = 24 × 60 × 60 s = 86400 s
Centripetal acceleration (ac) = `ω^2 R = R((2pi)/T)^2 = (4pi^2R)/T`
= `(4 xx (22/7)^2 xx 64 xx 10^6)/(24 xx 60 xx 60)^2`
= `(4 xx 484 xx 64 xx 10^6)/(49 xx (24 xx 3600)^2`
= 0.034 m/s2
At equator, latitude θ = 0°
∴ `a_c/g = 0.034/9.8 = 1/288`
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