Centripetal Acceleration?

Question

Find the centripetal accelerations at each of the following points due to the rotation of Earth about its axis.
(a) a point on the Equator of Earth
? m/s2
(b) the North Pole
? m/s2

Answer 1

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Centripetal acceleration at a point on the equator of earth=v^2/r

where 'v' is speed on circular path and 'r' is radius of circular path that is r adius of earth, r=6.4 x10^6 m

but v=distance / time

distance=circumference=2 pi r
time is one period 'T',here T is period of rotation of earth about its axis, T= 24 hour =86400 s

v =2 pi r /T

(a) Centripetal acceleration at a point on the equator of earth=v^2/r=4 (pi)^2r^2/T^2 r=4 (pi)^2r /T^2

Centripetal acceleration on equator 'a' = 4 (pi)^2r/T^2
'a' =4 x3.14 x3.14 x6.4 x 10^6 /[86400x86400]=0.0338 m/s^2

Centripetal acceleration at a point on the equator of earth is 0.0338 m/s^2
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(b)The point on north pole is not revolving ,at the north pole centripetal acceleration is zero
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Answer 2

Think about how the earth rotates...it turns about the axis to the east. at the North pole you're standing on the axis, and at the equator you're the radius of earth away.

oh yeah, and centripetal acceleration = (v^2)/r

Answer 3

Rotational velocity at the equator is 465m/s.
Equatorial Radius a = 6371000 m
a = v^2 / R = 0.34 m/s^2
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At the pole v = 0 and hence a = 0.