The radius of the Earth is about6.37 ×10^6 m.A) What is the centripetal acceleration of a point on the equator due to the rotation of earth about its axis?Answer in units of m/s^2
B) What is the centripetal acceleration of a point at the North Pole?Answer in units of m/s^2.
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A) On the surface the centripetal acceleration a = v^2/R
where v is the velocity of a point on the surface and R is the radius of the Earth at the equator.
Now v = w*R
where w is the angular velocity and w = 2*Pi / T where T is the period of rotation of the Earth in seconds ie 24*3600s.
So a = w^2*R = 4*Pi^2*R / T^2
a = 4*Pi^2*R / T^2
B) At north pole the answer is 0 as R = 0 in the last equation above as this radius is the distance from the axis of rotation and this passes through the North (and South) Pole.
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