why is the centripetal acceleration in the poles and equator if different? I know that it's related to Equatorial budge. But what I don't get it is that the centripetal force is zero at the poles. There's still radius and angular velocity at the pole right, so why is it zero? And at the equator, the centripetal force is acting in the opposite direction of the gravitational force. Centripetal force should always directed to the center. Hence, why wouldn't it be in the same direction as the gravitational force?
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Not that the centrifugall pseudoforce would be different at the poles than the equator even if the Earth was a perfect sphere (i.e. the centrifugal pseudoforce explains the bulge, but not the other way around).. – dmckee --- ex-moderator kitten Jan 29 '16 at 04:57
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2The question formulation (v1) seems to conflate the centripetal and the centrifugal force/acceleration. – Qmechanic Jan 29 '16 at 12:31
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1I don't understand what confuses you. Centripetal/Centrifugal force scales with the distance to the axis of rotation. At the poles, you have zero distance to the axis of rotation, hence no force. – ACuriousMind Jan 29 '16 at 14:07
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There's still radius and angular velocity at the pole right, so why is it zero?
Considering a point particle exactly at the pole, there is no angular velocity about the axis of rotation of the earth. This can easily be seen from the definition of angular momentum $$\mathbf L=\mathbf r\times\mathbf p$$ Since $\mathbf r=0$ at the poles, the angular momentum is $0$.
And at the equator, the centripetal force is acting in the opposite direction of the gravitational force. Centripetal force should always directed to the center. Hence, why wouldn't it be in the same direction as the gravitational force?
As pointed out in the comments, there seems to be a mix up between centripetal and centrifugal forces. If using an inertial reference frame, there is no centrifugal force, and there are two forces acting on someone on Earth's surface: gravity and a normal force. The centripetal force is not an additional force, but rather it is just the radial inward component of the net force. In our case $$F_{cent}=F_g-F_N$$ This composite component is in the same direction of gravity at the equator.
For a rotating reference frame a centrifugal force is present, and it always points radially away from the axis of rotation.
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The centrifugal acceleration is proportional to the perpendicular distance from the point mass to the spin axis of the Erath. Since there is no perpendicular distance at the pole from the point mass to the axis, therefore, the centrifugal acceleration is zero. However, the perpendicular distance reaches its maximum at the equator so does the centrifugal acceleration.
