What is the magnitude of the centripetal acceleration of an object on Earth's equator owing to the rotation of Earth?

1 Answer
Apr 27, 2018

#~~0.0338\ "ms"^-2#


On the equator, a point rotates in a circle of radius #R~~6400\ "km"=6.4 times 10^6\ "m"#.

The angular velocity of rotation is

#omega = (2 pi)/(1\ "day") = (2pi)/(24times 60times 60\ "s") = 7.27times 10^-5\ "s"^-1#

Thus the centripetal acceleration is

#omega^2R = (7.27times 10^-5\ "s"^-1)^2times 6.4 times 10^6\ "m" =0.0338\ "ms"^-2#