Question

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

Answer 1

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

Explanation:

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`