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

Apr 27, 2018

$\approx 0.0338 \setminus {\text{ms}}^{-} 2$

#### Explanation:

On the equator, a point rotates in a circle of radius $R \approx 6400 \setminus \text{km"=6.4 times 10^6\ "m}$.

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