Question

a. What is the magnitude of centripetal acceleration due to the daily rotation of an object at earth's equator. The equatorial radius is 6.38x10^6 m.

b. How does this acceleration affect a person's weight at the equator?

Answer 1

The magnitude of cenripetal acceleration is given by

$$a=\omega^2R=\frac{4\pi^2}{T^2}R$$

$$a=\frac{4\pi^2}{(24*3600\:\rm s)^2}*6378*10^3\:\rm m=0.034\: \rm m/s^2$$