Thanks for checking out my thread. Okay, first off, the question. The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 X 10^6 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 30.0° north of the equator. I'll show what i've done so far. a) V=2(pi)r/t plugged in the appropriate values where r is (6.38 x 10^6) and t is (86400). my results from that equation was 463.97 m/s. I then continued to calculate the centripetal acceleration: a=v^2/t a=(463.97^2)/6.38 x 10^6) a=0.03 m/s^2 (I wasn't expecting it to be this small. b) the troubling thing now is obviously the angle. Do I use the following formula? v=sq(rgtan(theta)) v=sq((6.38 x 10^6)(9.8)(tan30)) v=6,008 m/s then to caluculate the centripetal accceleration: a=v^2/r and I end up getting 5.6 m/s^2 I'm not sure if the answers are correct (most likely not) Any help would be greatly appreciated. Thanks for your time!!