physics question!!! please help immediately - circular motion?

[John S]

Assuming that the earth is a perfect sphere with a radius of 6.38Mm, calculate the speed and the centripetal acceleration of a person standing on the equator. Calculate the same for a person at latitude 45 degrees.

[jim m]

try v^2/R for equator.
(40000000m/24/3600)^2/6380000
=0.033594781m/s^2-9.81=9.78

At latitude 45 r=Rcoslat=.707*6380
40000000*.707/24/3600)^2/6380000*.707
=0.011872169m/s^2-9.81=9.80

[jim m]

try v^2/R for equator.
(40000000m/24/3600)^2/6380000
=0.033594781m/s^2-9.81=9.78

At latitude 45 r=Rcoslat=.707*6380
40000000*.707/24/3600)^2/6380000*.707
=0.011872169m/s^2-9.81=9.80

[jim m]

try v^2/R for equator.
(40000000m/24/3600)^2/6380000
=0.033594781m/s^2-9.81=9.78

At latitude 45 r=Rcoslat=.707*6380
40000000*.707/24/3600)^2/6380000*.707
=0.011872169m/s^2-9.81=9.80