Determine the magnitude of the centripetal acceleration of an object on the surface of the Earth at a latitude of 71.6°. Consider the Earth to be a sphere of radius
R = 6.38 ✕ 106 m.
2 answers
Correction: 6.38x10^6 m
r = 6.38 * 10^6 * cos 71.6
= 2.01 * 10^6 meters
angular velocity = w
= 2 pi radians/24 hours
*1 hour/3600 seconds
= 2 pi / (24*3600)
= 7.27 * 10^-5 radians/second
so
Ac = r w^2
= 2.01*10^6 (7.27)^2*10^-10 m/s^2
= 2.01 * 10^6 meters
angular velocity = w
= 2 pi radians/24 hours
*1 hour/3600 seconds
= 2 pi / (24*3600)
= 7.27 * 10^-5 radians/second
so
Ac = r w^2
= 2.01*10^6 (7.27)^2*10^-10 m/s^2
1 answer