Determine the acceleration of a particle located on the surface of the earth at the equator due to the earth's rotation. The diameter of the earth is 12800 km and it takes 24 hours for the earth to complete one revolution around its axis.

Step-by-step solution

Radius of the earth, R = D 2 = 12800 2 = 6400 km = 6.4 10^6 m Time period of rotation, T = 24 hours = 24 3600 s = 86400 s The acceleration of a particle at the equator due to the earth's rotation is the centripetal acceleration: a_c = ^2 R = ( 2 T )^2 R Substituting the values: a_c = ( 2 86400 )^2 6.4 10^6 a_c = 4 ^2 7464960000 6400000 a_c = 25.6 ^2 7464.96 Taking ^2 9.8 : a_c 25.6 9.8 7464.96 = 250.88 7464.96 0.0336 m/s ^2 Answer: 0.0336 m/s ^2