The angular speed of the Earth's rotation is 7.3 10⁻⁵ rad s ⁻¹ . Take the radius of the Earth at the equator to be 6400 km. Then the ratio (a_c/g) of the magnitude of the centripetal acceleration a_c at a point on the equator to g , is of the order

Step-by-step solution

Given angular speed = 7.3 10⁻⁵ rad s ⁻¹ and radius R = 6400 km = 6.4 10^6 m. The centripetal acceleration at the equator is: a_c = ^2 R a_c = (7.3 10⁻⁵)^2 (6.4 10^6) a_c 53.29 10⁻¹⁰ 6.4 10^6 3.41 10⁻² m s ⁻² Taking acceleration due to gravity g 9.8 m s ⁻² , the ratio is: a_c g = 3.41 10⁻² 9.8 3.48 10⁻³ The order of magnitude of this ratio is 10⁻³ . Answer: 10⁻³