Final answer:
The radial acceleration of an object at the Earth's equator is 0.0339 m/s², which is approximately 0.00346 of the acceleration due to gravity (g). For the radial acceleration to exceed g and cause objects to fly off into space, the Earth's rotation period would need to be significantly less than 24 hours.
Step-by-step explanation:
Calculating Radial Acceleration at the Earth's Equator
The period of rotation of Earth in seconds is 24 hours multiplied by 3600 seconds, which equals 86400 seconds. The angular velocity (ω) of Earth is calculated by the formula ω = 2π / T, where T is the period in seconds. Thus, ω is approximately 7.27 x 10-5 radians per second. The radius (r) of Earth at the equator is 6.4 × 106 meters.
To find out the radial acceleration (ar), we use the centripetal acceleration formula ar = ω2 × r. Substituting the values gives us the radial acceleration at the equator, which is approximately 0.0339 m/s2. To express this acceleration as a fraction of g, where g is the acceleration due to gravity (approximately 9.81 m/s2), we simply divide ar by g, resulting in approximately 0.00346 or 0.346% of g.
If the radial acceleration at the equator were to be greater than g, objects would indeed start to fly off into space. The period (T) for rotation to achieve this can be found by rearranging the centripetal acceleration formula to solve for T, given that ar must be equal to g. This gives us T = 2π √(r / g), which calculates to a period significantly less than Earth's current period of rotation.