taking you as the point of reference, the earth's gravity excerts a force of g.m (where g is the gravitational acceleration and m is your mass). so it has to have enough tangential speed in order for the centripetal force that you require to be standing in the ground to be greater than g.m (that way you would not be able to stick to the ground using your weight alone, and therefore you would get thrown off.

as you may know, force equals mass per acceleration, and since all the masses we're talking about get the same acceleration (g) we'll just work with acceleration in this segment

so, the earth's centripetal acceleration at seal level must be greater than g

centripetal acceleration in a circular motion is determined by the equation V^2/r where V is the tangential velocity, and r is the radius

the earth's radius is 6730,000 m, so V^2/6730000m must be greater than 9,8 m/s^2

we solve for V, and we get that V = square root of (9.8 * 6730000) m^2/s^2

using a calculator, V must be greater than 8121.2 m/s.

So, we need angular speed, which is V/r, so we get that it equals 1.207*10^-3 rad/s. Therefore frequency, or W/2pi, is 1.92*10^-4, and therefore period, or lengh or the day in our problem, is 1/frequency, or 5206 s (86.78 minutes).

Hope it helped. If you didn't understand a step, let me know

Source(s):
Engineering student.

Anonymous
· 9 years ago