If the earth has a radius of 6.4 x 10^6 and it spins once per day, what is the centripetal force on a 100 kg man who is on the equator? this would make him weigh (less/more) by

The spinning of the Earth would make him to weigh less.

Explanation:

Given data,

The radius of the Earth, R = 6.4 x 10⁶ m

The rotational period, T = 86400 s

The mass of the man, m = 100 kg

The centripetal force at equator,

                             f = m v²/R

Since,

                              T = 2π/ω   & v = Rω

                              v²/R = 4π²R/T²                  

Substituting in the equation for centripetal force,

                                    f =  4π²mR/T²      

Substituting the values,

                                    f = 4π² x 100 x 6.4 x 10⁶ / 86400²

                                       = 3.39 N

The centripetal force is directed along the radius towards the center, the centrifugal force acts opposite to it.

The gravitational force acting the man towards the center,

                                   F = mg

                                      = 100 x 9.8

                                       = 980 N            

The net force acting on the person,

                                   F'  = F - f

                                        = 980 N - 3.39 N  

                                        = 976.61 N

Hence, this would make him to weigh less.      

the reason is because the moon has some form of gravity, and that gravity pulls the ocean tides.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more