The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh 1 6 th as much as at present is ( g= gravitational acceleration, R= equatorial radius of the earth)

Step-by-step solution

The apparent weight at the equator under Earth's rotation is W' = m(g - ^2 R) , where m is mass, g is gravitational acceleration, and R is Earth's equatorial radius. Given that W' = 1 6 W and W = mg , substitution yields: m(g - ^2 R) = 1 6 mg Dividing both sides by m : g - ^2 R = 1 6 g Solving for ^2 : ^2 R = g - 1 6 g = 5 6 g ^2 = 5 6 g R Taking the square root: = 5 6 g R This corresponds to option A .