What should be the length of the day so that the weight of a body on the equator of earth becomes zero ? Given that radius of earth is 6400 km and acceleration due to gravity on its surface is `9.8 m//s^(2)`

Explanation

As, `W=m(g-Romega^(2))` so the weight of a body at equator will become zero if
`g-Romega^(2)=0 ["as m is finite"]`
or `omega=sqrt(g/R)=sqrt((9.8)/(6400xx10^(3)))=1/808 ("rad")/s`
So, `T=(2pi)/omega=2pisqrt(R/g)`
i.e, `T=6.28xx808s ~=84.6 min ~= 1.4hr`
Note: As the present length of the day is 24 hrs and `T prop(1//omega)`
`omega/omega_(0)=T_(0)/T=24/1.4~=17`
i.e., if the earth starts spinning about 17 times of its present value, the weight of a body on equator will become zero.