What would be the angular speed of earth, so that body lying on equator may appear weightless? ( g = 10 m//s^2 , radius of earth = 6400 km)

What would be the angular speed of earth, so that body lying on equato

1.	What would be the angular speed of earth, so that body lying on equator may appear weightless? ( g = 10 m//s^2 , radius of earth = 6400 km)
Answer» `1.25 xx10^(-3)` rad/sec `1.56 xx10^(-3)` rad/sec `1.25 xx10^(-1)` rad/sec `1.56 xx10^(-1)` rad/sec Solution :`implies g. = g - R_(e) omega^(2) COS^(2) LAMDA` In weightless state `g.=0` Taking for equator `lamda=0` `0 = g - Romega^2` `:. Romega^2 =g` `omega=sqrt(g/R_e)=sqrt(10/(6400xx10^3))=1/800 (rad)/s` `omega=1.25 xx10^(-3) (rad)/s`