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) 

Options

A. `1.25 xx10^(-3)` rad/sec
B. `1.56 xx10^(-3)` rad/sec
C. `1.25 xx10^(-1)` rad/sec
D. `1.56 xx10^(-1)` rad/sec

Answer

A

Explanation

`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`