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» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.25 xx10^(-3)` rad/sec <br/>`1.56 xx10^(-3)` rad/sec <br/>`1.25 xx10^(-1)` rad/sec <br/>`1.56 xx10^(-1)` rad/sec </p>Solution :`implies g. = g - R_(e) omega^(2) <a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a>^(2) <a href="https://interviewquestions.tuteehub.com/tag/lamda-536483" style="font-weight:bold;" target="_blank" title="Click to know more about LAMDA">LAMDA</a>` <br/> In weightless state `g.=0` <br/> Taking for equator `lamda=0` <br/> `0 = g - Romega^2` <br/> `:. Romega^2 =g` <br/> `omega=sqrt(g/R_e)=sqrt(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>/(6400xx10^3))=1/800 (rad)/s` <br/> `omega=1.25 xx10^(-3) (rad)/s`</body></html>