What is the time period of ratation of the earth around its axis so that the objects at the equator becomes weightless ? (g = 9.8 m//s^(2), Radius of earth= 6400 km)

What is the time period of ratation of the earth around its axis so th

1.	What is the time period of ratation of the earth around its axis so that the objects at the equator becomes weightless ? (g = 9.8 m//s^(2), Radius of earth= 6400 km)
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> at the equator is<br/> `g_(0) = g-g_(0) = g - R omega^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` <br/> If bodies are to <a href="https://interviewquestions.tuteehub.com/tag/become-389953" style="font-weight:bold;" target="_blank" title="Click to know more about BECOME">BECOME</a> <a href="https://interviewquestions.tuteehub.com/tag/weightless-3272644" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHTLESS">WEIGHTLESS</a> at the equator, `g_(0) = 0`. <br/> `0 = g - R omega^(2) rArr R omega^(2) = g""omega = sqrt((g)/(R))` <br/> Time period of rotation, `T = (2pi)/(omega) = 2pi sqrt((R)/(g))` <br/> `R = 6400 xx 10^(3)m, g = 9.8 m//s^(2)` <br/> `T = 2pi sqrt((<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>.4 xx 10^(6))/(9.8)) = 5078 s = 84` minute 38s.</body></html>