If bodies at a latitude of 45^@of the earth are to be weightless, what should be earth.s time period of rotation?

If bodies at a latitude of 45^@of the earth are to be weightless, what

1.	If bodies at a latitude of 45^@of the earth are to be weightless, what should be earth.s time period of rotation?
Answer» <html><body><p></p>Solution : The acceleration due to gravity at latitude `phi`is<br/>`g_phi =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> cos^2 phi`<br/>For bodies to be <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 a latitude of`45^@` then “ `g_phi` ” must be zero.<br/>` rArr <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a> = g - R omega^2 cos^2 45^@ rArr 0 = g - R omega^(2) 1/2 rArr R omega^2 = 2g rArr omega = sqrt((2g)/(R ))`<br/> We <a href="https://interviewquestions.tuteehub.com/tag/know-534065" style="font-weight:bold;" target="_blank" title="Click to know more about KNOW">KNOW</a> that , `omega = (<a href="https://interviewquestions.tuteehub.com/tag/2pi-1838601" style="font-weight:bold;" target="_blank" title="Click to know more about 2PI">2PI</a>)/(T) rArr (2pi)/(T) = sqrt((2g)/(R ))rArr T = 2pi sqrt((R)/(2g)) " or " T = pi sqrt((2R)/(g))`</body></html>