The rotation of the Earth (of radius R) about | vedclass

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(B) Given that,the angle of latitude is $\lambda = 45^{\circ}$.The effective acceleration due to gravity at a latitude $\lambda$ is given by $g_{\lamb

Vedclass · Accepted Answer

$\pi \sqrt{\frac{2 R}{g}}$

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(B) Given that,the angle of latitude is $\lambda = 45^{\circ}$.The effective acceleration due to gravity at a latitude $\lambda$ is given by $g_{\lambda} = g - \omega^2 R \cos^2 \lambda$.The apparent weight of a man of mass $m$ at this point is $w = m g_{\lambda} = m(g - \omega^2 R \cos^2 \lambda)$.According to the question,the man feels weightlessness,so $w = 0$.Therefore,$m(g - \omega^2 R \cos^2 45^{\circ}) = 0$.Since $m 
eq 0$,we have $g - \omega^2 R (\frac{1}{\sqrt{2}})^2 = 0$,which simplifies to $g - \frac{\omega^2 R}{2} = 0$.This gives $\omega^2 = \frac{2g}{R}$,or $\omega = \sqrt{\frac{2g}{R}}$.The duration of a day (time period $T$) is given by $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$,we get $T = \frac{2\pi}{\sqrt{2g/R}} = 2\pi \sqrt{\frac{R}{2g}} = \pi \sqrt{\frac{2R}{g}}$.

The rotation of the Earth (of radius $R$) about its axis speeds up to a value such that a man at latitude angle $45^{\circ}$ feels weightlessness. The duration of a day in such a case is

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