If the speed of rotation of the earth increases, then weight of the body at the equator\

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Professor PC Thomas Classes · Answer

Wt of a body at the equator is\ $=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}-\mathrm{mR} \omega^{2}$ where $\omega=$ angular\ velocity of rotation of the earth, $\mathrm{R}=$ equatorial radius, $\mathrm{M}=$ mass of the earth\ When increases the weight of the body at the equator decreases. For a certain angular velocity $\omega_{0}$, the weight will be zero.\ $0=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}-\mathrm{mR} \omega_{0}^{2}$\ $\omega_{0}^{2}=\frac{\mathrm{GM}}{\mathrm{R}^{3}}$\ $=\frac{\mathrm{g}}{\mathrm{R}}$\ $\omega_{0}=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}=\frac{1}{800} \mathrm{rad} / \mathrm{s}$\ At this speed of rotation a body at the equator will appear to be weightless. Let us look at the present speed of rotation $\omega$\ $\omega=\frac{2 \pi}{24 	imes 60 	imes 60} \mathrm{rad} / \mathrm{sec}$\ Find $\frac{\omega_{0}}{\omega}$\