Question
Find the angular velocity for an object to experience weightlessness at the equator of earth. Under this condition, also find the duration of a day.
✓
Answer
Due to the rotation $\text{g}'=\text{g}\Big(1-\frac{\text{R}\omega^2}{\text{g}}\cos^2\theta\Big)$ where $\theta$ is the latitude. For weightlessness, g' = 0. At the equator, $\theta=0^\circ$ $\therefore1-\frac{\text{R}\omega^2}{\text{g}}\cos^20^\circ=0$ $\text{or }\omega=\sqrt{\frac{\text{g}}{\text{R}}}$ $\Rightarrow\omega=\sqrt{\frac{10}{6400\times10^3}}=\frac{1}{800}$ $\omega=\frac{2\pi}{\text{T}}$ $\Rightarrow=\frac{2\text{x}}{\omega}=\frac{2\times3.14}{\frac{1}{800}}$ $=5024\text{sec.}$
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