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प्रश्न
At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?
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उत्तर
The apparent acceleration due to gravity at the equator becomes zero.
i.e., g' = g − ω2R = 0
⇒ g = ω2R
\[\Rightarrow \omega = \sqrt{\frac{g}{R}} = \sqrt{\frac{9 . 8}{6400 \times {10}^3}}\]
\[ \Rightarrow \omega = \sqrt{\frac{9 . 8 \times {10}^{- 5}}{6 . 4}} = \sqrt{1 . 5 \times {10}^{- 6}}\]
\[ \Rightarrow \omega = 1 . 2 \times {10}^{- 3} \text{ rad }/s\]
\[ \therefore T = \frac{2\pi}{\omega} = \frac{2 \times 3 . 14}{1 . 2 \times {10}^{- 3}}\]
\[ = \frac{6 . 28}{1 . 2 \times {10}^{- 3}}\]
= 1 . 41 h
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