Question

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?

Answer 1

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