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?
Solution:
Apparent acceleration due to gravity at equator becomes 0 .
$G^{\prime}=g-\omega^{2} R=0$
$g=\omega^{2} R$
$\omega=\sqrt{\frac{g}{R}}$
$=\sqrt{\frac{9.8}{6400 \times 10^{\mathrm{s}}}}$
Time period $=\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \times 3.14}{1.2 \times 10^{\mathrm{s}}}=1.41 \mathrm{hrs}$
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