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What would be the angular speed of earth, so that bodies lying on equator may appear weightless? $(g=10 m/s^2\; R=6400km)$

\[(a)\;1.25 \times 10^{-3} rad/s \quad (b)\;1.25 \times 10^{-2} rad/s \quad (c)\;1.25 \times 10^{-4}\;rad/s\quad (d)\;1.25 \times 10^{-1}\;rad/s \]

1 Answer

The apparent weight of a person on the equator (Lattitude $ \alpha =0$) is given by
$g'=g -R_e w^2$
$w=\sqrt {\large\frac{g}{R}}$
$\quad=\sqrt {\large\frac{10}{6400 \times 10^3}}$
$\quad=1.25 \times 10^{-3} rad /s$
Hence a is the correct answer.


answered Aug 23, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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