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

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

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