The angular speed of earth, so that the object on equator may appear weightless, is $(g = 10\,m/{s^2}$, radius of earth $6400\, km$)
The angular speed of earth, so that the object on equator may appear weightless, is $(g = 10\,m/{s^2}$, radius of earth $6400\, km$)
$1.25 \times {10^{ - 3}}\,rad/sec$
$1.56 \times {10^{ - 3}}\,rad/sec$
$1.25 \times {10^{ - 1}}\,rad/sec$
$1.56\; rad/sec$
The angular speed of earth, so that the object on equator may appear weightless, is $(g = 10\,m/{s^2}$, radius of earth $6400\, km$)
For condition of weightlessness of equator $\omega = \sqrt {\frac{g}{R}} = \frac{1}{{800}} = 1.25 \times {10^{ - 3}}\frac{{rad}}{s}$
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