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(D) For an object on the equator to feel weightless,the effective acceleration due to gravity must be zero.The effective acceleration due to gravity a

Vedclass · Accepted Answer

$1.41$

Vedclass · Answer

(D) For an object on the equator to feel weightless,the effective acceleration due to gravity must be zero.The effective acceleration due to gravity at the equator is given by $g_{eff} = g - \omega^2 R$,where $g$ is the acceleration due to gravity,$\omega$ is the angular velocity of the earth,and $R$ is the radius of the earth.Setting $g_{eff} = 0$,we get $g - \omega^2 R = 0$,which implies $\omega^2 R = g$.Thus,$\omega = \sqrt{\frac{g}{R}}$.The duration of the day (time period $T$) is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{R}{g}}$.Substituting the values $R \approx 6.4 	imes 10^6 \, m$ and $g \approx 9.8 \, m/s^2$:$T = 2 	imes 3.14 	imes \sqrt{\frac{6.4 	imes 10^6}{9.8}} \approx 2 	imes 3.14 	imes 808 \, s \approx 5074 \, s$.Converting to hours: $T \approx \frac{5074}{3600} \, hr \approx 1.41 \, hr$.

If all objects on the equator of the earth feel weightless,then the duration of the day will nearly become ....... $hr$.

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