If the radius of the earth is 6347 , km,then | vedclass

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(A) The acceleration due to gravity $(g)$ near the earth's surface is given by $g = \frac{GM}{R^2} \approx 9.8 \, m/s^2$.The acceleration of free fall

Vedclass · Accepted Answer

$0.0340$

Vedclass · Answer

(A) The acceleration due to gravity $(g)$ near the earth's surface is given by $g = \frac{GM}{R^2} \approx 9.8 \, m/s^2$.The acceleration of free fall $(g_{	ext{eff}})$ considering the rotation of the earth is given by $g_{	ext{eff}} = g - \omega^2 R \cos^2 \phi$. Near the equator $(\phi = 0)$,this becomes $g_{	ext{eff}} = g - \omega^2 R$.The difference between the acceleration due to gravity and the acceleration of free fall is $\Delta g = g - g_{	ext{eff}} = \omega^2 R$.Given $R = 6347 	imes 10^3 \, m$ and the angular velocity of the earth $\omega = \frac{2\pi}{T}$,where $T = 24 	imes 3600 \, s$.$\Delta g = \left(\frac{2\pi}{86400}ight)^2 	imes 6347 	imes 10^3$.$\Delta g = \left(\frac{6.283}{86400}ight)^2 	imes 6347000 \approx (7.27 	imes 10^{-5})^2 	imes 6347000$.$\Delta g \approx 5.285 	imes 10^{-9} 	imes 6347000 \approx 0.0335 \, m/s^2 \approx 0.0340 \, m/s^2$.

If the radius of the earth is $6347 \, km$,then what will be the difference between the acceleration of free fall and the acceleration due to gravity near the earth's surface?

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