The acceleration due to gravity on the earth' | vedclass

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Question

(D) The effective acceleration due to gravity at the equator $(g_e)$ is given by $g_e = g - R\omega^2$,where $g$ is the acceleration due to gravity at

Vedclass · Accepted Answer

$\frac{R^{2} \omega^{2}}{2g}$

Vedclass · Answer

(D) The effective acceleration due to gravity at the equator $(g_e)$ is given by $g_e = g - R\omega^2$,where $g$ is the acceleration due to gravity at the poles.The acceleration due to gravity at a height $h$ above the poles $(g_h)$ is given by $g_h = g(1 - \frac{2h}{R}) = g - \frac{2gh}{R}$.Given that the weights are the same at both locations,the effective acceleration due to gravity must be equal: $g_e = g_h$.Substituting the expressions: $g - R\omega^2 = g - \frac{2gh}{R}$.Simplifying the equation: $R\omega^2 = \frac{2gh}{R}$.Solving for $h$: $h = \frac{R^2\omega^2}{2g}$.

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