R is the radius of the Earth,omega is its ang | vedclass

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(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g_p - R\omega^2 \cos^2 \lambda$.Given that t

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$g_p - \frac{1}{4}R\omega^2$

Vedclass · Answer

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g_p - R\omega^2 \cos^2 \lambda$.Given that the latitude $\lambda = 60^\circ$,we substitute this value into the equation.Since $\cos 60^\circ = \frac{1}{2}$,we have $\cos^2 60^\circ = (\frac{1}{2})^2 = \frac{1}{4}$.Substituting this into the formula: $g' = g_p - R\omega^2 (\frac{1}{4})$.Therefore,the effective value of $g$ is $g_p - \frac{1}{4}R\omega^2$.

$R$ is the radius of the Earth,$\omega$ is its angular velocity,and $g_p$ is the value of acceleration due to gravity at the poles. The effective value of $g$ at latitude $\lambda = 60^\circ$ will be equal to:

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