If R is the radius of earth- omega is its angular velocity and g-p is the value of acceleration due to gravity the poles- then effective value of acceleration due to gravity the latitude lambda - 60-o will be equal tog-p - Romega-2g-p - dfrac-1-4-Romega-2g-p - dfrac-1-4-Romega-2g-p - dfrac-3-4-Romega-2

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Toppr Admin · Accepted Answer

G-gp-x2212-R-x3C9-2cos2-x3BB-g-gp-x2212-x3C9-2Rcos260-x2218-g-gp-x2212-14R-x3C9-2

If R is the radius of earth, ω is its angular velocity and gp is the value of acceleration due to gravity at the poles, then effective value of acceleration due to gravity at the latitude λ=60o will be equal togp−Rω2gp+14Rω2gp−14Rω2gp−34Rω2

g=gp−Rω2cos2λg=gp−ω2Rcos260∘g=gp−14Rω2

$g = g_{p} - R ω^{2} c o s^{2} λ g = g_{p} - ω^{2} R c o s^{2} 60^{\circ} g = g_{p} - \frac{1}{4} R ω^{2}$