First slide

Acceleration due to gravity and its variation

Question

R is the radius of the earth and $ω$ is its angular velocity and $g_{p}$ is the value of g at the poles. The effective value of g at the latitude $λ = 60 °$ will be equal to

Easy

A

$g_{p} - \frac{1}{4} R ω^{2}$
B

$g_{p} - \frac{3}{4} R ω^{2}$
C

$g_{p} - R ω^{2}$
D

$g_{p} + \frac{1}{4} R ω^{2}$

Solution

$g = g_{p} - R ω^{2} \cos^{2} λ = g_{p} - ω^{2} R \cos^{2} 60 ° = g_{p} - ω^{2} R {(\frac{1}{2})}^{2} = g_{p} - \frac{1}{4} R ω^{2} (∵ \cos 60 ° = \frac{1}{2})$

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