Question

If R is the radius of earth, $ω$ is its angular velocjty and g_p is the value of g at the poles. The effective value of g at a latitude $λ$ = 60⁰ is [JIPMER 2018]

Moderate

Solution

The acceleration due to gravity at a latitude $λ$ is
$g^{'} = g - R ω^{2} \cos^{2} λ$
At the poles,    $λ = 90 °$
   $∴ g = g = g_{p}$    $(∵ \cos 90 ° = 0)$
For    $λ = 60 °, g^{'} = g_{p} - R ω^{2} \cos^{2} λ$
or    $g^{'} = g_{p} - R ω^{2} \cos^{2} 60 °$
Hence,    $g^{'} = g_{p} - R ω^{2} {(\frac{1}{2})}^{2} = g_{p} - \frac{1}{4} R ω^{2}$

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