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Q.

On surface of the earth acceleration due to gravity at equator is g₁ and that poles is g₂. Acceleration due to gravity at a latitude of 45⁰ will be

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a

$\frac{g_{1} - g_{2}}{2}$

b

$\frac{g_{1} + g_{2}}{2}$

c

$g_{1} + g_{2}$

d

$g_{1} - \frac{g_{2}}{2}$

answer is C.

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Detailed Solution

$g_{ϕ} = g_{p o l e} - R ω^{2} \cos^{2} ϕ$
$\Rightarrow g_{1} = g_{2} - R ω^{2} \cos^{2} 0^{0}$
$g_{45^{0}} = g_{2} - R ω^{2} \cos 45^{0}$
$\begin{array}{l} \Rightarrow 2 (g_{2} - g_{45^{0}}) = (g_{2} - g_{1}) \\ \Rightarrow g_{45^{0}} = (\frac{g_{1} + g_{2}}{2}) \end{array}$

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