First slide

Acceleration due to gravity and its variation

Question

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60° latitude becomes zero

Easy

A

$2.5 \times 1 0^{- 3} r a d s^{- 1}$
B

$1.5 \times 10^{- 3} r a d s^{- 1}$
C

$4.5 \times 10^{- 3} r a d s^{- 1}$
D

$0.5 \times 10^{- 3} r a d s^{- 1}$

Solution

$g_{λ} = g - R ω^{2} \cos^{2} λ, g i v e n λ = 60^{0} \Rightarrow 0 = g - R ω^{2} \cos^{2} 60 \Rightarrow g = R ω^{2} {(\frac{1}{2})}^{2} \Rightarrow 10 = 6400 \times 1000 \times \frac{ω^{2}}{4} \Rightarrow ω^{2} = \frac{1}{160000} \Rightarrow ω = \sqrt{\frac{1}{160000}} = \frac{1}{400} = 0.25 \times 10^{- 2} = 2.5 \times 10^{- 3} r a d / s$

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