Question

Calculate angular velocity of earth so that acceleration due to gravity at $60^{\circ}$ latitude becomes zero. (Radius of earth $= 6400\, km$, gravitational acceleration at poles $= 2\, s\, 10\, m$ , $cos \,60^{\circ} = 0.5$)

• A. $7.8\times10^{-2}\,rad/s$
• B. $0.5 \times10^{-3}\,rad/s$
• C. $1 \times10^{-3}\,rad/s$
• D. $2.5 \times 10^{-3}\,rad/s$

Answer 1

Answer: (D)

New acceleration due to gravity $g'$ is given by
$g'=g-R_{e} \omega^{2} \cos ^{2} \lambda $
$0 =10-6.4 \times 10^{6} \omega^{2} \cos ^{2} 60^{\circ} $
$\Rightarrow \omega^{2} =\frac{10}{6.4 \times 10^{6}[0.5]^{2}}=2.5 \times 10^{-3} rad / s$