Question

The linear velocity of a point on the surface of earth at a latitude of $60^\circ $ is

• A. $\frac{800}{3}ms^{- 1}$
• B. $\frac{800 \pi }{3}ms^{- 1}$
• C. $800\times \frac{5}{18}ms^{- 1}$
• D. $\frac{2000 \pi }{27} \, m \, s^{- 1}$

Answer 1

Answer: (D)

$V=r\omega $
$r=Rcos \theta $
Where $R=6400km$ is radius of the earth; $\theta =60^\circ $ is the latitude angle.
$\omega =\frac{2 \pi }{T}$ $\left(\right.T=24hours=24\cdot 60\cdot 60s\left.\right)$
$\therefore V=Rcos \theta \frac{2 \pi }{T} \, =6400\times 10^{3}\times \frac{1}{2}\times \frac{2 \pi }{24 \cdot 60 \cdot 60}=\frac{2000 \pi }{27}ms^{- 1}$