Question

Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

• A. 0.0336 $m/s^2$
• B. $4{\pi }^{2}m/s^2$
• C. 336 $m/s^2$
• D. none of these

Answer 1

The correct option is A

0.0336 $m/s^2$

Earth completes one rotation in 24 hrs
i.e. angle swept =2$\pi$
in time=24 hrs
$\Rightarrow angularvelocity=\omega =\frac{2\pi }{24\times 60\times 60}$
Centripetal acceleration =${\omega }^{2}R$
$\Rightarrow {\left(\frac{2\pi }{24\times 60\times 60}\right)}^2\times 6400\times 1000$
$(\because R=\frac{Diameter}{2})$
$\Rightarrow 0.0336m/s^2$