Physics - Gravitation Question with Solution | TestHub

The linear speed of a particle at the equator due to the earth's spin is given by $V=R\omega$, where $R$ is the radius of the earth and $\omega$ is its angular velocity.

At a latitude $\theta$, the radius of the circular path described by the particle is $r=R\cos \theta$.

The linear speed of the particle at latitude $\theta$ is $v=r\omega =(R\cos \theta )\omega =(R\omega )\cos \theta =V\cos \theta$.

Given $\theta =30^{\circ }$, we have:

$v=V\cos 30^{\circ }$

Substituting $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$, we get:

$v=\frac{\sqrt{3}}{2}V$

Answer: $\frac{\sqrt{3}}{2}\mathrm{V}$