Question

Let $\omega$ be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed by a spring balance gives the same reading at the equator as at a height h above the poles $(h<<R)$. The value of $h$ is

• A. $\frac{{\omega }^2{R}^2}{g}$
• B. $\frac{{\omega }^2{R}^2}{2g}$
• C. $\frac{2{\omega }^2{R}^2}{g}$
• D. $\frac{\sqrt{Rg}}{\omega }$

Answer 1

The correct option is B $\frac{{\omega }^2{R}^2}{2g}$

At equator,

Apparent weight $=$ gravitational force $-$ centrifugal force

$\Rightarrow W_e=mg-m{\omega }^{2}R$

At pole at height $h$, centrifugal force is zero, as angular velocity is zero there.

So, Apparent weight $=$ gravitational force at height $h$.

$\Rightarrow W_p=mg(1-2h/R)$

Equating $W_e=W_p$

$\Rightarrow mg-m{\omega }^{2}R=mg(1-2h/R)$

$\Rightarrow h={\omega }^2{R}^2/2g$