Question

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

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

Answer 1

Answer: (C)

The value of acceleration due to gravity at the equator is $g_{E}=g-R\omega ^{2}$ and the acceleration due to gravity at a height h above the pole is $g_{h}=\left(1 - \frac{2 h}{R}\right)g$ . It is given that
$\because mg_{E}=mg_{h}$
$\therefore g_{E}=g_{h}$
$\therefore g-R\left(\omega \right)^{2}=\left(1 - \frac{2 h}{R}\right)g=g-\frac{2 h g}{R}$
$\therefore R\omega ^{2}=\frac{2 h g}{R}$ $\therefore h=\frac{R^{2} \omega ^{2}}{2 g}$