Question

The acceleration due to gravity on the earth's surface at the poles is $g$ and angular veocity of the earth about the axis passing through the pole is $\omega$. An object is weight at the equator and at a height $h$ above the poles by using a spring balance. If the weight are found to be same, then $h$ is
($h<<R$, where $R$, where $R$ is the radius of the earth)

• A. $\frac{R^2{\omega }^2}{2g}$
• B. $\frac{R^2{\omega }^2}{g}$
• C. $\frac{R^2{\omega }^2}{4g}$
• D. $\frac{R^2{\omega }^2}{8g}$

Answer 1

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

Value of $g$ at equator, $g_A=g-R{\omega }^2$
value of $g$ at height $h$ above the pole,

$g_B=g(1-\frac{2h}{R})$
As object weighed equally at the equator and poles, it means $g$ is same at these places.

$g_A=g_B$
$\Rightarrow g-R{\omega }^2=g(1-\frac{2h}{R})$
$\Rightarrow R{\omega }^2=\frac{2gh}{R}\Rightarrow h=\frac{R^2{\omega }^2}{2g}$