The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is (h << R, where R is the radius of the earth)
Solution
At equator, g<sub>1</sub> = g - R${\omega ^2}$
<br><br>At height h, g<sub>2</sub> = $g\left( {1 - {{2h} \over R}} \right)$ [as given h << R]
<br><br>$\because$ Weight same at poles and at h (so g<sub>1</sub>
= g<sub>2</sub>)
<br><br>$\therefore$ g - R${\omega ^2}$ = $g\left( {1 - {{2h} \over R}} \right)$
<br><br>$\Rightarrow$ ${R{\omega ^2} = {{2gh} \over R}}$
<br><br>$\Rightarrow$ ${h = {{{R^2}{\omega ^2}} \over {2g}}}$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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