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Q.
The weight of an object at the equator is the same as its weight at height $h$ above the pole. If $\omega$ is the rotational speed of Earth, then the value of $h$ is
Gravitation
Solution:
$g=g_{p}-\omega^{2} R \cos ^{2} \lambda$
At equator, $\lambda=0$
$\Rightarrow g _{e q}=g_{p}-\omega^{2} R \cos ^{2} 0=g_{p}-\omega^{2} \cdot R$
Now, $g^{\prime}$ above the pole is
$g^{\prime}=g_{p}\left(1-\frac{2 h}{R}\right)$
As $g_{e q}=g^{\prime}$
$g_{p}-\omega^{2} R \quad=g_{p}\left(1-\frac{2 h}{R}\right)$
So, $\omega^{2} R=\frac{2 h \cdot g_{p}}{R} $
$\Rightarrow h=\frac{\omega^{2} \cdot R^{2}}{2 g_{p}}=\frac{\omega^{2} \cdot R^{2}}{2 g}$