The value of $h$ will be

Question

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Answer 1

Gravity is related to the angular speed is given by
$g^{^{'}} = g - ω^{2} R$
So the weight at the equator is $m g^{^{'}} = m (g - ω^{2} R)$
Gravity above the earth's surface is given by
$g^{'} = g (1 - \frac{2 h}{R})$
Implies the weight at the equator above the earth's surface is $m g^{^{'}} = m g (1 - \frac{2 h}{R})$
It is given that these two weights are equal,
$\Rightarrow m (g - ω^{2} R) = m g (1 - \frac{2 h}{R})$
$\Rightarrow g - ω^{2} R = g - \frac{2 g h}{R}$

$\Rightarrow ω^{2} R = \frac{2 g h}{R}$

$\Rightarrow h = \frac{ω ^{2} R ^{2}}{2 g}$

Answer 2

Gravity is related to the angular speed is given by

g^{^{'}} = g - ω^{2} R

So the weight at the equator is

m g^{^{'}} = m (g - ω^{2} R)

Gravity above the earth's surface is given by

g^{'} = g (1 - \frac{2 h}{R})

Implies the weight at the equator above the earth's surface is

m g^{^{'}} = m g (1 - \frac{2 h}{R})

It is given that these two weights are equal,

\Rightarrow m (g - ω^{2} R) = m g (1 - \frac{2 h}{R})

\Rightarrow g - ω^{2} R = g - \frac{2 g h}{R}

\Rightarrow ω^{2} R = \frac{2 g h}{R}

\Rightarrow h = \frac{ω ^{2} R ^{2}}{2 g}