Let ω 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 give

Question

Let ω 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 ?

TestHub · Accepted Answer

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

$∵ m g_{E} = m g_{h}$
$∴ g_{E} = g_{h}$

$∴ g - R ω^{2} = (1 - \frac{2 h}{R}) g = g - \frac{2 h g}{R}$

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

Physics - Gravitation Question with Solution | TestHub

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JEE Main 2027 (Class 12)