The height above the surface of the earth where acceleration due to gravity becomes ( 9 g ) is ( R is radius of the earth, g is acceleration due to gravity)

g ′ = ( 1 + R h ) 2 g 9 g = ( 1 + R h ) 2 g h = 2 R

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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceEasy2 minPYQ_2022

PhysicsEasysingle choice

The height above the surface of the earth where acceleration due to gravity becomes $(\frac{g}{9})$ is ( $R$ is radius of the earth, $g$ is acceleration due to gravity)

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Solution:

$g^{'} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}} \frac{g}{9} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}} h = 2 R$

Stream:MHT_CETSubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2022

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