The height at which the weight of the body becomes ( 9 1 ) th its weight on the surface of earth is ( R = radius of earth)

∴ ∴ ∴ W h = 9 W mg h = 9 mg g h = 9 g ( R + h ) 2 GM = 9 R 2 GM R + h = 3 R h = 2 R

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PhysicsGravitationField & ForceMedium2 minPYQ_2023

PhysicsMediumsingle choice

The height at which the weight of the body becomes $(\frac{1}{9})^{th}$ its weight on the surface of earth is ( $R =$ radius of earth)

Answer:

Solution:

$∴ ∴ ∴ W_{h} = \frac{W}{9} mg_{h} = \frac{mg}{9} g_{h} = \frac{g}{9} \frac{GM}{( R + h ) ^{2}} = \frac{GM}{9 R ^{2}} R + h = 3 R h = 2 R$

Stream:MHT_CETSubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2023

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