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

g ′ = GM R + h 2 (acceleration due to gravity at height h) ⟹ g 9 = GM R 2 ⋅ R 2 R + h 2 = g R R + h 2 ⟹ 1 9 = R R + h 2 ⟹ R R + h = 1 3 ⟹ 3 R = R + h ⟹ 2 R = h

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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceEasy2 minPYQ_2020

PhysicsEasysingle choice

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

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Answer:

Solution:

$g^{'} = \frac{GM}{{(R + h)}^{2}}$ (acceleration due to gravity at height h)

$⟹ \frac{g}{9} = \frac{GM}{R^{2}} \cdot \frac{R^{2}}{{(R + h)}^{2}}$

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

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

$⟹ \frac{R}{R + h} = \frac{1}{3}$

$⟹ 3 R = R + h$

$⟹ 2 R = h$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2020

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