Value of acceleration due to gravity at Earth’s surface is 9.8, m,s^{-2} . altitude above its

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7.Gravitation

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The value of acceleration due to gravity at Earth’s surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$, is close to: (Radius of earth $= 6.4\times10^6\, m$)

A

$6.4\times10^6\, m$

B

$9.0\times10^6\, m$

C

$2.6\times10^6\, m$

D

$1.6\times10^6\, m$

(JEE MAIN-2019)

Solution

$\frac{{GM}}{{{{\left( {R + h} \right)}^2}}} = \frac{{GM}}{{2{R^2}}}$

$R + h = \sqrt 2 R$

$h = \left( {\sqrt 2 – 1} \right)R$

$ \simeq 2.6 \times {10^6}m$

Standard 11

Physics

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