The acceleration due to gravity at height h a | vedclass

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Question

For point outside the surface of earth

$g =\frac{ GM }{ r ^2}$

I = distance from center of earth

$\Rightarrow g ( h )=\frac{ GM }{( R + h )^2

Vedclass · Accepted Answer

$g^{\prime}=g\left(1-\frac{2 h}{R}\right)$

Vedclass · Answer

For point outside the surface of earth

$g =\frac{ GM }{ r ^2}$

I = distance from center of earth

$\Rightarrow g ( h )=\frac{ GM }{( R + h )^2} \Rightarrow g ( h )=\frac{ GM }{ R ^2\left(1+\frac{ h }{ R }ight)^2}$

$\Rightarrow g ( h )=\frac{ GM }{ R ^2}\left(1+\frac{ h }{ R }ight)^{-2}$

$	ext { If } h \ll R ,\left(1+\frac{ h }{ R }ight)^{-2} \approx 1-\frac{2 h }{ R }$

$\Rightarrow g ( h )=\frac{ GM }{ R ^2}\left(1-\frac{2 h }{ R }ight)$

$\Rightarrow g ( h )= g _{	ext {surface }}\left(1-\frac{2 h }{ R }ight), \frac{ GM }{ R ^2}= g _{	ext {surface }}$

The acceleration due to gravity at height $h$ above the earth if $h \ll R$ (radius of earth) is given by

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