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The acceleration due to gravity at height $h$ above the earth if $h \ll R$ (radius of earth) is given by
$g^{\prime}=g\left(1-\frac{2 h}{R}\right)$
$g^{\prime}=g\left(1-\frac{2 h^2}{R^2}\right)$
$g^{\prime}=g\left(1-\frac{h}{2 R}\right)$
$g^{\prime}=g\left(1-\frac{h^2}{2 R^2}\right)$
Solution
For point outside the surface of earth
$g =\frac{ GM }{ r ^2}$
I = distance from center of earth
$\Rightarrow g ( h )=\frac{ GM }{( R + h )^2} \Rightarrow g ( h )=\frac{ GM }{ R ^2\left(1+\frac{ h }{ R }\right)^2}$
$\Rightarrow g ( h )=\frac{ GM }{ R ^2}\left(1+\frac{ h }{ R }\right)^{-2}$
$\text { If } h \ll R ,\left(1+\frac{ h }{ R }\right)^{-2} \approx 1-\frac{2 h }{ R }$
$\Rightarrow g ( h )=\frac{ GM }{ R ^2}\left(1-\frac{2 h }{ R }\right)$
$\Rightarrow g ( h )= g _{\text {surface }}\left(1-\frac{2 h }{ R }\right), \frac{ GM }{ R ^2}= g _{\text {surface }}$
