The value of the acceleration due to gravity is g _ 1 at a height h...

Given that

${g}_{n} = {g}_{d}$

$\dfrac{GM}{( R + h{)}^{2}} = \dfrac{GM}{{R}^{3}} ( R - d )$

$\dfrac{GM}{( R + R / 2{)}^{2}} = \dfrac{GM}{{R}^{3}} ( R - d )$

$\dfrac{4GM}{9{R}^{2}} = \dfrac{GM}{{R}^{2}}\left(1 - \dfrac{d}{R}\right)$

$\dfrac{4}{9} = 1 - \dfrac{d}{R}$

$\dfrac{d}{R} = 1 - \dfrac{4}{9} = \dfrac{5}{9}$

$d = \dfrac{5}{9}R$