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(B) The acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$.At a height $h$ above the surface,the acceleration due to grav

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$(\sqrt{2}-1) R$

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(B) The acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$.At a height $h$ above the surface,the acceleration due to gravity $g'$ is given by $g' = \frac{GM}{(R+h)^2}$.Dividing the two expressions,we get $\frac{g'}{g} = \left(\frac{R}{R+h}\right)^2$.Given that $g' = \frac{g}{2}$,we substitute this into the equation:$\frac{1}{2} = \left(\frac{R}{R+h}\right)^2$.Taking the square root on both sides:$\frac{1}{\sqrt{2}} = \frac{R}{R+h}$.Rearranging the terms:$R+h = R\sqrt{2}$.$h = R(\sqrt{2}-1)$.

The acceleration due to gravity becomes $\left(\frac{g}{2}\right)$ ($g =$ acceleration due to gravity on the surface of the earth) at a height equal to

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