The value of acceleration due to gravity at height $h$ from earth surface will become half its value on the surface if (R $=$ radius of earth)

The gravitational force acting on mass m on the surface of Earth is

$F = \dfrac{GMm}{R^2}$

$mg = \dfrac{GMm}{R^2}$

$g = \dfrac{GM}{R^2}$

$gR^2 = GM$.............$(1)$

where, $g$ is acceleration due to gravity, $G$ is gravitational constant, $M$ is mass of Earth and R is radius of Earth.

Now, value of g at height h is

$g = \dfrac{GM}{(R+h)^2}$

As per the problem,

$\dfrac{g}{2} = \dfrac{GM}{(R+h)^2}$

$g(R+h)^2 = 2 gR^2$................from $(1)$

$(R+h)^2 = 2 R^2$

Taking roots from both sides, we get

$R+h = \sqrt 2 R$

$\therefore h = \sqrt 2 R - R$

$\therefore h = (\sqrt 2 - 1)R$