The acceleration due to gravity would be half of its value at sea level at an altitude

Acceleration due to gravity is given by:
$\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$

Let $g_h$ be the value of acceleration due to gravity at certain height h

${\text{g}}_{\text{h}}=\frac{\text{GM}}{{\left(\text{R}+\text{h}\right)}^2}$ 

${\text{g}}_{\text{h}}=\frac{\text{g}}{\text{2}}$

$\frac{\text{g}}{\text{2}}=\frac{\text{GM}}{{\left(\text{R}+\text{h}\right)}^2}$

$\frac{g}{2}=\frac{GM}{R^2{(1+{\displaystyle \frac{h}{R}})}^2}$

$\frac{\text{g}}{\text{2}}=\frac{{\text{gR}}^2}{{\left(\text{R}+\text{h}\right)}^2}$                                  

$\frac{\text{1}}{\text{2}}=\frac{{\text{R}}^2}{{\left(\text{R}+\text{h}\right)}^2}$

$\frac{\text{1}}{\text{2}}={\left(\frac{\text{R}}{\text{R}+\text{h}}\right)}^2$

$\frac{1}{\sqrt{2}}=\frac{\text{R}}{\text{R}+\text{h}}$

$\text{R}+\text{h}=\sqrt{2}\text{R}$

$\text{h}=\sqrt{2}\text{R}-\text{R}$

$\text{h}=\text{R}\left(\sqrt{2}-1\right)$