An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height (h) of the satellite above the earth’s surface is (Take radius of earth as Re).

Given that the satellite's velocity is half that of escape velocity, we will apply the formulas for escape velocity and satellite velocity to determine the relationship between the satellite's height and earth's radius.

Formulae used:
Escape velocity, $v_e=\sqrt{2g{R}_e}$
orbital speed of the satellite, $v_s=\frac{\sqrt{G{M}_e}}{R_e+h}$
Gravitational acceleration $g=\frac{G{M}_e}{R_e^2}$

The equation for the Earth's escape velocity is

${\mathrm{v}}_{\mathrm{e}}=\sqrt{2{\mathrm{gRR}}_{\mathrm{e}}\dots ..\left(\mathrm{i}\right)}$
A satellite's orbital velocity as it revolves around the planet is given by
${\mathrm{v}}_0=\frac{\sqrt{{\mathrm{GM}}_{\mathrm{e}}}}{\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}$
where, $M_e=$ mass of earth, $R_e=$ radius of earth, $h=$ height of satellite from surface of earth.
By the relation ${\mathrm{GM}}_e={\mathrm{gR}}_{\mathrm{e}}^2$
So, ${\mathrm{v}}_0=\frac{\sqrt{{\mathrm{gR}}_{\mathrm{e}}^2}}{\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}\dots .$. (ii)
The result of dividing equation I by (ii) is

$\frac{{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{v}}_0}=\frac{\sqrt{2\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}}{\left({\mathrm{R}}_{\mathrm{e}}\right)}$
Given, $v_0=\frac{v_e}{2}$
$\frac{2{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{v}}_{\mathrm{e}}}=\frac{\sqrt{2\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}}{{\mathrm{R}}_{\mathrm{e}}}$
By squaring both sides, we obtain $4=\frac{2\left({\mathrm{R}}_{\mathrm{e}}+\mathrm{h}\right)}{{\mathrm{R}}_{\mathrm{e}}}$ or $R_e+h=2R_e$.
$\Rightarrow h=R_e$

Hence, the correct answer is option B.