The orbital velocity of an artificial satellite in a circular orbit just above the earth’s surface is $v$. For a satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity will be

The correct option is C
$\sqrt{\frac{2}{3}}v$
We know that $v=\sqrt{\frac{GM}{R+h}}$ where $h$ is the height of the satellite, $v$ is the orbital velocity of satellite, $M$ is mass of Earth and $R$ is radius of Earth.
When the satellite is orbiting just above the surface of the earth $(h=0)$
$v=\sqrt{\frac{GM}{R}}$
When the satellite is orbiting at an altitude of half of the earth’s radius i.e $h=\frac{h}{2}$
$\Rightarrow v_o=\sqrt{\frac{GM}{R+\frac{R}{2}}}\Rightarrow v_o=\sqrt{\frac{2}{3}\frac{GM}{R}}\Rightarrow v_o=\sqrt{\frac{2}{3}}v$