An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. The height of the satellite above the earth’s surface will be $x\times {10}^6$, then the value of $x$ is (Radius of earth = 6400 km)

Given that,

      The orbital velocity of satellite = Escape velocity                 

                                                                                          2

We know that

$v_o=\sqrt{\frac{g{R}^2}{R+h}}$ and $v_e=\sqrt{2gR}$

Putting these value in equation  (i)

$\sqrt{\frac{g{R}^2}{R+h}}=\frac{\sqrt{2gR}}{2}$     

Now, squaring both sides

$\frac{g{R}^2}{R+h}=\frac{2gR}{4}$$\Rightarrow 2g{R}^2=gR(R+h) \Rightarrow 2R=R+h$                                       

R = h ( $\because$R_e=6400 km)

H=R=6400 km