An artificial satellite is moving in a circular orbit around the...

The escape velocity from earth is given by $v_{e}=\sqrt{2 \mathrm{ g} R_{e}} \ldots$ (i) The orbital velocity of a satellite revolving around earth is given by v_{0}=$\frac{\sqrt{\mathrm{G} M_{e}}}{\left(R_{e}+h\right)}$ where, $M_{e}=$ mass of earth, $R_{e}=$ radius of earth, $h=$ height of satellite from surface of earth. By the relation $\mathrm{G} M_{e}=\mathrm{g} R_e^{2}\text{ So, } \quad$ v_{0}=$\frac{\sqrt{g R_e^{2}}}{\left(R_{e}+h\right)} \ldots$ Dividing equation (i) by (ii), we get $\frac{v_{e}}{v_{0}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{\left(R_{e}\right)}$ Given, $v_{0}=\frac{v_{e}}{2}\frac{2 v_{e}}{v_{e}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{R_{e}}$ Squaring on both side, we get 4=$\frac{2\left(R_{e}+h\right)}{R_{e}}$ or $R_{e}+h=2 R_{e} \quad$ i.e., $h=R_{e}$