BITSAT 2015 Paper

The escape velocity from earth is given by $v_e$ = $\sqrt{2gR_e}$ ... (i) The orbital velocity of a satellite revolving around earth is given by $v_0$ = $\frac{\sqrt{GM^e}}{R_e+h}$ where, $M_e$ = mass of earth, $R_e$ = radius of earth, h = height of satellite from surface of earth. By the relation $GM_e$ = $gR_e^2$ So, $v_0$ = $\frac{\sqrt{gR^2_e}}{R_e+h}$ ... (ii) Dividing equation (i) by (ii), we get $\frac{v-e}{v_0}$ = $\frac{\sqrt{2(R_e+h)}}{R_e}$ Given, $v_0$ = $\frac{v_e}{2}$ $\frac{2v_e}{v_e}$ = $\frac{\sqrt{2(R_e+h)}}{R_e}$ Squaring on both side, we get 4 = $\frac{2(R_e+h)}{R_e}$ or $R_e$ + h = $2R_e$ , i.e. h = $R_e$