Physics - Gravitation Question with Solution | TestHub

The escape velocity from earth is given by $v_e=\sqrt{2\mathrm{g}{R}_e}\dots$ (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 $\mathrm{G}{M}_e=\mathrm{g}{R}_e^2$ $$\text{ So, }{v}_0=\frac{\sqrt{g{R}_e^2}}{\left(R_e+h\right)}\dots$$ 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, $v_0=\frac{v_e}{2}$ $$\frac{2v_e}{v_e}=\frac{\sqrt{2\left(R_e+h\right)}}{R_e}$$ Squaring on both side, we get $$4=\frac{2\left(R_e+h\right)}{R_e}$$ or $R_e+h=2R_e$ i.e., $h=R_e$