$A$ rocket is fired vertically with a speed of $5 \; km/s$ from the Earth's surface. How far from the Earth's surface does the rocket go before returning? (Mass of the Earth $M_e = 6.0 \times 10^{24} \; kg$,mean radius of the Earth $R_e = 6.4 \times 10^{6} \; m$,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$)

(A) Let the mass of the rocket be $m$ and the height reached be $h$.
At the Earth's surface,the total energy $E_i$ is the sum of kinetic and potential energy:
$E_i = \frac{1}{2} m v^2 - \frac{G M_e m}{R_e}$
At the highest point $h$,the velocity $v = 0$,so the total energy $E_f$ is purely potential:
$E_f = -\frac{G M_e m}{R_e + h}$
By the law of conservation of energy,$E_i = E_f$:
$\frac{1}{2} m v^2 - \frac{G M_e m}{R_e} = -\frac{G M_e m}{R_e + h}$
Dividing by $m$ and rearranging:
$\frac{v^2}{2} = G M_e \left( \frac{1}{R_e} - \frac{1}{R_e + h} \right) = G M_e \left( \frac{h}{R_e(R_e + h)} \right)$
Using $g = \frac{G M_e}{R_e^2}$,we have $G M_e = g R_e^2$:
$\frac{v^2}{2} = \frac{g R_e^2 h}{R_e(R_e + h)} = \frac{g R_e h}{R_e + h}$
Solving for $h$:
$v^2(R_e + h) = 2 g R_e h \implies v^2 R_e = h(2 g R_e - v^2)$
$h = \frac{R_e v^2}{2 g R_e - v^2}$
Substituting values $v = 5 \times 10^3 \; m/s$,$R_e = 6.4 \times 10^6 \; m$,$g = 9.8 \; m/s^2$:
$h = \frac{6.4 \times 10^6 \times (5 \times 10^3)^2}{2 \times 9.8 \times 6.4 \times 10^6 - (5 \times 10^3)^2}$
$h = \frac{6.4 \times 10^6 \times 25 \times 10^6}{125.44 \times 10^6 - 25 \times 10^6} = \frac{160 \times 10^{12}}{100.44 \times 10^6} \approx 1.593 \times 10^6 \; m \approx 1.6 \times 10^6 \; m$.