Height reached by rocket mass, $m=h$

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

$=\frac{1}{2} m v^{2}+\left(\frac{-G M_{c} m}{R_{e}}\right)$

At highest point $h$

$v=0$

And, Potential energy $=-\frac{ G M_{e} m}{R_{e}+h}$

$=0+\left(-\frac{ G M_{e} m}{ R _{e}+h}\right)=-\frac{ G M_{e} m}{ R _{e}+h}$

Total energy of the rocketi

From the law of conservation of energy, we have

Total energy of the rocket at the Earth's surface = Total energy at height $h$

$\frac{1}{2} m v^{2}+\left(-\frac{G M_{e} m}{R_{e}}\right)=-\frac{G M_{e} m}{R_{e}+h}$

$\frac{1}{2} v^{2}= G M_{e}\left(\frac{1}{R_{e}}-\frac{1}{R_{e}+h}\right)$

$= G M_{e}\left(\frac{R_{e}+h-R_{e}}{R_{e}\left(R_{e}+h\right)}\right)$

$\frac{1}{2} v^{2}=\frac{G M_{e} h}{R_{e}\left(R_{e}+h\right)} \times \frac{R_{e}}{R_{e}}$

$\frac{1}{2} \times v^{2}=\frac{g R_{e} h}{R_{e}+h}$

Where $g=\frac{G M}{R_{c}^{2}}=9.8 m / s ^{2}$ (Acceleration due to gravity on the Earth's surface) $\therefore v^{2}\left(R_{r}+h\right)=2 g R, h$

$v^{2} R_{t}=h\left(2 g R_{t}-v^{2}\right)$

$h=\frac{R_{e} v^{2}}{2 g R_{e}-v^{2}}$

$=\frac{6.4 \times 10^{6} \times\left(5 \times 10^{3}\right)^{2}}{2 \times 9.8 \times 6.4 \times 10^{6}-\left(5 \times 10^{3}\right)^{2}}$

$h=\frac{6.4 \times 25 \times 10^{12}}{100.44 \times 10^{6}}=1.6 \times 10^{6} m$

Height achieved by the rocket with respect to the centre of the Earth

$=R_{e}+h$

$=6.4 \times 10^{6}+1.6 \times 10^{6}$

$=8.0 \times 10^{6} m$