# A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface.

Question:

A rocket is fired vertically with a speed of $5 \mathrm{~km} \mathrm{~s}^{-1}$ from the earth's surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $=6.0 \times 10^{24} \mathrm{~kg}$; mean radius of the earth $=6.4 \times 10^{6} \mathrm{~m}$; $\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.

Solution:

$8 \times 10^{6} \mathrm{~m}$ from the centre of the Earth

Velocity of the rocket, $v=5 \mathrm{~km} / \mathrm{s}=5 \times 10^{3} \mathrm{~m} / \mathrm{s}$

Mass of the Earth, $M_{e}=6.0 \times 10^{24} \mathrm{~kg}$

Radius of the Earth, $R_{e}=6.4 \times 10^{6} \mathrm{~m}$

Height reached by rocket mass, $m=h$

At the surface of the Earth,

Total energy of the rocket $=$ Kinetic energy $+$ Potential energy

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

At highest point $h$,

$v=0$

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

Total energy of the rocket $=0+\left(-\frac{\mathrm{G} M_{e} m}{\mathrm{R}_{e}+h}\right)=-\frac{\mathrm{G} M_{e} m}{\mathrm{R}_{e}+h}$

From the law of conservation of energy, we have

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

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

$\frac{1}{2} v^{2}=\mathrm{GM}_{e}\left(\frac{1}{R_{e}}-\frac{1}{R_{e}+h}\right)$

$=\mathrm{GM}_{e}\left(\frac{R_{e}+h-R_{c}}{R_{c}\left(R_{c}+h\right)}\right)$

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

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

Where $\mathrm{g}=\frac{\mathrm{G} M}{R^{2}}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ (Acceleration due to gravity on the Earth's surface)

$\therefore v^{2}\left(R_{c}+h\right)=2 \mathrm{~g} R h$

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

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

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

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

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

$=R_{e}+h$

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

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