A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1.

Question:

A rocket is fired 'vertically' from the surface of mars with a speed of $2 \mathrm{~km} \mathrm{~s}-1$. If $20 \%$ of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars $=6.4 \times 1023 \mathrm{~kg}$; radius of mars $=3395$ $\mathrm{km} ; \mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.

Solution:

Initial velocity of the rocket, $v=2 \mathrm{~km} / \mathrm{s}=2 \times 10^{3} \mathrm{~m} / \mathrm{s}$

Mass of Mars, $M=6.4 \times 10^{23} \mathrm{~kg}$

Radius of Mars, $R=3395 \mathrm{~km}=3.395 \times 10^{6} \mathrm{~m}$

Universal gravitational constant, $\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$

Mass of the rocket $=m$

Initial kinetic energy of the rocket $=\frac{1}{2} m v^{2}$

Initial potential energy of the rocket $=\frac{-\mathrm{GMm}}{R}$

Total initial energy $=\frac{1}{2} m v^{2}-\frac{\mathrm{GMm}}{R}$

If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.

Total initial energy available $=\frac{80}{100} \times \frac{1}{2} m v^{2}-\frac{\mathrm{G} M m}{R}=0.4 m v^{2}-\frac{\mathrm{G} M m}{R}$

Maximum height reached by the rocket = h

At this height, the velocity and hence, the kinetic energy of the rocket will become zero.

Total energy of the rocket at height $h=-\frac{\mathrm{GMm}}{(R+h)}$

Applying the law of conservation of energy for the rocket, we can write:

$0.4 m v^{2}-\frac{\mathrm{G} M m}{R}=\frac{-\mathrm{G} M m}{(R+h)}$

$0.4 v^{2}=\frac{\mathrm{G} M}{R}-\frac{\mathrm{G} M}{R+h}$

$=\mathrm{GM}\left(\frac{1}{R}-\frac{1}{R+h}\right)$

$=\mathrm{G} M\left(\frac{R+h-R}{R(R+h)}\right)$

$=\frac{G M h}{R(R+h)}$

$\frac{R+h}{h}=\frac{\mathrm{GM}}{0.4 v^{2} R}$

$\frac{R}{h}+1=\frac{\mathrm{GM}}{0.4 v^{2} R}$

$\frac{R}{h}=\frac{G M}{0.4 v^{2} R}-1$

$h=\frac{R}{\mathrm{GM}}$

$\frac{R}{h}=\frac{\mathrm{G} M}{0.4 v^{2} R}-1$

$h=\frac{R}{\frac{\mathrm{GM}}{0.4 v^{2} R}-1}$

$=\frac{0.4 R^{2} v^{2}}{G M-0.4 v^{2}}$

$=\frac{0.4 R^{2} v^{2}}{\mathrm{GM}-0.4 v^{2} R}$

$=\frac{0.4 \times\left(3.395 \times 10^{6}\right)^{2} \times\left(2 \times 10^{3}\right)^{2}}{6.67 \times 10^{-11} \times 6.4 \times 10^{23}-0.4 \times\left(2 \times 10^{3}\right)^{2} \times\left(3.395 \times 10^{6}\right)}$

$=\frac{18.442 \times 10^{18}}{42.688 \times 10^{12}-5.432 \times 10^{12}}=\frac{18.442}{37.256} \times 10^{6}$

$=495 \times 10^{3} \mathrm{~m}=495 \mathrm{~km}$