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A rocket is fired 'vertically' from the surface of Mars with a speed of $2 \mathrm{~km} / \mathrm{s}$. If $20 \%$ of its initial K.E. 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 $=6.4 \times 10^{23} \mathrm{~kg}$; radius of Mars $=3395 \mathrm{~km} . \mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2$
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Verified Answer
Let $m=$ mass of the rocket; $M=$ mass of Mars;
$R=$ Radius of Mars
$v=$ initial velocity of the rocket.
Total initial energy of the rocket $=$ K.E. $+$ P.E.
$$
=\frac{1}{2} m v^2-\frac{G M m}{R}
$$
$\because \quad 20 \%$ of the K.E. is lost
$\therefore$ Energy available to the rocket $=\frac{80}{100} \times \frac{1}{2} m v^2-\frac{G M m}{R}=0.4 m v^2-\frac{G M m}{R}$
If the rocket reaches at a height $h$ above the surface of Mars, at this point K.E. $=0$ and P.E. $=\frac{-G M m}{R+h}$
$\therefore$ From law of conservation of energy,
$$
\begin{aligned}
&0.4 m v^2-\frac{G M m}{R}=-\frac{G M m}{R+h} \\
&\frac{G M}{R+h}=\frac{G M}{R}-0.4 v^2=\frac{1}{R}\left[G M-0.4 R v^2\right] \\
&\frac{R+h}{G M}=\frac{R}{G M-0.4 R v^2}
\end{aligned}
$$
$\Rightarrow \quad R+h=\frac{G M R}{G M-0.4 R v^2}$
$\Rightarrow \quad h=\frac{G M R}{G M-0.4 R v^2}-R$
$h=R\left[\frac{G M}{G M-0.4 R v^2}-1\right]=3395 \times 10^3$
$\left[\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{6.67 \times 10^{-11} \times 6.4 \times 10^{23}-0.4 \times 3395 \times 10^3 \times\left(2 \times 10^3\right)^2}-1\right]$
$=42$
$=3395 \times 10^3 \times 0.1458$
$\Rightarrow h=495 \times 10^3 \mathrm{~m}=495 \mathrm{~km}$.
$R=$ Radius of Mars
$v=$ initial velocity of the rocket.
Total initial energy of the rocket $=$ K.E. $+$ P.E.
$$
=\frac{1}{2} m v^2-\frac{G M m}{R}
$$
$\because \quad 20 \%$ of the K.E. is lost
$\therefore$ Energy available to the rocket $=\frac{80}{100} \times \frac{1}{2} m v^2-\frac{G M m}{R}=0.4 m v^2-\frac{G M m}{R}$
If the rocket reaches at a height $h$ above the surface of Mars, at this point K.E. $=0$ and P.E. $=\frac{-G M m}{R+h}$
$\therefore$ From law of conservation of energy,
$$
\begin{aligned}
&0.4 m v^2-\frac{G M m}{R}=-\frac{G M m}{R+h} \\
&\frac{G M}{R+h}=\frac{G M}{R}-0.4 v^2=\frac{1}{R}\left[G M-0.4 R v^2\right] \\
&\frac{R+h}{G M}=\frac{R}{G M-0.4 R v^2}
\end{aligned}
$$
$\Rightarrow \quad R+h=\frac{G M R}{G M-0.4 R v^2}$
$\Rightarrow \quad h=\frac{G M R}{G M-0.4 R v^2}-R$
$h=R\left[\frac{G M}{G M-0.4 R v^2}-1\right]=3395 \times 10^3$
$\left[\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{6.67 \times 10^{-11} \times 6.4 \times 10^{23}-0.4 \times 3395 \times 10^3 \times\left(2 \times 10^3\right)^2}-1\right]$
$=42$
$=3395 \times 10^3 \times 0.1458$
$\Rightarrow h=495 \times 10^3 \mathrm{~m}=495 \mathrm{~km}$.
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