# Q: 8.25 A rocket is fired ‘vertically’ from the surface of mars with a speed of $\small 2\hspace{1mm}km\hspace{1mm}s^-^1$. If  $\small 20\%$​​​ 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 $\small =6.4\times 10^2^3\hspace{1mm}kg$; radius of mars $\small =3359\hspace{1mm}km$;              $\small G=6.67\times 10^-^1^1\hspace{1mm}Nm^2kg^-^2$.

The kinetic energy of the rocket is (initial):-

$=\ \frac{1}{2}mv^2$

And the initial potential energy is :

$=\ \frac{-GMm}{R}$

Thus total initial energy is given by :

$=\ \frac{1}{2}mv^2\ +\ \frac{-GMm}{R}$

Further, it is given that 20 per cent of kinetic energy is lost.

So the net initial energy is :

$=\ 0.4mv^2\ -\ \frac{GMm}{R}$

The final energy is given by :

$=\ \frac{GMm}{R\ +\ h}$

Using the law of energy conservation we get :

$0.4mv^2\ -\ \frac{GMm}{R}\ =\ \frac{GMm}{R\ +\ h}$

Solving the above equation we get :

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

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

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

or                                               $=\ 495\times 10^3\ m$

or                                              $=\ 495\ Km$

Thus the required distance is 495 Km.

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