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.

Answers (1)

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.