Question

A rocket is fired 'vertically' from the surface of mars with a speed of 2 km/sec. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far (in km) will the rocket go from the surface of Mars before returning to it?
[Given Mass of mars =$ 6.4 \times 1023 kg$; radius of mars = $3395 km$; $G=6.67 \times 10^{−11} Nm^2kg^{−2}$.]

Answer 1

Hint: The given problem can be solved in a similar manner as we use to calculate the altitude or height of any satellite (rocket) above the earth’s surface.
While solving these types of problems kinetic energy of the satellite (rocket), gravitational energy of the planet system, and gravitational energy of the satellite (rocket) at some height due to its position w.r.t. earth.

Complete step by step answer:
Step 1: The total initial energy of the rocket (satellite) and the planet system can be defined as the sum of the kinetic energy of the rocket and the gravitational energy of the system i.e.,
\[\mathop E\nolimits_i = \dfrac{{80}}{{100}}\dfrac{1}{2}m\mathop V\nolimits^2 + \left( { - \dfrac{{GMm}}{r}} \right)\] ……….. (1)
Where \[\mathop K\nolimits_r = \dfrac{{80}}{{100}}\dfrac{1}{2}m\mathop V\nolimits^2 \] i.e. kinetic energy of the rocket which is reduced by \[20\% \]
\[m = \]mass of rocket, and \[V = \]speed of rocket
And \[\mathop U\nolimits_P = \left( { - \dfrac{{GMm}}{r}} \right)\] i.e. potential energy of the planet system
\[G = 6.67 \times \mathop {10}\nolimits^{ - 11} \] $Nm^2kg^{−2}$ =Universal gravitational constant;
\[M = 6.5 \times \mathop {10}\nolimits^{23} kg\] = mass of Mars, and \[r = 3395km\]= radius of Mars
And, the total final energy will be the sum of gravitational energy of the rocket at height \[H\]from the surface and the kinetic energy of the rocket at the moment when the rocket just starts returning back towards mass.
This kinetic energy at that instant will be zero because the moment when the speed of the rocket will be zero it will start returning back.
So, \[\mathop E\nolimits_t = 0 + \left( { - \dfrac{{GMm}}{{r + H}}} \right)\] ….. (2)
where \[H = \]Height of rocket from the surface of Mars that is to be calculated

Step 2: We know that energy will be conserved in this whole process so total initial energy will be equal to the total final energy i.e., from equation (1) and (2) can be seen-
\[\Rightarrow \dfrac{{80}}{{100}}\dfrac{1}{2}m\mathop V\nolimits^2 + \left( { - \dfrac{{GMm}}{r}} \right) = 0 + \left( { - \dfrac{{GMm}}{{r + H}}} \right)\]
\[\Rightarrow \dfrac{{80}}{{100}}\dfrac{1}{2}m\mathop V\nolimits^2 = GMm\left( {\dfrac{1}{r} - \dfrac{1}{{r + H}}} \right)\]; rearranging the terms in above equation
\[\Rightarrow \dfrac{4}{{10}}\mathop V\nolimits^2 = GM\left( {\dfrac{1}{r} - \dfrac{1}{{r + H}}} \right)\]; mass of rocket \[m\] will cancel out from both the sides
\[\Rightarrow 0.4\dfrac{{\mathop V\nolimits^2 }}{{GM}} = \dfrac{1}{r}\left( {1 - \dfrac{r}{{r + H}}} \right)\]; rearranging the known terms at one side
\[\Rightarrow 0.4\dfrac{{\mathop V\nolimits^2 r}}{{GM}} = \left( {1 - \dfrac{r}{{r + H}}} \right)\]
\[\Rightarrow \dfrac{r}{{r + H}} = 1 - \left( {0.4\dfrac{{\mathop V\nolimits^2 r}}{{GM}}} \right)\]
\[\Rightarrow \dfrac{r}{{r + H}} = 1 - \left( {\dfrac{{0.4 \times \mathop {\left( {2 \times \mathop {10}\nolimits^3 } \right)}\nolimits^2 \times 3.395 \times \mathop {10}\nolimits^6 }}{{6.67 \times \mathop {10}\nolimits^{ - 11} \times 6.4 \times \mathop {10}\nolimits^{23} }}} \right)\]; putting the given values in the equation
\[\Rightarrow \dfrac{r}{{r + H}} = 1 - \left( {\dfrac{{1.6 \times \mathop {10}\nolimits^6 \times 3.395 \times \mathop {10}\nolimits^6 }}{{6.67 \times \mathop {10}\nolimits^{ - 11} \times 6.4 \times \mathop {10}\nolimits^{23} }}} \right)\]
\[\Rightarrow \dfrac{r}{{r + H}} = 1 - \left( {\dfrac{{1.6 \times 3.395}}{{6.67 \times 6.4}}} \right)\]
\[\Rightarrow \dfrac{r}{{r + H}} = 0.873\]; on solving the R.H.S.
\[\Rightarrow r + H = \dfrac{r}{{0.873}}\]
\[\Rightarrow H = \dfrac{r}{{0.873}} - r\]; after rearranging the equation
\[\Rightarrow H = \dfrac{{r - 0.873r}}{{0.873}}\]
\[\Rightarrow H = \dfrac{{0.127 \times 3395 \times \mathop {10}\nolimits^3 }}{{0.873}}\]; putting the value or Mars radius in the equation
\[H = 495.02 \times \mathop {10}\nolimits^3 \]m or \[H = 495.02\]km.

$\therefore $The distance traveled by the rocket is 495.02 km

Note:
(i) The rocket will go up to \[H = 495.02km\] before returning to the surface of Mars.
(ii) While calculating the total final energy it should be remembered that the rocket will return to Mars when its final speed will become zero so kinetic energy at this instant will be zero. If this point is not taken into consideration then the solution will be wrong.