Question

A rocket is fired 'vertically' from the surface of mars with a speed of $2km{s}^{-1}$. 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 ? Mass of mars = $6.4\times {10}^{23}kg$; radius of mars = 3395 km; $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer 1

The total initial energy of the rocket and the planet system is the sum of the kinetic energy of the rocket and the gravitational energy of the system.

whereas the final energy will be the gravitational energy at a height H from the surface.

The total energy can be written as:

$\frac{-GMm}{R}+\left(\frac{80}{100}\right)\frac{1}{2}m{V}^2=\frac{-GMm}{R+h}+0$

Simplify the above expression:

$GMm(\frac{1}{R}-\frac{1}{R+h})=0.4m(2\times {10}^3{)}^2$

$\Rightarrow \frac{GM}{R}(1-\frac{R}{R+h})=1.6\times {10}^6$

$=1-\frac{3.395\times {10}^6\times 1.6\times {10}^6}{6.67\times {10}^{-11}\times 6.4\times {10}^{23}}$

$R+h=\frac{R}{0.873}=3,\frac{395}{0.873}=3,888.9km$

So, the height attained by the rocket from the surface will be

$H=3,888.9-3,395=493.9km$