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 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 $=3395km;G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer 1

Total initial energy.

Let us assume, $m=$ Mass of rocket

$M=$ Mass of mars, and $R=$ Radius of mars
Initial $KE$ of the rocket $=\frac{1}{2}m{v}^2$
Initial $PE$ of the rocket $=-\frac{GMm}{R}$


Therefore, the total energy is given by, $TE=\frac{1}{2}m{v}^2-\frac{GMm}{R}$


Total energy available.

Since $20\%$ of the $KE$ is lost, only $80\%$ can be used to reach the height.

Total energy available $=\frac{4}{5}\times \frac{1}{2}m{v}^2-\frac{GMm}{R}$
$=0.4m{v}^2-\frac{GMm}{R}$



Total energy at highest point.

At the highest point $h$ above the surface,
its $KE$ is zero and $PE=-\frac{GMm}{R+h}$


Apply principle of conservation of energy

$-\frac{GMm}{R+h}=0.4m{v}^2-\frac{GMm}{R}$
$\Rightarrow \frac{GM}{R+h}=\frac{GM}{R}-0.4v^2$
$\Rightarrow \frac{GM}{R+h}=\frac{1}{R}(GM-0.4R{v}^2)$


$\Rightarrow \frac{R+h}{R}=\frac{GM}{(GM-0.4R{v}^2)}$

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

$\Rightarrow \frac{h}{R}=\frac{0.4R{v}^2}{(GM-0.4R{v}^2)}$

$\Rightarrow h=\frac{0.4R^2{v}^2}{(GM-0.4R{v}^2)}$

$\Rightarrow h=495km$

$\text{Final Answer}:495km$