91Ó°ÊÓ

Given that a rocket is fired from rest at a space station in deep space, where gravity is negligible.

$\mathrm{u}=0$

The relative speed of gas ejection, ${\mathrm{v}}_{\mathrm{ex}}=2000\mathrm{m}/\mathrm{s}$

In rocket propulsion, the mass of a rocket changes as the fuel is used up and ejected from the rocket. Analysis of the motion must contain the momentum carried away by the spent fuel as well as the momentum of the rocket itself.

The initial velocity is u and exhaust speed is v_x.

Formula for change in speed in terms of final mass as the fraction of initial mass is

$\mathrm{v}-\mathrm{u}={\mathrm{v}}_{\mathrm{ex}}\ln \left({\mathrm{m}}_{\mathrm{o}}/\mathrm{m}\right)$

The desired speed of rocket

$\begin{array}{rcl}\mathrm{v}& =& 1×{10}^{-3}\mathrm{c}\\ & =& 3×{10}^5\mathrm{m}/\mathrm{s}\end{array}$

Now,

$\begin{array}{rcl}\mathrm{v}-\mathrm{u}& =& {\mathrm{v}}_{\mathrm{ex}}\ln \left({\mathrm{m}}_{\mathrm{o}}/\mathrm{m}\right)\\ \mathrm{m}/{\mathrm{m}}_{\mathrm{o}}& =& {\mathrm{e}}^{-\mathrm{v}/{\mathrm{v}}_{\mathrm{ex}}}\end{array}$

Where $\mathrm{m}/{\mathrm{m}}_{\mathrm{o}}$ is the initial mass that is not fuel.

$\begin{array}{rcl}\mathrm{m}/{\mathrm{m}}_{\mathrm{o}}& =& {\mathrm{e}}^{-\left(3×{10}^5\right)/\left(2000\right)}\\ & =& 7.2×{10}^{-66}\end{array}$

Hence, the fraction of the initial mass of the rocket is $7.2×{10}^{-66}$.

Since, $\mathrm{m}/{\mathrm{m}}_{\mathrm{o}}={\mathrm{e}}^{-\mathrm{v}/{\mathrm{v}}_{\mathrm{ex}}}$

$\begin{array}{rcl}\mathrm{m}/{\mathrm{m}}_{\mathrm{o}}& =& {\mathrm{e}}^{-\left(3000\right)/\left(2000\right)}\\ & =& {\mathrm{e}}^{-1.5}\\ & =& 0.223\end{array}$

Therefore, the fraction of the initial mass of the rocket and fuel is not fuel is $0.223$.