A spaceship is stationed on Mars.

Solution:

Mass of the spaceship, $m_{\mathrm{s}}=1000 \mathrm{~kg}$

Mass of the Sun, $M=2 \times 10^{30} \mathrm{~kg}$

Mass of Mars, $m_{m}=6.4 \times 10^{23} \mathrm{~kg}$

Orbital radius of Mars, $R=2.28 \times 10^{8} \mathrm{~kg}=2.28 \times 10^{11} \mathrm{~m}$

Radius of Mars, $r=3395 \mathrm{~km}=3.395 \times 10^{6} \mathrm{~m}$

Universal gravitational constant, $\mathrm{G}=6.67 \times 10^{-11} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$

Potential energy of the spaceship due to the gravitational attraction of the Sun $=\frac{-G M m_{s}}{R}$

Potential energy of the spaceship due to the gravitational attraction of Mars $=\frac{-G M_{m} m_{\mathrm{s}}}{r}$

Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.

Total energy of the spaceship $=\frac{-\mathrm{G} M m_{\mathrm{s}}}{R}-\frac{-\mathrm{G} M_{\mathrm{s}} m_{\mathrm{m}}}{r}$

$=-\mathrm{G} m_{\mathrm{s}}\left(\frac{M}{R}+\frac{m_{\mathrm{m}}}{r}\right)$

The negative sign indicates that the system is in bound state.

Energy required for launching the spaceship out of the solar system

= – (Total energy of the spaceship)

$=\mathrm{G} m_{\mathrm{s}}\left(\frac{M}{R}+\frac{m_{m}}{r}\right)$

$=6.67 \times 10^{-11} \times 10^{3} \times\left(\frac{2 \times 10^{30}}{2.28 \times 10^{11}}+\frac{6.4 \times 10^{23}}{3.395 \times 10^{6}}\right)$

$=6.67 \times 10^{-8}\left(87.72 \times 10^{17}+1.88 \times 10^{17}\right)$

$=596.97 \times 10^{9}$

$=6 \times 10^{11} \mathrm{~J}$