A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship $=1000 \mathrm{~kg}$; mass of the Sun $=2 \times 10^{30} \mathrm{~kg}$; mass of mars $=6.4 \times 10^{23} \mathrm{~kg}$; radius of mars $=3395 \mathrm{~km}$; radius of the orbit of mars $=2.28 \times 10^{8} \mathrm{~kg} ; \mathrm{G}=6.67 \times 10^{-11} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$.
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)
$=\mathrm{G} m_{\mathrm{s}}\left(\frac{M}{R}+\frac{m_{m}}{r}\right)$
$=6.67 \times 10^{-11} \times 10^{3} \times\left(\frac{2 \times 10^{30}}{2.28 \times 10^{11}}+\frac{6.4 \times 10^{23}}{3.395 \times 10^{6}}\right)$
$=6.67 \times 10^{-8}\left(87.72 \times 10^{17}+1.88 \times 10^{17}\right)$
$=596.97 \times 10^{9}$
$=6 \times 10^{11} \mathrm{~J}$