Potential energy of spaceship due to gravitational attraction of the Sun $= -\frac{G M_{S} m}{R}$ Here, $M_{S}=$ Mass of Sun, m = Mass of spaceship, R = radius of orbit of mars Total potential energy of spaceship, $U = -\frac{G M_{S} m}{R} - \frac{G M_{m} m}{R_{m}}$ $= -G m \left[ \frac{M_{S}}{R} + \frac{M_{m}}{R_{m}} \right]$ Here, $M_{m}=$ mass of Mars $R_{m}=$ radius of mars $= \left[ -G m \left[ \frac{M_{S}}{R} + \frac{M_{m}}{R_{m}} \right] \right]$ Let $v$ be the orbital velocity of Mars, $\frac{m v^{2}}{r_{0}} = \frac{G M m}{r_{0}^{2}}$ or $v^{2} = \frac{G M}{r_{0}}$ Since, velocity of spaceship is same as that of Mars, therefore, K.E. of spaceship $= \frac{1}{2}(m) v^{2} = \frac{1}{2}(m) \frac{G M}{r_{0}}$ $= \frac{G M (m)}{2 r_{0}}$ or, Total energy of the spaceship, i.e., Negative energy denotes that the spaceship is bound to the solar system.Thus, energy needed by the spaceship to escape from the solar system $= 3.1 \times 10^{11} \, \text{J}$