### Q.24:- 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 kg; mass of the Sun = 2 × 10^{30} kg; mass of mars = 6.4 × 10^{23} kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10^{8}kg; G= 6.67 × 10^{–11} m^{2}kg^{–2}.

**Answer:-**Mass of the spaceship, *m*_{s }= 1000 kg

Mass of the Sun, *M* = 2 × 10^{30} kg

Mass of Mars, *m*_{m} = 6.4 × 10 ^{23} kg

Orbital radius of Mars, *R* = 2.28 × 10^{8 }kg =2.28 × 10^{11}m

Radius of Mars, *r *= 3395 km = 3.395 × 10^{6} m

Universal gravitational constant, G = 6.67 × 10^{–11} m^{2}kg^{–2}

Potential energy of the spaceship due to the gravitational attraction of the Sun = -G*Mm*_{s} / *R*

Potential energy of the spaceship due to the gravitational attraction of Mars = -G*M*_{m}*m*_{s} / *r*

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

Total energy of the spaceship = -G*Mm*_{s} / *R* **–** G*M*_{m}*m*_{s} / *r*

= -G*m*_{s}[ (*M* / *R*) + (*m*_{m} / *r*) ]

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)

= G*m*_{s}[ (*M* / *R*) + (*m*_{m} / *r*) ]

= 6.67 × 10^{-11 }× 10^{3} × [ (2 × 10^{30} / 2.28 × 10^{11}) + (6.4 × 10^{23} / 3.395 × 10^{6} ) ]

= 596.97 × 10^{9} = 6 × 10^{11} J.