__Question 8.24__:^{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}.

__Solution__:*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 = -GMm

_{s}/ R

Potential energy of the spaceship due to the gravitational attraction of Mars = -GM

_{}

_{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 = -GMm

_{s}/ R - GM

_{m}m

_{s}/ r

= -Gm

_{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)

= Gm

_{s}[ (M / R) + (m

_{m}/ r) ]

= 6.67 X 10

^{-11 }X 10

^{3}X [ (2 X 10

^{30}/ 2.28 X 10

^{11}) + (6.4 X 10

^{23}/ 3.395 X 10

^{6}) ]

=

^{}

^{}596.97 X 10

^{9}= 6 X 10

^{11}J

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