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Motion In Straight Line

Question
CBSEENPH11020165

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 × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108kg; G= 6.67 × 10–11 m2kg–2.

Solution
Mass of the spaceship, m= 1000 kg
Mass of the Sun, M = 2 × 1030 kg 
Mass of Mars, mm = 6.4 × 10 23 kg 
Orbital radius of Mars, R = 2.28 × 10kg
                                    =2.28 × 1011
Radius of Mars, = 3395 km
                          = 3.395 × 106 m 
Universal gravitational constant, G = 6.67 × 10–11 m2kg–2 

Potential energy of the spaceship due to the gravitational attraction of the Sun = negative GMm subscript straight s over straight R

Potential energy of the spaceship due to the gravitational attraction of Mars = -fraction numerator GM subscript straight m straight m subscript straight s over denominator straight r end fraction 

Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero. 
Total energy of the spaceship =  fraction numerator negative G M m subscript straight s over denominator straight R end fraction space minus space fraction numerator G M subscript straight m straight m subscript straight s over denominator straight r end fraction
                                            = negative space Gm subscript straight s space open square brackets straight M over straight R plus straight m subscript straight m over straight r close square brackets 
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) 
= space Gm subscript straight s space open square brackets straight M over straight R plus straight m subscript straight m over straight r close square brackets 
= 6.67 × 1011 × 103 ×open square brackets open parentheses fraction numerator 2 space straight x space 10 to the power of 30 over denominator 2.28 space straight x space 10 to the power of 11 end fraction close parentheses space plus space open parentheses fraction numerator 6.4 space x space 10 to the power of 23 over denominator 3.395 space x space 10 to the power of 6 end fraction close parentheses close square brackets
= 596.97 × 109 
=  6 × 1011 J.