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.
Answers
Answer:
Mass of the spaceship, ms = 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 × 108 kg =2.28 × 1011m
Radius of Mars, r = 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 = -GMms / R
Potential energy of the spaceship due to the gravitational attraction of Mars = -GMmms / r
Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.
Total energy of the spaceship = -GMms / R - GMmms / r
= -Gms[ (M / R) + (mm / 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)
= Gms[ (M / R) + (mm / r) ]
= 6.67 × 10-11 × 103 × [ (2 × 1030 / 2.28 × 1011) + (6.4 × 1023 /3.395 × 106 ) ]
= 596.97 × 109 = 6 × 1011 J.
Answer:
Explanation:
PE of spaceship due to sun = -GMm(s) / R
PE of spaceship die to Mars = -GM(m) m(s)/r
KE = 0
Total Enrgy = -GMm(s) + Gm(s) M(m) /r
= -GM(s)[ M/R + M(m) / r]
Energy = Gm(s) [M/R + M(m) /r]
= 6.67 x 10^-11 x 10^3 [ 2 x 10^30 / 2.28 x 10^11 + 6.4 x 10^23 / 3.395 x 10^6]
= 6.67 x 10^-8 x 89.56 x 10^17
= 596.6 x 10^9
= 6.0 x 10^11 J (aprox)