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.
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
Potential energy of the spaceship due to the gravitational attraction of Mars
Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.
Total energy of the spaceship = `(-GMm)/r = (-GM_sm_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)`
`=6.67xx10^(-11)xx 10^3 xx (2xx10^30)/(2.28xx10^(11))+ (6.4xx10^(23))/(3.395xx10^6)`
Let R be the radius of orbit of Mars and R’ be the radius of the Mars. M be the mass of the Sun and M’ be the mass of Mars. If m is the mass of the space-ship, then Potential energy of space-ship due to gravitational attraction of the Sun = – GM m/R Potential energy of space-ship due to gravitational attraction of Mars = – G M’ m/R’ Since the K.E. of space ship is zero, therefore
total energy of spaceship =` -(GMm)/R - (GM'm)/R = -Gm(M/R + (M')/(R'))`
:. Energy required to rocket out the spaceship from the solar system = -(total energy of spaceship)
`=-[-Gm(M/R + (M')/R')] = Gm [M/R+(M')/R]`
`= 6.67xx10^(-11)xx1000xx [(2xx10^(30))/(2.28xx10^(11) )+(6.4xx10^(23))/(3395xx10^(3))]`
`= 6.67xx10^(-8) [20/2.28 + 6.4/33.95] xx 10^(18) J = 5.98 xx 10^(11) J`
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