# 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 - Physics

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 1

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 × 1011m

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

= (-GMm)/R

Potential energy of the spaceship due to the gravitational attraction of Mars

= (-GM_mm_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)/r = (-GM_sm_m)/r

= -Gm_s(M/R+m_m/r)

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)

=6.67xx10^(-8)(87.72xx10^(17)+1.88xx10^(17))

=6.67xx10^(-8)xx89.50xx10^(17)

=596.97xx10^9

=6xx10^(11)J

#### Solution 2

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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Chapter 8: Gravitation - Exercises [Page 202]

#### APPEARS IN

NCERT Class 11 Physics
Chapter 8 Gravitation
Exercises | Q 24 | Page 202

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