8.24 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 ×108 km; G = 6.67×10-11 N m2 kg–2.

Mass of the spaceship, ms = 1000 kgMass of the Sun, M = 2 × 1030 kgMass of Mars, mm = 6.4×1023 kgOrbital radius of Mars, R = 2.28 ×108 km = 2.28 ×1011 mRadius of Mars, r = 3395 km = 3.395 ×106mUniversal Gravitational constant, G = 6.67×10-11 N m2 kg–2Potential energy of the spaceship due to the gravitational attraction of the Sun = -GMmsRPotential energy of the spaceship due to the gravitational attraction of Mars= -GmsmmrSince the spaceship is stationed on Mars, its velocity and hence its kinetic energy will be zeroTotal energy of the spaceship = -GMmsR --Gmsmmr = -Gms(MR + mmr )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(MR + mmr )= 6.67 ×10-11×103×(2×10302.28×1011 + 6.4×10233.395×106)= 6.67 ×10-8×(8.77×1018 + 1.88 ×1017 )= 6.67 ×10-8×8.958×1018= 5.97 ×1011 J

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8.24 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 ×10⁸ km; G = 6.67×10^-11 N m2 kg^–2.

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Vishal Baghel | Contributor-Level 10

5 months ago

Mass of the spaceship, $m_{s}$ = 1000 kg

Mass of the Sun, M = 2 * 10³⁰ kg

Mass of Mars, $m_{m}$ = 6.4*10²³ kg

Orbital radius of Mars, R = 2.28 *10⁸ km = 2.28 *10¹¹ m

Radius of Mars, r = 3395 km = 3.395 $* 10^{6} m$

Universal Gravitational constant, G = 6.67*10^-11 N m2 kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun = $\frac{- G M m_{s}}{R}$

Potential energy of the spaceship due to the gravitational attraction of Mars= $\frac{- G m_{s} m_{m}}{r}$

Since the spaceship is stationed on Mars, its velocity and hence its kinetic energy will be zero

Total energy of the spaceship = $\frac{- G M m_{s}}{R}$ $- \frac{- G m_{s} m_{m}}{r}$ = $- G m_{s} (\frac{M}{R}$ + $\frac{m_{m}}{r}$ )

The negative sign indicates that th

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Mass of the spaceship, $m_{s}$ = 1000 kg

Mass of the Sun, M = 2 * 10³⁰ kg

Mass of Mars, $m_{m}$ = 6.4*10²³ kg

Orbital radius of Mars, R = 2.28 *10⁸ km = 2.28 *10¹¹ m

Radius of Mars, r = 3395 km = 3.395 $* 10^{6} m$

Universal Gravitational constant, G = 6.67*10^-11 N m2 kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun = $\frac{- G M m_{s}}{R}$

Potential energy of the spaceship due to the gravitational attraction of Mars= $\frac{- G m_{s} m_{m}}{r}$

Since the spaceship is stationed on Mars, its velocity and hence its kinetic energy will be zero

Total energy of the spaceship = $\frac{- G M m_{s}}{R}$ $- \frac{- G m_{s} m_{m}}{r}$ = $- G m_{s} (\frac{M}{R}$ + $\frac{m_{m}}{r}$ )

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)

= $G m_{s} (\frac{M}{R}$ + $\frac{m_{m}}{r}$ )

= 6.67 $* 10^{- 11} * 10^{3} * (\frac{2 * 10^{30}}{2.28 * 10^{11}}$ + $\frac{6.4 * 10^{23}}{3.395 * 10^{6}})$

= 6.67 $* 10^{- 8} * (8.77 * 10^{18}$ + 1.88 $* 10^{17}$ )

= 6.67 $* 10^{- 8} * 8.958 * 10^{18}$

= 5.97 $* 10^{11}$ J

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