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

Ask & Answer Question

Gravitation ncert solutions physics class 11th physics ncert solutions class 11th Physics Class 11th Physics Gravitation

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.

3 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 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

...more

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

less

0 Upvotes 0 Downvotes

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question