Question

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 spaceship\[ = 1000\,{\text{kg}}\]; mass of the sun\[ = 2 \times {10^{30}}\,{\text{kg}}\]; mass of mars\[ = 6.4 \times {10^{23}}\,{\text{kg}}\]; Radius of mars\[ = 3395\,{\text{km}}\]; Radius of the orbit of mars\[ = 2.28 \times {10^8}\,{\text{km}}\]; \[G = 6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2} \cdot {\text{k}}{{\text{g}}^{{\text{ - 2}}}}\]

Answer 1

Hint: Use the formula for the potential energy of an object on the surface of a plane. The energy required to launch the spaceship out of the solar system is equal to the energy of the spaceship on the surface of Mars. The spaceship also has the force of attraction of the sun on it. Hence, the total energy of the spaceship is equal to the potential energy of the spaceship due to Mars and the sun.

Formula used:
The potential energy \[U\] of an object on the surface of a plane is given by
\[U = - \dfrac{{GMm}}{r}\] …… (1)
Here, \[G\] is the universal gravitational constant, \[M\] is the mass of the planet, \[m\] is the mass of the object and \[r\] is the radius of the planet.

Complete step by step answer:
We have given that a spaceship is stationed on Mars.We are asked to calculate the energy required to launch the spaceship from the surface of Mars out of the solar system.We have given that mass of the spaceship, Mars and the sun are \[1000\,{\text{kg}}\], \[6.4 \times {10^{23}}\,{\text{kg}}\] and \[2 \times {10^{30}}\,{\text{kg}}\] respectively.
\[m = 1000\,{\text{kg}}\]
\[ \Rightarrow{M_M} = 6.4 \times {10^{23}}\,{\text{kg}}\]
\[ \Rightarrow{M_S} = 2 \times {10^{30}}\,{\text{kg}}\]
The radius of Mars and the radius of the solar orbit are \[3395\,{\text{km}}\] and \[2.28 \times {10^8}\,{\text{km}}\] respectively.
\[{R_M} = 3395\,{\text{km}}\]
\[ \Rightarrow{M_S} = 2.28 \times {10^8}\,{\text{km}}\]
The energy required to launch the spaceship out of the solar system is equal to the energy of the spaceship on the surface of Mars.The potential energy \[{U_M}\] of the spaceship on the surface of Mars is
\[{U_M} = - \dfrac{{G{M_M}m}}{{{R_M}}}\]
The potential energy \[{U_S}\] of the spaceship on the surface of Mars due to attraction of the Sun is
\[{U_S} = - \dfrac{{G{M_S}m}}{{{R_S}}}\]

On the surface of Mars, the kinetic energy of the spaceship is zero as its velocity is zero.The total energy \[E\] of the spaceship on the surface of Mars is the sum of the potential energy \[{U_M}\] of spaceship due to Mars and the potential energy \[{U_S}\] of spaceship due to the sun.
\[E = {U_M} + {U_S}\]
\[ \Rightarrow E = - \dfrac{{G{M_M}m}}{{{R_M}}} - \dfrac{{G{M_S}m}}{{{R_S}}}\]
\[ \Rightarrow E = - Gm\left( {\dfrac{{{M_M}}}{{{R_M}}} + \dfrac{{{M_S}}}{{{R_S}}}} \right)\]
Substitute \[6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2} \cdot {\text{k}}{{\text{g}}^{{\text{ - 2}}}}\] for \[G\], \[1000\,{\text{kg}}\] for \[m\], \[6.4 \times {10^{23}}\,{\text{kg}}\] for \[{M_M}\], \[2 \times {10^{30}}\,{\text{kg}}\] for \[{M_S}\], \[3395\,{\text{km}}\] for \[{R_M}\] and \[2.28 \times {10^8}\,{\text{km}}\] for \[{R_S}\] in the above equation.
\[ \Rightarrow E = - \left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2} \cdot {\text{k}}{{\text{g}}^{{\text{ - 2}}}}} \right)\left( {1000\,{\text{kg}}} \right)\left( {\dfrac{{6.4 \times {{10}^{23}}\,{\text{kg}}}}{{3395\,{\text{km}}}} + \dfrac{{2 \times {{10}^{30}}\,{\text{kg}}}}{{2.28 \times {{10}^8}\,{\text{km}}}}} \right)\]
\[ \Rightarrow E = - \left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2} \cdot {\text{k}}{{\text{g}}^{{\text{ - 2}}}}} \right)\left( {1000\,{\text{kg}}} \right)\left( {\dfrac{{6.4 \times {{10}^{23}}\,{\text{kg}}}}{{3395 \times {{10}^3}\,{\text{m}}}} + \dfrac{{2 \times {{10}^{30}}\,{\text{kg}}}}{{2.28 \times {{10}^{11}}\,{\text{m}}}}} \right)\]
\[ \therefore E = - 5.97 \times {10^{11}}\,{\text{J}}\]

Hence, the energy expended to launch the spaceship out of the solar system is \[5.97 \times {10^{11}}\,{\text{J}}\].The negative sign indicates that the spaceship is bound to the solar system.

Note: The students should not forget to use the gravitational potential energy of the sun on the spaceship on the surface of Mars. Also the gravitational potential energy of the other planets on this spaceship is neglected as the values of these energies are negligible. The students should not forget to convert the units of the radii of Mars and the orbit of Mars around the sun to the SI system of units.