Problem 32 - Master Resource Book in JEE Main Physics

Question

A spaceship is stationed on mars. How much energy must be expended on the spaceship to rocket it out of the solar system? Mass of the spaceship \(=1000 \mathrm{~kg}\); mass of sun \(=2 \times 10^{30} \mathrm{~kg} ;\) mass of mars \(=6.4 \times 10^{23} \mathrm{~kg} ;\) radius of mars \(=3395 \mathrm{~km}\); radius of orbit of mars \(=228 \times 10^{8} \quad \mathrm{~km}, \quad G=6.67 \times 10^{-11}\) \(\mathrm{N}-\mathrm{m}^{2} / \mathrm{kg}^{2}\). (a) \(3 \times 10^{11} \mathrm{~J}\) (b) \(4 \times 10^{11} \mathrm{~J}\) (c) \(33 \times 10^{11} \mathrm{~J}\) (d) None of these

Step-by-Step Solution

Answer

None of these

Step 1: Calculate the Gravitational Potential Energy on Mars

The gravitational potential energy on the surface of Mars is given by the formula: \[U_{Mars} = - \frac{G imes M_{Mars} imes m}{R_{Mars}}\]where:- \( U_{Mars} \) is the gravitational potential energy on Mars,- \( G \) is the gravitational constant \(6.67 \times 10^{-11} \, \mathrm{N}-\mathrm{m}^{2}/\mathrm{kg}^{2}\),- \( M_{Mars} \) is the mass of Mars \(6.4 \times 10^{23} \, \mathrm{kg}\),- \( m \) is the mass of the spaceship \(1000 \, \mathrm{kg}\),- \( R_{Mars} \) is the radius of Mars \(3395 \, \mathrm{km} = 3.395 \times 10^{6} \, \mathrm{m}\).Substitute the values to find \( U_{Mars} \):\[U_{Mars} = - \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23} \times 1000}{3.395 \times 10^{6}}\]\[U_{Mars} = -1.88 \times 10^{10} \, \mathrm{J}\]

Step 2: Calculate the Gravitational Potential Energy Due to the Sun

The gravitational potential energy due to the Sun at Mars' orbit is given by:\[U_{Sun} = - \frac{G \times M_{Sun} \times m}{d_{Mars-Sun}}\]where:- \( M_{Sun} \) is the mass of the Sun \(2 \times 10^{30} \, \mathrm{kg}\),- \( d_{Mars-Sun} \) is the distance from Mars to the Sun \(228 \times 10^{8} \, \mathrm{km} = 2.28 \times 10^{11} \, \mathrm{m}\).Substitute the values:\[U_{Sun} = - \frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 1000}{2.28 \times 10^{11}}\]\[U_{Sun} = -5.84 \times 10^{10} \, \mathrm{J}\]

Step 3: Calculate the Total Gravitational Potential Energy at Mars Orbit

The total gravitational potential energy \( U_{total} \) at the orbit of Mars is:\[U_{total} = U_{Mars} + U_{Sun}\]\[U_{total} = -1.88 \times 10^{10} + (-5.84 \times 10^{10})\]\[U_{total} = -7.72 \times 10^{10} \, \mathrm{J}\]

Step 4: Calculate the Escape Energy Required to Leave the Solar System

To escape the solar system, the spaceship must be completely free of the Sun's gravitational pull. Therefore, the escape energy \( E_{escape} \) is:\[E_{escape} = -U_{total} = 7.72 \times 10^{10} \, \mathrm{J}\]This calculation demonstrates the energy needed to escape. Checking against given answers, this is equivalent to \(7.72 \times 10^{11} \) if we consider scaling errors or misprints, but none exactly match options provided. Hence, the correct choice is:- \(\text{None of these}\)

Key Concepts

Escape Velocity

Escape velocity is a fundamental concept in physics, especially in the context of astrophysics. It refers to the minimum speed an object, like a spaceship, must reach to break free from the gravitational pull of a celestial body, such as a planet or star, without falling back. This concept is crucial when thinking about space missions and launches.

For a given celestial body, escape velocity depends on two main factors:

• The mass of the celestial body.
• The distance from the center of the celestial body to the object trying to escape (typically its surface).

Escape velocity is derived from the idea that an object's kinetic energy must at least equal the gravitational potential energy that holds it in orbit. The formula for escape velocity (\[v_{e}\]) is:\[v_{e} = \sqrt{\frac{2GM}{r}}\]where \[G\] is the gravitational constant,\[M\] is the mass of the celestial body, and \[r\] is the distance from the center of the mass.
Understanding escape velocity helps us calculate how much energy is needed to launch objects into space, making it essential for efficient space travel and exploration.

Gravitational Constant

The gravitational constant, often denoted as \(G\), is one of the most fundamental constants in physics. It appears in Newton's law of universal gravitation and plays a crucial role in calculating forces between masses. This constant is key to understanding how masses attract each other in the universe.

\[G\] has a value of approximately \(6.67 \times 10^{-11} \mathrm{N}-\mathrm{m}^{2}/\mathrm{kg}^{2}\). This small value reflects gravity's relative weakness compared to other fundamental forces, like electromagnetism. This constant enables us to calculate gravitational forces and potential energy accurately.

When dealing with problems involving gravitational force or potential energy, \[G\] allows us to factor in the masses involved and the distance between them. The formula for gravitational force between two masses \(m_1\) and \(m_2\) is:\[F = \frac{G \times m_1 \times m_2}{r^2}\]where \[r\] is the distance between the centers of the two masses. This understanding is vital when calculating interactions in space or other gravitational systems.

Potential Energy Calculation

Potential energy due to gravity is the energy possessed by an object because of its position in a gravitational field. In this problem, it’s essential for determining how much energy is required to move the spaceship out of the solar system.

The potential energy (\[U\]) related to gravity is calculated using the formula:\[U = -\frac{G \times M \times m}{r}\]where:

• \[M\] is the mass of the celestial body (such as a planet or star).
• \[m\] is the mass of the object (like the spaceship).
• \[r\] is the distance from the center of the celestial body to the object.

The negative sign indicates that gravitational force is attractive.
In this problem, potential energy from both the planet Mars and the Sun determines the total gravitational pull on the spaceship. By adding these energies, we find the overall energy needed to escape the celestial bodies. This calculation guides us in understanding energy budgets necessary for missions that venture into deep space.