Problem 72 - Master Resource Book in JEE Main Physics

Question

The escape velocity on the surface of earth is \(11.2 \mathrm{kms}^{-1} .\) If mass and radius of a planet is 4 and 2 times respectively, than that of earth, the escape velocity on the planet (a) \(11.2 \mathrm{kms}^{-1}\) (b) \(1.12 \mathrm{kms}^{-1}\) (c) \(22.4 \mathrm{kms}^{-1}\) (d) \(15.8 \mathrm{kms}^{-1}\)

Step-by-Step Solution

Answer

The escape velocity on the planet is 22.4 km/s, which is option (c).

Step 1: Understanding Escape Velocity

The escape velocity is the minimum velocity an object must have to break free from a planet or moon's gravitational pull without further propulsion. It is given by the formula \( v_e = \sqrt{\frac{2GM}{R}} \) where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

Step 2: Earth's Escape Velocity Equation

For Earth, the escape velocity \( v_e \) is 11.2 km/s. Its relation with its mass \( M_e \) and radius \( R_e \) can be expressed as \( 11.2 = \sqrt{\frac{2GM_e}{R_e}} \).

Step 3: Escape Velocity Formula for Another Planet

For a planet with mass \( M = 4M_e \) and radius \( R = 2R_e \), the escape velocity \( v_e \) can be written as \( v_e = \sqrt{\frac{2G(4M_e)}{2R_e}} \).

Step 4: Simplify the Expression

Substitute and simplify: \( v_e = \sqrt{\frac{8GM_e}{2R_e}} = \sqrt{4 \cdot \frac{2GM_e}{R_e}} = 2 \sqrt{\frac{2GM_e}{R_e}} \). Since \( \sqrt{\frac{2GM_e}{R_e}} = 11.2 \), it follows that \( v_e = 2 \times 11.2 = 22.4 \mathrm{km/s} \).

Step 5: Determine Correct Answer

The escape velocity on the new planet is therefore \( 22.4 \mathrm{km/s} \). Thus, the correct answer is option (c).

Key Concepts

Gravitational Force

Gravitational force is one of the fundamental forces of nature. It is the force that attracts two bodies towards each other, and it plays a critical role in planetary science. This force is determined by two main factors:

• The masses of the two objects involved
• The distance between the centers of the two objects

Gravitational force can be calculated using Newton's law of universal gravitation, which is expressed mathematically as follows:\[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]Here, \(F\) is the gravitational force, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between the centers of the two objects. Understanding this force is essential when calculating parameters like escape velocity, as it dictates how strong the gravitational pull is on a given object. If a planet's mass increases, as in our scenario, the gravitational force becomes stronger, making it harder for objects to escape its pull.

Planetary Science

Planetary science is the study of planets, moons, and planetary systems. It involves understanding their properties, compositions, and the forces that affect them. The escape velocity of a planet is an essential topic in planetary science, as it reveals a lot about the planet's gravitational pull and its potential to retain an atmosphere.
When studying planets, scientists consider factors such as:

• Mass - A key factor affecting gravitational force and escape velocity
• Radius - Influences the gravitational pull experienced by objects on or near the planetary surface
• Density and Composition - Offer insights into the planet's internal make-up and potential habitability

In the context of this exercise, the planet in question has both its mass and radius increased—4 times the Earth's mass and 2 times its radius—significantly affecting its escape velocity. Such changes imply that the gravitational environment is different from that on Earth. These concepts help scientists in determining characteristics like atmospheric composition and potential spacecraft trajectories.

Kinematic Equations

Kinematic equations are a set of equations that describe the motion of objects. They are especially useful in physics for solving problems related to motion with constant acceleration. Though these equations don't directly compute escape velocity, they aid in understanding the broader dynamics of motion and forces involved.
Escape velocity computation effectively results from balancing gravitational forces and motion principles. It is derived from kinetic and potential energy considerations, not the kinematic equations directly. However, understanding these equations can help interpret how an object moves once it achieves escape velocity:\[v_e = \sqrt{\frac{2GM}{R}}\]Here, kinetic energy equals gravitational potential energy at the threshold of escape. Calculating escape velocity is akin to solving a motion problem where an object is aiming to 'escape' a gravitational field rather than accelerating under it. Thus, while the core formulas differ, the concepts like velocity, speed, and forces still relate intimately with kinematic principles.