Problem 69 - Master Resource Book in JEE Main Physics

Question

The escape velocity of a body on the surface of earth is \(11.2 \mathrm{kms}^{-1}\). If the mass of the earth is doubled and its radius halved, the escape velocity becomes (a) \(5.6 \mathrm{kms}^{-1}\) (b) \(11.2 \mathrm{kms}^{-1}\) (c) \(22.4 \mathrm{kms}^{-1}\) (d) \(44.8 \mathrm{kms}^{-1}\)

Step-by-Step Solution

Answer

The escape velocity becomes 22.4 km/s, option (c).

Step 1: Understanding Escape Velocity Formula

The escape velocity \( v_e \) is given by the formula \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet (Earth, in this case), and \( R \) is the radius of the planet.

Step 2: Original Escape Velocity Context

For Earth, with the original mass \( M \) and radius \( R \), the escape velocity is given as \( 11.2 \ \text{km/s} \). This translates to the equation \( 11.2 = \sqrt{\frac{2GM}{R}} \).

Step 3: New Scenario Calculation

If the mass of the Earth is doubled and the radius is halved, the new mass \( M' = 2M \) and the new radius \( R' = \frac{R}{2} \). The new escape velocity \( v'_e \) will be \( v'_e = \sqrt{\frac{2G(2M)}{R/2}} \).

Step 4: Simplifying the New Escape Velocity Formula

Simplifying, we have: \( v'_e = \sqrt{\frac{4GM}{R/2}} = \sqrt{\frac{8GM}{R}} = \sqrt{4} \cdot \sqrt{\frac{2GM}{R}} = 2 \times 11.2 = 22.4 \).

Step 5: Conclusion

Therefore, the escape velocity in the new scenario becomes \( 22.4 \ \text{km/s} \), which corresponds to option (c) \( 22.4 \ \text{km/s} \).

Key Concepts

Understanding Gravitational Constant

The gravitational constant, denoted as \( G \), is a fundamental constant in physics that measures the strength of gravity. It is a key factor in calculating the gravitational force between two bodies.
The value of \( G \) is known to be approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \). This constant appears in Newton's law of universal gravitation, which can be expressed as:\[F = G \frac{m_1 m_2}{r^2}\]where:

• \( F \) is the gravitational force between two masses
• \( m_1 \) and \( m_2 \) are the masses of the two objects
• \( r \) is the distance between the centers of their masses

This constant helps us to analyze how massive objects interact with one another due to gravity. The escape velocity formula also utilizes \( G \), illustrating how the gravitational pull depends on it.

Mass of the Earth and Its Role in Escape Velocity

The mass of the Earth, symbolized as \( M \), plays a crucial part in determining escape velocity. In simple terms, escape velocity is the minimum speed needed for an object to break free from a planet's gravitational pull without any further propulsion.
When calculating escape velocity, the mass of the Earth provides insight into the gravitational force exerted by Earth. In the escape velocity formula \( v_e = \sqrt{\frac{2GM}{R}} \), \( M \) represents the Earth's mass and identifies how much force is needed for an object to leave Earth's gravity.If Earth's mass doubles while other factors remain constant, the gravitational force also doubles. This relationship is apparent in the solution for the new escape velocity: \( v'_e = \sqrt{\frac{2G(2M)}{R/2}} \), highlighting that doubling mass increases escape velocity proportionally.
Hence, understanding Earth's mass is vital in overcoming its gravitational hold.

The Significance of Earth's Radius in Calculating Escape Velocity

The radius of the Earth plays a pivotal role in the escape velocity calculation. Denoted by \( R \), it influences how gravitational force changes with distance from the Earth's center.
In the formula for escape velocity, \( v_e = \sqrt{\frac{2GM}{R}} \), \( R \) serves as the denominator, indicating that a larger radius means a weaker gravitational pull at the surface—thus requiring a lower escape velocity. Conversely, a smaller radius implies a stronger pull, necessitating a higher speed to escape.In the problem scenario, halving the Earth's radius intensifies the gravitational force experienced by an object on its surface. This alteration impacts the new escape velocity calculation as shown: \( v'_e = \sqrt{\frac{8GM}{R}} \). The reduced radius magnifies the gravitational force, which in turn increases the escape velocity significantly.
Therefore, the radius is essential in modifying how far and how quickly we must travel to leave a planet's influence.