Problem 59 - Master Resource Book in JEE Main Physics

Question

The escape velocity from the earth is \(11 \mathrm{kms}^{-1}\).The escape velocity from a planet having twice the radius and same mean density as that of earth is [UP SEE 2009] (a) \(5.5 \mathrm{kms}^{-1}\) (b) \(11 \mathrm{kms}^{-1}\) (c) \(22 \mathrm{kms}^{-1}\) (d) None of these

Step-by-Step Solution

Answer

The escape velocity from the new planet is 22 km/s (option c).

Step 1: Understanding Escape Velocity

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a massive body without further propulsion. It is given by the formula \( v = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

Step 2: Relating Earth and the Planet

For Earth, the escape velocity is given as \( v_e = 11 \text{ km/s} \). For the planet, we have twice the radius \( R' = 2R_e \) and the same mean density, which implies \( M' = \rho \times \frac{4}{3}\pi (2R_e)^3 = 8M_e \). Therefore, the formula for escape velocity for the new planet is: \( v_p = \sqrt{\frac{2G \times 8M_e}{2R_e}} \).

Step 3: Simplifying to Find the Planet's Escape Velocity

Simplifying the expression from Step 2, we have \( v_p = \sqrt{\frac{8GM_e}{2R_e}} = \sqrt{4} \times \sqrt{\frac{2GM_e}{R_e}} \). Using the escape velocity of earth, \( \sqrt{\frac{2GM_e}{R_e}} = v_e \). Hence, \( v_p = 2v_e = 2 \times 11 = 22 \text{ km/s} \).

Step 4: Conclusion

Since the escape velocity of the planet is calculated to be 22 km/s, the correct answer is \( v_p = 22 \text{ km/s} \), which corresponds to option (c).

Key Concepts

Gravitational Constant

The gravitational constant, denoted as \( G \), is a key element in the calculation of gravitational forces between objects in the universe. This constant is vital when calculating the escape velocity of any celestial body, such as a planet or moon. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \). It directly influences the strength of the gravitational pull between two bodies.
The formula for escape velocity is derived using \( G \), showing its essential role:

• Escape Velocity Formula: \( v = \sqrt{\frac{2GM}{R}} \)

Here,

• \( v \) is the escape velocity,
• \( M \) is the mass of the celestial body,
• \( R \) is the radius of the body.

Understanding the gravitational constant allows students to comprehend the unseen force that binds us to Earth and other celestial bodies. Its value ensures consistent calculations of gravitational interactions across various planetary bodies.

Planetary Mass and Radius Relationship

The relationship between a planet's mass and radius is crucial for understanding its gravitational characteristics. Mass \( M \) and radius \( R \) determine the gravitational pull experienced at the planet's surface. For calculating escape velocity, both of these factors play a significant role.
If a planet has the same mean density as Earth and a radius that is twice as large, its mass will increase significantly. This is because the mass is directly related to the volume of the planet, which can be calculated using the formula:

• Volume \( V = \frac{4}{3} \pi R^3 \)

Knowing that volume increases with the cube of the radius, if the radius doubles, the volume (and thus mass for constant density) becomes eight times greater:

• Mass \( M' \) for a planet with twice the radius is \( 8M_e \).

This relationship explains why a planet with greater radius and mass also requires higher escape velocity, as its gravitational pull is stronger.

Mean Density of Planetary Bodies

Mean density is another critical concept that links mass to volume within a celestial body. The density \( \rho \) of a planetary body is given by the mass divided by its volume:

• Density Formula: \( \rho = \frac{M}{V} \)

In the problem, we assume that the new planet has the same mean density as Earth. This assumption helps understand how the mass changes with radius while keeping density constant.
For the new planet where radius \( R' = 2R_e \), the mean density remains the same, ensuring that any change in mass is because of the change in volume. This conceptual link is helpful:

• For higher radius with same density, mass increases significantly.
• The volume increases by a factor of eight when the radius is doubled.

These changes help calculate escape velocity accurately as shown in the exercise, providing insight into how celestial bodies work in terms of their structure and gravitational influence.