Problem 85 - Master Resource Book in JEE Main Physics

Question

The mass of the moon is \(1 / 81\) of earth's mass and its radius \(1 / 4\) th that of the earth. If the escape velocity from the earth's surface is \(11.2 \mathrm{kms}^{-1}\), its value for the moon will be (a) \(0.15 \mathrm{kms}^{-1}\) (b) \(5 \mathrm{kms}^{-1}\) (c) \(2.5 \mathrm{kms}^{-1}\) (d) \(0.5 \mathrm{kms}^{-1}\)

Step-by-Step Solution

Answer

The escape velocity for the moon is approximately \(2.5\mathrm{kms}^{-1}\), answer (c).

Step 1: Understanding Escape Velocity

Escape velocity is determined by the formula \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius. For the moon, we need to adjust the formula according to the moon's mass and radius in comparison to Earth's.

Step 2: Substitute Moon's Relative Values

Given that the moon's mass \( M_{moon} = \frac{1}{81}M_{earth} \) and its radius \( R_{moon} = \frac{1}{4}R_{earth} \), substitute these values into the escape velocity formula for the moon: \( v_{e,moon} = \sqrt{\frac{2G \times (\frac{1}{81}M_{earth})}{\frac{1}{4}R_{earth}}} \).

Step 3: Simplify the Expression

Extract common factors in the escape velocity equation: \( v_{e,moon} = \sqrt{\frac{2G \times M_{earth}}{R_{earth}} \times \frac{1}{81} \times \frac{4}{1}} \). Simplify this to \( v_{e,moon} = \sqrt{\frac{8}{81}} \times v_{e,earth} \).

Step 4: Calculate Moon's Escape Velocity

Calculate the expression \( \sqrt{\frac{8}{81}} \approx 0.3149 \). Thus, \( v_{e,moon} = 0.3149 \times 11.2 \mathrm{kms}^{-1} \approx 3.52 \mathrm{kms}^{-1} \).

Step 5: Identify Closest Multiple-Choice Answer

The calculated escape velocity for the moon is approximately \(3.52 \mathrm{kms}^{-1}\). Compare this with the provided options: (a) \(0.15\), (b) \(5\), (c) \(2.5\), (d) \(0.5\) \(\mathrm{kms}^{-1}\). The closest answer is option (c) \(2.5\mathrm{kms}^{-1}\).

Key Concepts

Gravitational Constant

Gravitational constant, symbolized as \( G \), is a key value in the realm of physics, particularly in the study of gravitation. It quantifies the inherent strength of the gravitational force between two point masses. The constant \( G \) is universal, meaning it is the same throughout the cosmos. Its value is approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
This constant is integral in calculations involving gravity, such as determining escape velocity, which requires overcoming the gravitational pull of celestial bodies.
Some practical uses of \( G \) include:

• Predicting planetary orbits in our solar system.
• Calculating forces acting on satellites.
• Understanding the dynamics of binary star systems.

The constant \( G \), though small, is essential for understanding the vast forces that govern the motion of planets and stars.

Mass of Celestial Bodies

The mass of celestial bodies like planets, moons, and stars significantly affects gravitational attraction. In the formula for escape velocity, \( M \) represents the mass of a celestial body.
A larger mass results in a stronger gravitational pull, making the escape velocity higher, as the body exerts more force on nearby objects trying to break free from its gravitational field.

• For instance, Earth's greater mass compared to the moon's means objects require greater initial speed to escape Earth's gravity than the moon's.
• The moon's mass is only \( \frac{1}{81} \) of Earth's mass, explaining its comparatively lower escape velocity.
• This also affects tidal forces, satellite orbits, and even potential colonization like on the moon or Mars.

Understanding mass distribution helps astronomers model planetary formations and dynamics accurately.

Radius of Celestial Bodies

The radius of a celestial body plays a crucial role in determining the escape velocity. The radius, denoted as \( R \) in calculations, reflects the distance from the body's center to its surface, affecting the gravitational potential energy.
A larger radius means a lower gravitational force at the surface, reducing the necessary speed to overcome the body's gravitational influence. In the escape velocity equation, as the radius increases, less energy is needed to escape.

• The moon's radius is only \( \frac{1}{4} \) of Earth's, meaning gravity at the moon's surface is less intense than on Earth.
• This is why astronauts on the moon can take long strides and jump higher than on Earth.
• Alterations in celestial bodies' radius can also impact atmospheres and potential habitability.

Grasping how radius affects gravitational pull helps us design space missions and develop technologies to explore beyond Earth.