Question

An unknown gas in which each molecule has mass \( 0.69 \times 10^{-26} \, \text{kg} \) escaped from a planet at a temperature \( T \). The radius of the planet is \( 18 \times 10^5 \, \text{m} \), acceleration due to gravity is \( 10 \, \text{m/s}^2 \), and the Boltzmann constant is \( 1.38 \times 10^{-23} \, \text{J/K}^{-1} \). The temperature \( T \) is:

• A. \(4.5 \times 10^3 \, {K}\)
• B. \(4.8 \times 10^3 \, {K}\)
• C. \(5.4 \times 10^3 \, {K}\)
• D. \(6 \times 10^3 \, {K}\)

Hint:

When calculating planetary escape velocities, consider the kinetic theory of gases, which relates the molecular kinetic energy to the gas's temperature.

Answer 1

The Correct Option is D

Solution:

This problem involves the escape velocity of gas molecules from a planet. We'll use the following concepts:

• Escape velocity: \( v_e = \sqrt{\frac{2GM}{R}} \)
• Kinetic energy: \( KE = \frac{1}{2}mv^2 \)
• Average kinetic energy of a gas molecule: \( KE = \frac{3}{2}kT \)

where:

• \( G \) is the universal gravitational constant,
• \( M \) is the mass of the planet,
• \( R \) is the radius of the planet,
• \( m \) is the mass of the molecule,
• \( v \) is the velocity of the molecule,
• \( k \) is the Boltzmann constant,
• \( T \) is the temperature.

To find the minimum temperature for escape, we equate the kinetic energy to the gravitational potential energy:

\[ \frac{1}{2}mv_e^2 = \frac{3}{2}kT \]

Substitute the escape velocity formula and simplify:

\[ \frac{1}{2}m \left( \sqrt{\frac{2GM}{R}} \right)^2 = \frac{3}{2}kT \] \[ \frac{mGM}{R} = 3kT \]

Since we have the acceleration due to gravity (\( g = \frac{GM}{R^2} \)), we can write:

\[ T = \frac{mgR}{3k} \]

Now, plug in the given values:

\[ T = \frac{(0.69 \times 10^{-26} \, \text{kg}) \times (10 \, \text{m/s}^2) \times (18 \times 10^5 \, \text{m})}{3 \times 1.38 \times 10^{-23} \, \text{J/K}^{-1}} \] \[ T \approx 6 \times 10^3 \, \text{K} \]

Therefore, the temperature \( T \) is approximately \( 6 \times 10^3 \) K.

The correct answer is (4) \( 6 \times 10^3 \) K.