Question

The escape speed of the moon when compared with escape speed of the earth is approximately:

• A. Twice smaller
• B. Thrice smaller
• C. 4 times smaller
• D. 5 times smaller
• E. 6 times smaller

Hint:

The escape speed depends on both the mass and the radius of the celestial body. The Moon's escape speed is approximately 5 times smaller than that of Earth because it has much less mass and a smaller radius.

Answer 1

The Correct Option is D

The escape speed \( v_e \) for any celestial body is given by the formula: \[ v_e = \sqrt{\frac{2GM}{R}}, \] where \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.
The escape speed depends on both the mass of the body and its radius. Let's compare the escape speeds of the Earth and the Moon. 
Let \( v_e^{{Earth}} \) and \( v_e^{{Moon}} \) represent the escape speeds of the Earth and the Moon, respectively.
The ratio of the escape speeds is: \[ \frac{v_e^{{Moon}}}{v_e^{{Earth}}} = \sqrt{\frac{2GM_{{Moon}}/R_{{Moon}}}{2GM_{{Earth}}/R_{{Earth}}}} = \sqrt{\frac{M_{{Moon}} R_{{Earth}}}{M_{{Earth}} R_{{Moon}}}}. \] Given that: 
- \( M_{{Moon}} \approx 0.012 \, M_{{Earth}} \), 
- \( R_{{Moon}} \approx 0.27 \, R_{{Earth}} \), the ratio becomes: \[ \frac{v_e^{{Moon}}}{v_e^{{Earth}}} = \sqrt{\frac{0.012 \, M_{{Earth}} \times R_{{Earth}}}{M_{{Earth}} \times 0.27 \, R_{{Earth}}}} = \sqrt{\frac{0.012}{0.27}} \approx \sqrt{\frac{1}{22.5}} \approx \frac{1}{5}. \] Thus, the escape speed of the Moon is approximately 5 times smaller than that of the Earth.
Thus, the correct answer is option (D), 5 times smaller.