Question

The mass of the moon is 1/144 times the mass of a planet and its diameter 1/16 times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:

• A. \(\frac{v}{3}\)
• B. \(\frac{v}{4}\)
• C. \(\frac{v}{12}\)
• D. \(\frac{v}{16}\)

Answer 1

The Correct Option is A

The escape velocity \( v_{\text{escape}} \) for a celestial body is given by:

\[ v_{\text{escape}} = \sqrt{\frac{2GM}{R}}, \]

where \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.

Given that the escape velocity on the planet is \( v \):

\[ v = \sqrt{\frac{2GM}{R}}. \]

For the moon:

• Mass \( M_{\text{moon}} = \frac{M}{144} \)
• Radius \( R_{\text{moon}} = \frac{R}{16} \)

Substitute these values into the escape velocity formula for the moon:

\[ v_{\text{moon}} = \sqrt{\frac{2G \cdot \frac{M}{144}}{\frac{R}{16}}} = \sqrt{\frac{2GM \cdot 16}{144R}} = \sqrt{\frac{2GM}{9R}} = \frac{1}{3} \sqrt{\frac{2GM}{R}} = \frac{v}{3}. \]

Thus, the escape velocity on the moon is:

\[ \frac{v}{3}. \]