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

To find the escape velocity on the moon in terms of the given escape velocity on the planet \(v\), let's use the formula for escape velocity:

\(v_{\text{escape}} = \sqrt{\frac{2GM}{R}}\)

where:

• \(G\) is the gravitational constant.
• \(M\) is the mass of the celestial body.
• \(R\) is the radius of the celestial body.

We are given:

• The mass of the moon is \(\frac{1}{144}\) times the mass of the planet.
• The diameter of the moon is \(\frac{1}{16}\) times the diameter of the planet, hence the radius of the moon is also \(\frac{1}{16}\) times the radius of the planet.

If we denote the mass of the planet as \(M\) and the radius of the planet as \(R\), then:

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

The escape velocity on the planet is given by:

\(v = \sqrt{\frac{2GM}{R}}\)

For the moon, the escape velocity \(v_{\text{moon}}\) is:

\(v_{\text{moon}} = \sqrt{\frac{2G \cdot \frac{M}{144}}{\frac{R}{16}}}\)

Simplify the expression for the escape velocity on the moon:

\(v_{\text{moon}} = \sqrt{\frac{2G \cdot \frac{M}{144}}{\frac{R}{16}}} = \sqrt{\frac{2G \cdot M}{144 \cdot \frac{R}{16}}} = \sqrt{\frac{2G \cdot M \cdot 16}{144 \cdot R}}\)

\(v_{\text{moon}} = \sqrt{\frac{16}{144}} \cdot \sqrt{\frac{2G \cdot M}{R}} = \frac{1}{3} \cdot \sqrt{\frac{2G \cdot M}{R}}\)

Since \(\sqrt{\frac{2G \cdot M}{R}} = v\), we have:

\(v_{\text{moon}} = \frac{v}{3}\)

Thus, the escape velocity on the moon is \(\frac{v}{3}\). Therefore, the correct answer is:

\(\frac{v}{3}\)

Answer 2

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:

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