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 determine the escape velocity on the moon relative to the given escape velocity on the planet \(v\), we will use the escape velocity formula: \(v_{\text{escape}} = \sqrt{\frac{2GM}{R}}\), where \(G\) is the gravitational constant, \(M\) is the mass of the celestial body, and \(R\) is its radius.

We are provided with the following information:

• The mass of the moon is \(\frac{1}{144}\) of the planet's mass.
• The diameter of the moon is \(\frac{1}{16}\) of the planet's diameter, implying its radius is also \(\frac{1}{16}\) of the planet's radius.

Let the planet's mass be \(M\) and its radius be \(R\). 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 \(v = \sqrt{\frac{2GM}{R}}\).

The escape velocity on the moon, \(v_{\text{moon}}\), is calculated as follows:

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

Simplifying this expression:

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

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

Since \(\sqrt{\frac{2G \cdot M}{R}} = v\), we can substitute this into the equation for \(v_{\text{moon}}\):

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

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