1. Use the formula for escape velocity

The escape velocity from the surface of a planet is

$v_e = \sqrt{\frac{2GM}{R}}$

So, escape velocity is proportional to:

$v_e \propto \sqrt{\frac{M}{R}}$

2. Compare the planet with Earth

Given:

• Earth escape velocity: $v_{eE} = 11.2\ \text{km/s}$
• Planet mass: $M_p = \frac{1}{6}M_E$
• Planet radius: $R_p = \frac{1}{3}R_E$

Thus,

$\frac{v_{ep}}{v_{eE}} = \sqrt{\frac{M_p/R_p}{M_E/R_E}}$

Substitute the ratios:

$\frac{v_{ep}}{11.2} = \sqrt{\frac{\left(\frac{1}{6}M_E\right)/\left(\frac{1}{3}R_E\right)}{M_E/R_E}}$

3. Simplify the expression

Inside the square root:

$\frac{\left(\frac{1}{6}M_E\right)}{\left(\frac{1}{3}R_E\right)} = \frac{1}{6}M_E \cdot \frac{3}{R_E} = \frac{1}{2}\frac{M_E}{R_E}$

So,

$\frac{v_{ep}}{11.2} = \sqrt{\frac{\frac{1}{2}\frac{M_E}{R_E}}{\frac{M_E}{R_E}}} = \sqrt{\frac{1}{2}}$

Hence,

$v_{ep} = 11.2 \times \frac{1}{\sqrt{2}}$

$v_{ep} \approx 11.2 \times 0.707$

$v_{ep} \approx 7.92\ \text{km/s}$

4. Match with the options

$v_{ep} \approx 7.9\ \text{km/s}$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.