To determine the escape velocity from a moon, we can use the formula for escape velocity:

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

Where:
- \( v_{\text{escape}} \) is the escape velocity
- \( G \) is the gravitational constant (\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \))
- \( M \) is the mass of the moon
- \( R \) is the radius of the moon

We are given that the mass of the moon (\( M \)) is one-sixth the mass of Earth, and the radius of the moon (\( R \)) is one-fourth the radius of Earth. Let's assume the mass of Earth \( M_E = 5.972 \times 10^{24} \, \text{kg} \) and its radius \( R_E = 6.371 \times 10^6 \, \text{m} \).

Therefore, the mass of the moon (\( M_m \)) is \( \frac{1}{6} M_E \) and the radius (\( R_m \)) is \( \frac{1}{4} R_E \).

Substituting these values into the escape velocity formula:

\[ v_{\text{escape}} = \sqrt{\frac{{2G \left(\frac{1}{6} M_E\right)}}{{\left(\frac{1}{4} R_E\right)}}} \]

Simplifying further:

\[ v_{\text{escape}} = \sqrt{\frac{8G M_E}{3 R_E}} \]

Now, we can substitute the known values:

- \( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)
- \( M_E = 5.972 \times 10^{24} \, \text{kg} \)
- \( R_E = 6.371 \times 10^6 \, \text{m} \)

\[ v_{\text{escape}} = \sqrt{\frac{8 \times 6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{3 \times 6.371 \times 10^6}} \]

Evaluating this expression gives the escape velocity from the moon.

Got a question with my answer?

Message Me

Cheapmath.com is completely free!

Asking question is free. Getting answers is free.

+99

Continue