The ratio of `g_("moon")"to" g_("earth")` is

Options

A. 6
B. `(1)/(6)`
C. 4
D. `(1)/(4)`

Explanation

To find the ratio of the acceleration due to gravity on the Moon (\(g_{\text{moon}}\)) to that on the Earth (\(g_{\text{earth}}\)), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept**:
- The acceleration due to gravity (\(g\)) varies from one celestial body to another. For the Earth, it is approximately \(9.81 \, \text{m/s}^2\), and for the Moon, it is about \(1.62 \, \text{m/s}^2\).

2. **Use the Known Ratio**:
- It is known that the acceleration due to gravity on the Moon is approximately \(1/6\) of that on Earth. This can be expressed mathematically as:
\[ g_{\text{moon}} = \frac{1}{6} g_{\text{earth}} \]

3. **Set Up the Ratio**:
- To find the ratio of \(g_{\text{moon}}\) to \(g_{\text{earth}}\), we can write:
\[ \frac{g_{\text{moon}}}{g_{\text{earth}}} = \frac{1/6 \cdot g_{\text{earth}}}{g_{\text{earth}}} \]

4. **Simplify the Ratio**:
- When we simplify the above expression, the \(g_{\text{earth}}\) in the numerator and denominator cancels out:
\[ \frac{g_{\text{moon}}}{g_{\text{earth}}} = \frac{1}{6} \]

5. **Conclusion**:
- Therefore, the ratio of the acceleration due to gravity on the Moon to that on the Earth is:
\[ \frac{g_{\text{moon}}}{g_{\text{earth}}} = \frac{1}{6} \]

### Final Answer:
The ratio of \(g_{\text{moon}}\) to \(g_{\text{earth}}\) is \( \frac{1}{6} \).

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