The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is

Options

A. 0.15
B. 0.04
C. 1
D. 6

Answer

To find the ratio of the acceleration due to gravity at the surface of the moon (g_moon) to that at the surface of the earth (g_earth), we can use the formula for acceleration due to gravity:

\[ g = \frac{GM}{R^2} \]

where \( G \) is the universal gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius.

### Step-by-Step Solution:

1. **Define Mass and Radius**:
- Let the mass of the moon be \( m \).
- Then, the mass of the earth is \( M_{earth} = 81m \).
- Let the radius of the moon be \( r \).
- Then, the radius of the earth is \( R_{earth} = 3.5r \).

2. **Write the Formula for Acceleration Due to Gravity**:
- For the moon:
\[ g_{moon} = \frac{Gm}{r^2} \]
- For the earth:
\[ g_{earth} = \frac{G(81m)}{(3.5r)^2} \]

3. **Simplify the Expression for \( g_{earth} \)**:
- Substitute \( R_{earth} = 3.5r \) into the equation:
\[ g_{earth} = \frac{G(81m)}{(3.5r)^2} = \frac{G(81m)}{12.25r^2} \]

4. **Calculate the Ratio \( \frac{g_{moon}}{g_{earth}} \)**:
- Now, we can find the ratio:
\[ \frac{g_{moon}}{g_{earth}} = \frac{\frac{Gm}{r^2}}{\frac{G(81m)}{12.25r^2}} \]
- The \( G \) and \( m \) cancel out:
\[ \frac{g_{moon}}{g_{earth}} = \frac{1}{\frac{81}{12.25}} = \frac{12.25}{81} \]

5. **Calculate the Numerical Value**:
- Now calculate \( \frac{12.25}{81} \):
\[ \frac{12.25}{81} \approx 0.1515 \]
- Rounding off, we can say:
\[ \frac{g_{moon}}{g_{earth}} \approx 0.15 \]

### Final Answer:
The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is approximately \( 0.15 \).

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