The mass of the moon is ` (1)/(8)` of the earth but the gravitational pull is ` (1)/(6)` earth It is due to the fact that .

Options

A. moon is the satellite of the earth
B. the radius of the earth is (8/6) of the moon
C. the radius of the earth is `(sqrt(8//6))` of the moon
D. the radius of the moon is (6/8) of the earth

Explanation

To solve the problem, we need to analyze the relationship between the mass of the moon, the mass of the earth, and the gravitational pull of both celestial bodies.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The mass of the moon (M_moon) is \( \frac{1}{8} \) of the mass of the earth (M_earth).
- The gravitational pull (acceleration due to gravity) on the moon (g_moon) is \( \frac{1}{6} \) of the gravitational pull on the earth (g_earth).

2. **Using the Formula for Gravitational Force**:
The gravitational force (F) exerted by a celestial body is given by:
\[ F = m \cdot g \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

3. **Setting Up the Equation for Moon and Earth**:
For an object of mass \( m \):
- The gravitational force on the moon:
\[ F_{moon} = m \cdot g_{moon} \]
- The gravitational force on the earth:
\[ F_{earth} = m \cdot g_{earth} \]

4. **Relating the Gravitational Pulls**:
Given that \( g_{moon} = \frac{1}{6} g_{earth} \), we can express this as:
\[ g_{moon} = \frac{1}{6} g_{earth} \]

5. **Using the Formula for Acceleration due to Gravity**:
The acceleration due to gravity is given by:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius.

6. **Setting Up the Ratios**:
For the moon:
\[ g_{moon} = \frac{G \cdot M_{moon}}{R_{moon}^2} \]
For the earth:
\[ g_{earth} = \frac{G \cdot M_{earth}}{R_{earth}^2} \]

7. **Substituting the Mass of the Moon**:
Since \( M_{moon} = \frac{1}{8} M_{earth} \), we can substitute this into the equation for \( g_{moon} \):
\[ g_{moon} = \frac{G \cdot \left(\frac{1}{8} M_{earth}\right)}{R_{moon}^2} \]

8. **Setting the Two Equations Equal**:
Now we can set the two equations for \( g_{moon} \) and \( g_{earth} \) equal to each other:
\[ \frac{G \cdot \left(\frac{1}{8} M_{earth}\right)}{R_{moon}^2} = \frac{1}{6} \cdot \frac{G \cdot M_{earth}}{R_{earth}^2} \]

9. **Canceling Common Terms**:
We can cancel \( G \) and \( M_{earth} \) from both sides:
\[ \frac{1}{8 R_{moon}^2} = \frac{1}{6 R_{earth}^2} \]

10. **Cross Multiplying**:
Cross multiplying gives:
\[ 6 R_{moon}^2 = 8 R_{earth}^2 \]
Simplifying this gives:
\[ \frac{R_{earth}}{R_{moon}} = \sqrt{\frac{6}{8}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

11. **Conclusion**:
Therefore, the relationship between the radius of the earth and the radius of the moon confirms that the moon's gravitational pull is less than that of the earth due to its smaller mass and radius.