The moon's radius is `1//4` that of the earth and its mass `1//80` times that of the earth. If generates the acceleration due to gravity on the surface of the earth, that on the surface of the moon is

Options

A. `g/4`
B. `g/5`
C. `g/6`
D. `g/8`

Explanation

To find the acceleration due to gravity on the surface of the moon, we can use the formula for gravitational acceleration:

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

2. **Acceleration due to Gravity on Earth**:
For Earth, we denote:
- Mass of Earth = \( M_e \)
- Radius of Earth = \( R_e \)
- Acceleration due to gravity on Earth = \( g_e \)

Thus, we have:
\[ g_e = \frac{GM_e}{R_e^2} \]

3. **Given Values for the Moon**:
We know from the problem statement:
- Radius of the Moon \( R_m = \frac{1}{4} R_e \)
- Mass of the Moon \( M_m = \frac{1}{80} M_e \)

4. **Acceleration due to Gravity on the Moon**:
Now, we can write the expression for the acceleration due to gravity on the Moon \( g_m \):
\[ g_m = \frac{GM_m}{R_m^2} \]

5. **Substituting the Values**:
Substitute \( M_m \) and \( R_m \) into the equation:
\[ g_m = \frac{G \left(\frac{1}{80} M_e\right)}{\left(\frac{1}{4} R_e\right)^2} \]

Simplifying the denominator:
\[ \left(\frac{1}{4} R_e\right)^2 = \frac{1}{16} R_e^2 \]

Therefore, we can rewrite \( g_m \):
\[ g_m = \frac{G \left(\frac{1}{80} M_e\right)}{\frac{1}{16} R_e^2} = \frac{G M_e}{R_e^2} \cdot \frac{1}{80} \cdot 16 \]

6. **Relating to Earth's Gravity**:
Since \( g_e = \frac{GM_e}{R_e^2} \), we can substitute this into our equation:
\[ g_m = g_e \cdot \frac{16}{80} \]

Simplifying \( \frac{16}{80} \):
\[ g_m = g_e \cdot \frac{1}{5} \]

7. **Conclusion**:
Therefore, the acceleration due to gravity on the surface of the moon is:
\[ g_m = \frac{g_e}{5} \]

### Final Answer:
If the acceleration due to gravity on the surface of the Earth is \( g_e \), then on the surface of the Moon it is \( \frac{g_e}{5} \).

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