The acceleration due to gravity near the surface of moon is-

Options

A. `1/6` of the acceleration due to gravity of earth
B. almost equal to acceleration due to gravity of earth
C. 6 times the acceleration due to gravity of earth
D. `1/12` of the acceleration due to gravity of earth.

Explanation

To find the acceleration due to gravity near the surface of the Moon, we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The formula for acceleration due to gravity (g) is given by:
\[ g = \frac{GM}{R^2} \]
where:
- \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the celestial body (Earth or Moon)
- \( R \) is the radius of the celestial body

### Step 2: Calculate the acceleration due to gravity on Earth
Using the known values for Earth:
- Mass of Earth, \( M_{Earth} \approx 5.97 \times 10^{24} \, \text{kg} \)
- Radius of Earth, \( R_{Earth} \approx 6378 \, \text{km} = 6378000 \, \text{m} \)

Substituting these values into the formula:
\[ g_{Earth} = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{(6378000)^2} \]
Calculating this gives \( g_{Earth} \approx 9.8 \, \text{m/s}^2 \).

### Step 3: Calculate the acceleration due to gravity on the Moon
Now, using the values for the Moon:
- Mass of Moon, \( M_{Moon} \approx 7.3 \times 10^{22} \, \text{kg} \)
- Radius of Moon, \( R_{Moon} \approx 1738 \, \text{km} = 1738000 \, \text{m} \)

Substituting these values into the formula:
\[ g_{Moon} = \frac{(6.67 \times 10^{-11}) \times (7.3 \times 10^{22})}{(1738000)^2} \]
Calculating this gives \( g_{Moon} \approx 1.63 \, \text{m/s}^2 \).

### Step 4: Determine the relationship between Moon's and Earth's gravity
To find how the Moon's gravity compares to Earth's gravity, we can take the ratio:
\[ \frac{g_{Moon}}{g_{Earth}} = \frac{1.63}{9.8} \approx \frac{1}{6} \]

### Conclusion
Thus, the acceleration due to gravity near the surface of the Moon is approximately \( \frac{1}{6} \) of the acceleration due to gravity on Earth.

### Final Answer
The acceleration due to gravity near the surface of the Moon is \( \frac{1}{6} \) of the acceleration due to gravity on Earth.

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