The acceleration due to gravity on the moon is

Options

A. `((1)/(6))^(th)` that of the earth
B. same that of the earth
C. `((1)/(3))^(rd)` that of the earth
D. `((1)/(5))^(th)` that of the earth

Explanation

To determine the acceleration due to gravity on the moon, we can follow these steps:

### Step 1: Understand the Formula
The formula for acceleration due to gravity (g) at a distance from the center of a celestial body is given by:

\[ g = \frac{G \cdot M}{R^2} \]

where:
- \( G \) is the universal gravitational constant (\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the celestial body (for the moon, \( 7.35 \times 10^{22} \, \text{kg} \)),
- \( R \) is the radius of the celestial body (for the moon, \( 1.738 \times 10^6 \, \text{m} \)).

### Step 2: Substitute Values for the Moon
Using the values for the moon, we can substitute them into the formula:

\[ g_m = \frac{(6.67 \times 10^{-11}) \cdot (7.35 \times 10^{22})}{(1.738 \times 10^6)^2} \]

### Step 3: Calculate the Denominator
First, calculate \( R^2 \):

\[ R^2 = (1.738 \times 10^6)^2 = 3.022244 \times 10^{12} \, \text{m}^2 \]

### Step 4: Calculate the Numerator
Now calculate the numerator:

\[ G \cdot M = (6.67 \times 10^{-11}) \cdot (7.35 \times 10^{22}) = 4.90145 \times 10^{12} \, \text{N m}^2/\text{kg} \]

### Step 5: Calculate \( g_m \)
Now substitute the numerator and denominator back into the equation for \( g_m \):

\[ g_m = \frac{4.90145 \times 10^{12}}{3.022244 \times 10^{12}} \approx 1.62 \, \text{m/s}^2 \]

### Step 6: Compare with Earth's Gravity
The acceleration due to gravity on Earth (\( g_e \)) is approximately \( 9.81 \, \text{m/s}^2 \). To find the ratio of \( g_e \) to \( g_m \):

\[ \frac{g_e}{g_m} = \frac{9.81}{1.62} \approx 6.06 \approx 6 \]

### Conclusion
Thus, we conclude that the acceleration due to gravity on the moon is approximately one-sixth that of the Earth.

### Final Answer
The acceleration due to gravity on the moon is **1/6th of that of the Earth**.

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