The acceleration due to gravity on the surface of the moon is `(1)/(6)th` of that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be

Options

A. `(sqrt6)/(2)`
B. `sqrt(24)`
C. `3`
D. `(sqrt3)/(2)`

Explanation

To find the ratio of escape velocities on Earth and the Moon, we can use the formula for escape velocity, which is given by:

\[ v_e = \sqrt{2gr} \]

where \( g \) is the acceleration due to gravity and \( r \) is the radius of the celestial body.

### Step 1: Define the variables for Earth and Moon
Let:
- \( g \) = acceleration due to gravity on Earth
- \( r \) = radius of Earth
- \( g' \) = acceleration due to gravity on Moon
- \( r' \) = radius of Moon

From the question, we know:
- \( g' = \frac{g}{6} \) (the acceleration due to gravity on the Moon is one-sixth that of Earth)
- The diameter of the Moon is one-fourth that of Earth, so the radius of the Moon is:
\[ r' = \frac{r}{4} \]

### Step 2: Write the escape velocity formulas for Earth and Moon
The escape velocity for Earth (\( v_e \)) is:
\[ v_e = \sqrt{2gr} \]

The escape velocity for Moon (\( v_{e'} \)) is:
\[ v_{e'} = \sqrt{2g'r'} = \sqrt{2 \left(\frac{g}{6}\right) \left(\frac{r}{4}\right)} = \sqrt{\frac{2gr}{24}} = \sqrt{\frac{gr}{12}} \]

### Step 3: Find the ratio of escape velocities
Now, we can find the ratio of the escape velocities:
\[ \frac{v_e}{v_{e'}} = \frac{\sqrt{2gr}}{\sqrt{\frac{gr}{12}}} \]

### Step 4: Simplify the ratio
This simplifies to:
\[ \frac{v_e}{v_{e'}} = \sqrt{\frac{2gr}{\frac{gr}{12}}} = \sqrt{\frac{2 \cdot 12}{1}} = \sqrt{24} \]

### Step 5: Finalize the ratio
Thus, we can express this as:
\[ \frac{v_e}{v_{e'}} = 2\sqrt{6} \]

### Conclusion
The ratio of escape velocities on Earth and the Moon is:
\[ \frac{v_e}{v_{e'}} = 2\sqrt{6} \]