Find the velocity of escape at the moon. Given that its radius is `1.7xx10^(6)` m and the value of 'g' is `1.63 ms^(-2)`.

Answer

To find the velocity of escape from the Moon, we can use the formula for escape velocity, which is given by:

\[ V_E = \sqrt{2gR} \]

where:
- \( V_E \) is the escape velocity,
- \( g \) is the acceleration due to gravity,
- \( R \) is the radius of the celestial body (in this case, the Moon).

### Step 1: Identify the values
From the problem, we have:
- Radius of the Moon, \( R = 1.7 \times 10^6 \) m
- Acceleration due to gravity on the Moon, \( g = 1.63 \, \text{m/s}^2 \)

### Step 2: Substitute the values into the formula
Now we substitute the values of \( g \) and \( R \) into the escape velocity formula:

\[ V_E = \sqrt{2 \times 1.63 \, \text{m/s}^2 \times 1.7 \times 10^6 \, \text{m}} \]

### Step 3: Calculate the product inside the square root
First, we calculate the product:

\[ 2 \times 1.63 \times 1.7 \times 10^6 = 5.543 \times 10^6 \]

### Step 4: Take the square root
Now, we take the square root of the result:

\[ V_E = \sqrt{5.543 \times 10^6} \]

Calculating the square root gives:

\[ V_E \approx 2353.5 \, \text{m/s} \]

### Step 5: Final answer
Thus, the velocity of escape from the Moon is approximately:

\[ V_E \approx 2.35 \times 10^3 \, \text{m/s} \]

### Summary
The escape velocity from the Moon is approximately \( 2.35 \times 10^3 \, \text{m/s} \).

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