The escape velocity of a body from the surface of earth is

Options

A. 11.2 km/s
B. 11.4 km/s
C. 11.6 km/s
D. 11.0 km/s

Answer

To find the escape velocity of a body from the surface of the Earth, we can use the formula for escape velocity, which is given by:

\[ v_e = \sqrt{\frac{2GM}{R}} \]

Where:
- \( v_e \) = escape velocity
- \( G \) = universal gravitational constant \( \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) = mass of the Earth \( \approx 5.972 \times 10^{24} \, \text{kg} \)
- \( R \) = radius of the Earth \( \approx 6.4 \times 10^{6} \, \text{m} \)

### Step 1: Substitute the values into the formula

We start by substituting the known values into the escape velocity formula:

\[ v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (5.972 \times 10^{24})}{6.4 \times 10^{6}}} \]

### Step 2: Calculate the numerator

First, we calculate the numerator:

\[ 2 \times 6.67 \times 10^{-11} \times 5.972 \times 10^{24} \]

Calculating this gives:

\[ = 79.63 \times 10^{13} \, \text{(after adjusting the powers)} \]

### Step 3: Divide by the radius of the Earth

Now, we divide this result by the radius of the Earth:

\[ \frac{79.63 \times 10^{13}}{6.4 \times 10^{6}} = 12.42 \times 10^{7} \]

### Step 4: Take the square root

Now we take the square root of the result:

\[ v_e = \sqrt{12.42 \times 10^{7}} = \sqrt{124.2 \times 10^{6}} = \sqrt{124.2} \times 10^{3} \]

Calculating \( \sqrt{124.2} \) gives approximately \( 11.14 \).

Thus, we have:

\[ v_e \approx 11.14 \times 10^{3} \, \text{m/s} = 11.14 \, \text{km/s} \]

### Step 5: Conclusion

Therefore, the escape velocity from the surface of the Earth is approximately:

\[ v_e \approx 11.2 \, \text{km/s} \]