The escape velocity of a projectile from the earth is approximately

Options

A. 11.2 m / sec
B. 112 km / sec
C. 11.2 km / sec
D. 11200 km / sec

Answer

To find the escape velocity of a projectile from the Earth, we can use the formula for escape velocity, which is derived from the principles of energy conservation. Here's a step-by-step solution:

### Step 1: Understand the Concept of Escape Velocity
Escape velocity is the minimum velocity required for an object to break free from the gravitational attraction of a celestial body without any additional propulsion.

### Step 2: Use the Escape Velocity Formula
The formula for escape velocity \( v \) from the surface of a planet is given by:
\[ v = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 5.972 \times 10^{24} \, \text{kg} \)),
- \( R \) is the radius of the Earth (\( 6.371 \times 10^6 \, \text{m} \)).

### Step 3: Substitute the Values
Now, we can substitute the values of \( G \), \( M \), and \( R \) into the formula:
\[ v = \sqrt{\frac{2 \times (6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{6.371 \times 10^6}} \]

### Step 4: Calculate the Value Inside the Square Root
Calculating the numerator:
\[ 2 \times (6.674 \times 10^{-11}) \times (5.972 \times 10^{24}) \approx 7.964 \times 10^{14} \]
Now, divide by the radius:
\[ \frac{7.964 \times 10^{14}}{6.371 \times 10^6} \approx 1.250 \times 10^8 \]

### Step 5: Take the Square Root
Now, take the square root of the result:
\[ v \approx \sqrt{1.250 \times 10^8} \approx 11,180 \, \text{m/s} \approx 11.2 \, \text{km/s} \]

### Final Answer
Thus, the escape velocity from the Earth is approximately:
\[ \boxed{11.2 \, \text{km/s}} \]