The escape velocity of a projectile on the earth's surface is `11.2 kms^(-1)` . A body is projected out with 4 times this speed. What is the speed of the body far way from the earth?

Explanation

To solve the problem, we will use the concept of conservation of energy. The total mechanical energy (kinetic + potential) at the surface of the Earth will equal the total mechanical energy when the body is far away from the Earth.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Escape velocity (V) = 11.2 km/s = 11,200 m/s
- The body is projected with a speed of 4 times the escape velocity.
- Therefore, initial speed (u) = 4 * V = 4 * 11,200 m/s = 44,800 m/s.

2. **Write the Conservation of Energy Equation:**
The conservation of energy states:
\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]
Mathematically:
\[ KE_i + PE_i = KE_f + PE_f \]

3. **Calculate Initial Kinetic Energy (KE_i):**
The initial kinetic energy (KE_i) is given by:
\[ KE_i = \frac{1}{2} m u^2 \]
where \( u = 44,800 \, \text{m/s} \).

4. **Calculate Initial Potential Energy (PE_i):**
The initial potential energy (PE_i) at the surface of the Earth is given by:
\[ PE_i = -\frac{GMm}{R} \]
where:
- \( G \) = gravitational constant = \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) = mass of the Earth = \( 6 \times 10^{24} \, \text{kg} \)
- \( R \) = radius of the Earth = \( 6.4 \times 10^6 \, \text{m} \)

5. **Calculate Final Potential Energy (PE_f):**
As the body moves far away from the Earth, the potential energy approaches zero:
\[ PE_f = 0 \]

6. **Set Up the Energy Equation:**
Now substituting the values into the conservation of energy equation:
\[ \frac{1}{2} m (44,800)^2 - \frac{GMm}{R} = \frac{1}{2} m v^2 + 0 \]

7. **Simplify the Equation:**
Since mass \( m \) appears in all terms, it cancels out:
\[ \frac{1}{2} (44,800)^2 - \frac{GM}{R} = \frac{1}{2} v^2 \]

8. **Calculate \( \frac{GM}{R} \):**
\[ \frac{GM}{R} = \frac{(6.67 \times 10^{-11})(6 \times 10^{24})}{6.4 \times 10^6} \approx 9.81 \, \text{m/s}^2 \times 6.4 \times 10^6 \approx 62.7 \times 10^6 \, \text{m}^2/\text{s}^2 \]

9. **Substitute and Solve for \( v \):**
\[ \frac{1}{2} (44,800)^2 - 62.7 \times 10^6 = \frac{1}{2} v^2 \]
\[ \frac{1}{2} (2,008,640,000) - 62.7 \times 10^6 = \frac{1}{2} v^2 \]
\[ 1,004,320,000 - 62.7 \times 10^6 = \frac{1}{2} v^2 \]
\[ 1,004,257,300 = \frac{1}{2} v^2 \]
\[ v^2 = 2 \times 1,004,257,300 \]
\[ v^2 \approx 2,008,514,600 \]
\[ v \approx \sqrt{2,008,514,600} \approx 44,800 \, \text{m/s} \]

10. **Final Speed Far Away from Earth:**
The speed of the body far away from the Earth is approximately \( 44,800 \, \text{m/s} \).