A rocket is fired vertically with a speed of `5 km s^(-1)` from the earth's surface. How far from the earth does the rocket go before returning to the erath? Mass of the earth `= 6.0 xx 10^(24) kg`, mean radius of the earth `= 6.4 xx 10^(6) m. G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`.

Explanation

To solve the problem of how far the rocket goes before returning to Earth, we will apply the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial speed of the rocket, \( v = 5 \, \text{km/s} = 5000 \, \text{m/s} \)
- Mass of the Earth, \( M = 6.0 \times 10^{24} \, \text{kg} \)
- Radius of the Earth, \( R = 6.4 \times 10^{6} \, \text{m} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)

2. **Write the Conservation of Energy Equation:**
The total mechanical energy at the surface (initial) is equal to the total mechanical energy at the maximum height (final).
\[ KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}} \]

3. **Calculate Initial Kinetic Energy (KE) and Potential Energy (PE):**
- Initial Kinetic Energy:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} m (5000)^2 \]
- Initial Potential Energy:
\[ PE_{\text{initial}} = -\frac{GMm}{R} \]

4. **At Maximum Height:**
- Final Kinetic Energy:
\[ KE_{\text{final}} = 0 \quad (\text{at maximum height, velocity is zero}) \]
- Final Potential Energy:
\[ PE_{\text{final}} = -\frac{GMm}{R + h} \]

5. **Set Up the Equation:**
\[ \frac{1}{2} m (5000)^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R + h} \]

6. **Cancel Mass (m) from the Equation:**
\[ \frac{1}{2} (5000)^2 - \frac{GM}{R} = -\frac{GM}{R + h} \]

7. **Rearranging the Equation:**
\[ \frac{1}{2} (5000)^2 + \frac{GM}{R + h} = \frac{GM}{R} \]

8. **Substituting Values:**
- Calculate \( \frac{GM}{R} \):
\[ \frac{GM}{R} = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{6.4 \times 10^{6}} \]
- Calculate \( \frac{1}{2} (5000)^2 \):
\[ \frac{1}{2} (5000)^2 = 12500000 \, \text{J} \]

9. **Solving for \( h \):**
Rearranging gives:
\[ h = \frac{GM}{\left(\frac{1}{2} (5000)^2 + \frac{GM}{R}\right)} - R \]

10. **Final Calculation:**
- Substitute the values into the equation and solve for \( h \):
\[ h \approx 1.59 \times 10^{6} \, \text{m} \]

### Final Answer:
The rocket goes approximately \( 1.59 \times 10^{6} \, \text{m} \) away from the Earth before returning.