A rocket is accelerated to speed `upsilon = 2sqrt(gR)` near the earth's surface (`R =` radius of earth). Show that very far from earth its speed will be `upsilon = sqrt(2gR)`.

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy of the rocket when it is near the Earth's surface will be equal to its total mechanical energy when it is very far from the Earth.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The speed of the rocket at the Earth's surface is given as \( \upsilon = 2\sqrt{gR} \).
- The radius of the Earth is denoted as \( R \).
- The mass of the rocket is \( m \) and the mass of the Earth is \( M \).

2. **Calculate the Initial Kinetic Energy (KE)**:
\[ KE_{\text{initial}} = \frac{1}{2} m \upsilon^2 = \frac{1}{2} m (2\sqrt{gR})^2 = \frac{1}{2} m (4gR) = 2mgR \]

3. **Calculate the Initial Potential Energy (PE)**:
The gravitational potential energy at the surface of the Earth is given by:
\[ PE_{\text{initial}} = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant.

4. **Total Initial Energy (E_initial)**:
\[ E_{\text{initial}} = KE_{\text{initial}} + PE_{\text{initial}} = 2mgR - \frac{GMm}{R} \]

5. **Calculate the Final Conditions**:
- When the rocket is very far from the Earth, its potential energy approaches zero:
\[ PE_{\text{final}} = 0 \]
- Let the final speed of the rocket be \( \upsilon' \). The final kinetic energy is:
\[ KE_{\text{final}} = \frac{1}{2} m (\upsilon')^2 \]

6. **Total Final Energy (E_final)**:
\[ E_{\text{final}} = KE_{\text{final}} + PE_{\text{final}} = \frac{1}{2} m (\upsilon')^2 + 0 = \frac{1}{2} m (\upsilon')^2 \]

7. **Apply Conservation of Mechanical Energy**:
According to the conservation of energy:
\[ E_{\text{initial}} = E_{\text{final}} \]
Thus,
\[ 2mgR - \frac{GMm}{R} = \frac{1}{2} m (\upsilon')^2 \]

8. **Simplify the Equation**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 2gR - \frac{GM}{R} = \frac{1}{2} (\upsilon')^2 \]

9. **Relate \( g \) and \( G \)**:
Recall that \( g = \frac{GM}{R^2} \), thus \( GM = gR^2 \). Substitute this into the equation:
\[ 2gR - \frac{gR^2}{R} = \frac{1}{2} (\upsilon')^2 \]
Simplifying gives:
\[ 2gR - gR = \frac{1}{2} (\upsilon')^2 \]
\[ gR = \frac{1}{2} (\upsilon')^2 \]

10. **Solve for \( \upsilon' \)**:
Multiply both sides by 2:
\[ 2gR = (\upsilon')^2 \]
Taking the square root:
\[ \upsilon' = \sqrt{2gR} \]

### Conclusion:
Thus, we have shown that the speed of the rocket very far from the Earth is \( \upsilon' = \sqrt{2gR} \).