The initial velocity `v_(i)` required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity `v_(e )` such that `v_(i)= sqrt((x)/(y)) xx v_(e )`. The value of x will be __________ .

Answer

To solve the problem of finding the initial velocity \( v_i \) required to project a body vertically upward from the surface of the Earth to reach a height of \( 10R \) (where \( R \) is the radius of the Earth), we will use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the initial velocity \( v_i \) such that a body reaches a height of \( 10R \). At this height, the body's velocity will be zero.

2. **Using Conservation of Energy**:
The total mechanical energy at the surface of the Earth (initial) must equal the total mechanical energy at the height of \( 10R \) (final).

\[ \text{Initial Energy} = \text{Final Energy} \]

The initial energy consists of kinetic energy and gravitational potential energy, while the final energy at height \( 10R \) consists only of gravitational potential energy since the kinetic energy is zero at the maximum height.

3. **Writing the Energy Equations**:
- **Initial Energy**:
\[ E_i = \frac{1}{2} m v_i^2 - \frac{GMm}{R} \]
- **Final Energy**:
\[ E_f = -\frac{GMm}{11R} \]
(At height \( 10R \), the distance from the center of the Earth is \( R + 10R = 11R \))

4. **Setting Up the Equation**:
Setting the initial energy equal to the final energy:
\[ \frac{1}{2} m v_i^2 - \frac{GMm}{R} = -\frac{GMm}{11R} \]

5. **Simplifying the Equation**:
- Multiply through by \( 11R \) to eliminate the denominators:
\[ 11R \left( \frac{1}{2} m v_i^2 \right) - 11GMm = -GMm \]
- Rearranging gives:
\[ 11R \left( \frac{1}{2} m v_i^2 \right) = 10GMm \]

6. **Canceling Mass**:
Since \( m \) appears in all terms, we can cancel it out:
\[ 11R \left( \frac{1}{2} v_i^2 \right) = 10GM \]

7. **Solving for \( v_i^2 \)**:
\[ \frac{1}{2} v_i^2 = \frac{10GM}{11R} \]
\[ v_i^2 = \frac{20GM}{11R} \]

8. **Relating to Escape Velocity**:
The escape velocity \( v_e \) is given by:
\[ v_e = \sqrt{2gR} = \sqrt{\frac{2GM}{R}} \]
Therefore, \( GM = \frac{v_e^2 R}{2} \).

9. **Substituting for \( GM \)**:
Substitute \( GM \) in the equation for \( v_i^2 \):
\[ v_i^2 = \frac{20 \cdot \frac{v_e^2 R}{2}}{11R} \]
\[ v_i^2 = \frac{10 v_e^2}{11} \]

10. **Finding \( v_i \)**:
Taking the square root gives:
\[ v_i = \sqrt{\frac{10}{11}} v_e \]

11. **Identifying \( x \) and \( y \)**:
The problem states that \( v_i = \sqrt{\frac{x}{y}} v_e \). From our equation:
\[ \sqrt{\frac{10}{11}} = \sqrt{\frac{x}{y}} \]
Thus, \( x = 10 \) and \( y = 11 \).

### Final Answer:
The value of \( x \) will be **10**.