A satelite is projected from surface of earth so that it can attain 10R height from surface of earth . its speed is `V= V_e((root(x/11))` find x

Answer

To solve the problem, we need to determine the value of \( x \) in the expression for the velocity \( V \) of a satellite projected from the Earth's surface to reach a height of \( 10R \) above the surface of the Earth. The expression is given as:

\[ V = V_e \sqrt{\frac{x}{11}} \]

where \( V_e \) is the escape velocity from the surface of the Earth.

### Step-by-step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( V_e \) from the surface of the Earth is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Total Distance from the Center of the Earth**:
When the satellite reaches a height of \( 10R \) above the Earth's surface, the total distance from the center of the Earth becomes:
\[ d = R + 10R = 11R \]

3. **Potential Energy at Maximum Height**:
At the maximum height, the kinetic energy of the satellite becomes zero, and the potential energy \( U \) at this height is given by:
\[ U = -\frac{GMm}{d} = -\frac{GMm}{11R} \]
where \( m \) is the mass of the satellite.

4. **Initial Kinetic Energy**:
The initial kinetic energy \( K \) when the satellite is projected is:
\[ K = \frac{1}{2} m V^2 \]

5. **Conservation of Energy**:
According to the conservation of mechanical energy, the initial kinetic energy plus the initial potential energy must equal the final potential energy at the maximum height (where kinetic energy is zero):
\[ \frac{1}{2} m V^2 - \frac{GMm}{R} = -\frac{GMm}{11R} \]

6. **Simplifying the Equation**:
We can cancel \( m \) from the equation (assuming \( m \neq 0 \)):
\[ \frac{1}{2} V^2 - \frac{GM}{R} = -\frac{GM}{11R} \]
Rearranging gives:
\[ \frac{1}{2} V^2 = \frac{GM}{R} - \frac{GM}{11R} \]
\[ \frac{1}{2} V^2 = \frac{GM}{R} \left(1 - \frac{1}{11}\right) = \frac{GM}{R} \left(\frac{10}{11}\right) \]

7. **Finding V**:
Multiplying both sides by 2:
\[ V^2 = \frac{20GM}{11R} \]
Taking the square root:
\[ V = \sqrt{\frac{20GM}{11R}} \]

8. **Relating to Escape Velocity**:
We know \( V_e = \sqrt{\frac{2GM}{R}} \). Therefore, we can express \( V \) in terms of \( V_e \):
\[ V = \sqrt{\frac{20}{11}} \cdot V_e \]

9. **Comparing with Given Expression**:
From the problem statement, we have:
\[ V = V_e \sqrt{\frac{x}{11}} \]
Setting the two expressions for \( V \) equal gives:
\[ \sqrt{\frac{20}{11}} \cdot V_e = V_e \sqrt{\frac{x}{11}} \]
Dividing both sides by \( V_e \) (assuming \( V_e \neq 0 \)):
\[ \sqrt{\frac{20}{11}} = \sqrt{\frac{x}{11}} \]

10. **Squaring Both Sides**:
Squaring both sides results in:
\[ \frac{20}{11} = \frac{x}{11} \]
Therefore, multiplying both sides by 11 gives:
\[ x = 20 \]

### Final Answer:
\[ x = 20 \]