An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth.
(i) Determine the height of the satellite above the earth's surface.
(ii) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed with which it hits the surface of the earth.

Explanation

To solve the problem step by step, we will break it down into two parts as specified in the question.

### Part (i): Determine the height of the satellite above the Earth's surface.

1. **Understanding Escape Velocity**:
The escape velocity \( V_e \) from the Earth is given by the formula:
\[ V_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity at the Earth's surface, and \( R \) is the radius of the Earth.

2. **Given Velocity of the Satellite**:
The satellite is moving with a speed equal to half the escape velocity:
\[ V = \frac{1}{2} V_e = \frac{1}{2} \sqrt{2gR} \]

3. **Centripetal Force and Gravitational Force**:
For a satellite in circular orbit, the centripetal force is provided by the gravitational force. Thus, we can set up the equation:
\[ \frac{mv^2}{R + H} = \frac{GMm}{(R + H)^2} \]
where \( H \) is the height of the satellite above the Earth's surface, \( m \) is the mass of the satellite, and \( M \) is the mass of the Earth.

4. **Substituting the Velocity**:
Substitute \( V \) into the centripetal force equation:
\[ \frac{m\left(\frac{1}{2} \sqrt{2gR}\right)^2}{R + H} = \frac{GMm}{(R + H)^2} \]
Simplifying the left side:
\[ \frac{m \cdot \frac{1}{4} \cdot 2gR}{R + H} = \frac{GMm}{(R + H)^2} \]
This simplifies to:
\[ \frac{mgR}{2(R + H)} = \frac{GM}{(R + H)^2} \]

5. **Cancelling Mass and Rearranging**:
Cancel \( m \) and rearranging gives:
\[ gR(R + H) = 2GM \]
Using \( g = \frac{GM}{R^2} \):
\[ \frac{GM}{R^2} R(R + H) = 2GM \]
Simplifying:
\[ R(R + H) = 2R^2 \]
\[ R + H = 2R \]
\[ H = R \]

6. **Calculating Height**:
Given that the radius of the Earth \( R \approx 6400 \) km, the height \( H \) is:
\[ H = 6400 \text{ km} \]

### Part (ii): Find the speed with which it hits the surface of the Earth.

1. **Energy Conservation**:
When the satellite is stopped and allowed to fall, we can use the conservation of energy principle. The potential energy at height \( H \) will convert into kinetic energy just before it hits the Earth.

2. **Potential Energy at Height \( H \)**:
The potential energy \( PE \) at height \( H \) is:
\[ PE = -\frac{GMm}{R + H} \]
At the surface of the Earth (height = 0), the potential energy is:
\[ PE_{surface} = -\frac{GMm}{R} \]

3. **Change in Potential Energy**:
The change in potential energy as it falls from height \( H \) to the surface is:
\[ \Delta PE = PE_{surface} - PE = -\frac{GMm}{R} + \frac{GMm}{R + H} \]

4. **Kinetic Energy Just Before Impact**:
The kinetic energy \( KE \) just before hitting the surface is given by:
\[ KE = \frac{1}{2} mv^2 \]

5. **Setting Up the Equation**:
By conservation of energy:
\[ \Delta PE = KE \]
Thus,
\[ -\frac{GMm}{R} + \frac{GMm}{R + H} = \frac{1}{2} mv^2 \]

6. **Solving for \( v \)**:
Cancel \( m \) and rearranging gives:
\[ \frac{GM}{R + H} - \frac{GM}{R} = \frac{1}{2} v^2 \]
Substituting \( H = R \):
\[ \frac{GM}{2R} = \frac{1}{2} v^2 \]
Thus,
\[ v^2 = \frac{GM}{R} \]
Therefore,
\[ v = \sqrt{\frac{GM}{R}} \]

7. **Calculating the Speed**:
Using \( g = \frac{GM}{R^2} \):
\[ v = \sqrt{gR} \]
Substituting \( g \approx 10 \, \text{m/s}^2 \) and \( R = 6400 \times 10^3 \, \text{m} \):
\[ v \approx \sqrt{10 \times 6400 \times 10^3} \approx 7920 \, \text{m/s} \approx 7.92 \, \text{km/s} \]

### Final Answers:
(i) The height of the satellite above the Earth's surface is \( 6400 \) km.
(ii) The speed with which it hits the surface of the Earth is approximately \( 7.92 \, \text{km/s} \).