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 given problem, we will break it down into two parts as stated in the question.

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

1. **Understanding the relationship between orbital speed and escape velocity**:
The escape velocity \( v_e \) from the Earth is given by:
\[ v_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity at the Earth's surface and \( R \) is the radius of the Earth.

According to the problem, the speed of the satellite \( v \) is half of the escape velocity:
\[ v = \frac{1}{2} v_e = \frac{1}{2} \sqrt{2gR} = \frac{\sqrt{gR}}{\sqrt{2}} \]

2. **Using the formula for orbital velocity**:
The orbital velocity \( v \) of a satellite in a circular orbit at height \( h \) above the Earth's surface is given by:
\[ v = \sqrt{\frac{gM}{R + h}} \]
where \( M \) is the mass of the Earth.

3. **Setting the two expressions for velocity equal**:
Equating the two expressions for \( v \):
\[ \frac{\sqrt{gR}}{\sqrt{2}} = \sqrt{\frac{gM}{R + h}} \]

4. **Squaring both sides**:
\[ \frac{gR}{2} = \frac{gM}{R + h} \]

5. **Cross-multiplying and simplifying**:
\[ gR(R + h) = 2gM \]
\[ R(R + h) = 2M \]

6. **Substituting \( M \) in terms of \( R \)**:
Using \( g = \frac{GM}{R^2} \), we can express \( M \) as \( M = \frac{gR^2}{G} \):
\[ R(R + h) = 2 \left(\frac{gR^2}{G}\right) \]

7. **Solving for \( h \)**:
Rearranging gives:
\[ R + h = \frac{2gR}{g} = 2R \]
\[ h = 2R - R = R \]

8. **Calculating height**:
The radius of the Earth \( R \) is approximately \( 6400 \) km, thus:
\[ h = 6400 \text{ km} \]

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

1. **Using conservation of energy**:
When the satellite is stopped and allowed to fall freely, its potential energy at height \( h \) converts into kinetic energy just before it hits the surface.

2. **Potential energy at height \( h \)**:
The potential energy \( PE \) at height \( h \) is:
\[ PE = -\frac{GMm}{R + h} \]
where \( m \) is the mass of the satellite.

3. **Potential energy at the Earth's surface**:
The potential energy at the Earth's surface is:
\[ PE_{surface} = -\frac{GMm}{R} \]

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

5. **Setting the change in potential energy equal to kinetic energy**:
The kinetic energy \( KE \) just before hitting the surface is:
\[ KE = \frac{1}{2} mv^2 \]
Thus,
\[ \Delta PE = KE \]

6. **Solving for \( v \)**:
\[ \frac{GMm}{R + h} - \frac{GMm}{R} = \frac{1}{2} mv^2 \]
Canceling \( m \) and simplifying gives:
\[ v^2 = \frac{2GM}{R} \left(\frac{h}{R(R + h)}\right) \]

7. **Substituting \( h = R \)**:
\[ v^2 = \frac{2GM}{R} \left(\frac{R}{R(R + R)}\right) = \frac{GM}{R} \]

8. **Calculating final speed**:
The speed with which it hits the surface is:
\[ v = \sqrt{gR} \]
Substituting \( g \approx 10 \, \text{m/s}^2 \) and \( R = 6400 \times 10^3 \, \text{m} \):
\[ v \approx \sqrt{10 \times 6400 \times 10^3} \approx 7920 \, \text{m/s} \]

### Final Answers:
(i) The height of the satellite above the Earth's surface is approximately **6400 km**.
(ii) The speed with which it hits the surface of the Earth is approximately **7920 m/s**.