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.

Options

A. `sqrt(gR)`
B. `sqrt(1.5gR)`
C. `sqrt((gR)/(2))`
D. `sqrt((gR)/(sqrt2))`

Answer

To solve the problem, we will break it down into two parts as given 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 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. **Given Speed of the Satellite**:
The speed of the satellite \( v \) is given as half of the escape velocity:
\[ v = \frac{1}{2} v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} \]

3. **Centripetal Force and Gravitational Force**:
For a satellite in circular motion, the gravitational force provides the necessary centripetal force:
\[ \frac{GMm}{(R+h)^2} = \frac{mv^2}{R+h} \]
where \( m \) is the mass of the satellite and \( h \) is the height above the Earth's surface.

4. **Substituting the Speed**:
Substitute \( v \) into the equation:
\[ \frac{GMm}{(R+h)^2} = \frac{m\left(\sqrt{\frac{GM}{2R}}\right)^2}{R+h} \]
This simplifies to:
\[ \frac{GM}{(R+h)^2} = \frac{GM}{2R(R+h)} \]

5. **Cross Multiplying**:
Cross-multiplying gives:
\[ 2R(R+h)^2 = (R+h)^2 \]
Simplifying this leads to:
\[ 2R = R + h \implies h = R \]

6. **Final Height**:
Therefore, the height of the satellite above the Earth's surface is:
\[ h = R \]

### 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, we can use the conservation of mechanical energy. Initially, the satellite has gravitational potential energy and no kinetic energy (since it is stopped):
\[ U_i = -\frac{GMm}{R+h} \]
Final potential energy when it hits the ground:
\[ U_f = -\frac{GMm}{R} \]
The change in potential energy will equal the kinetic energy just before it hits the ground.

2. **Setting Up the Equation**:
The change in potential energy is:
\[ \Delta U = U_f - U_i = -\frac{GMm}{R} - \left(-\frac{GMm}{R+h}\right) \]
This simplifies to:
\[ \Delta U = -\frac{GMm}{R} + \frac{GMm}{R+h} \]

3. **Equating to Kinetic Energy**:
The kinetic energy just before impact is:
\[ K = \frac{1}{2} mv^2 \]
Setting the change in potential energy equal to the kinetic energy gives:
\[ \frac{GMm}{R+h} - \frac{GMm}{R} = \frac{1}{2} mv^2 \]
Canceling \( m \) and rearranging gives:
\[ v^2 = \frac{2GM}{R} \left(\frac{1}{R+h} - \frac{1}{R}\right) \]

4. **Substituting \( h = R \)**:
Substitute \( h = R \):
\[ v^2 = \frac{2GM}{R} \left(\frac{1}{2R} - \frac{1}{R}\right) = \frac{2GM}{R} \left(\frac{1}{2R} - \frac{2}{2R}\right) = \frac{2GM}{R} \left(-\frac{1}{2R}\right) \]
This simplifies to:
\[ v^2 = \frac{GM}{R} \]

5. **Final Speed**:
Thus, the speed with which it hits the surface of the Earth is:
\[ v = \sqrt{\frac{GM}{R}} = \sqrt{gR} \]

### Summary of Results:
- Height of the satellite above the Earth's surface: \( h = R \)
- Speed with which it hits the surface of the Earth: \( v = \sqrt{gR} \)