An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height (h) of the satellite above the earth’s surface is (Take radius of earth as `R_(e )`)

Options

A. `h=R_(e )^(2)`
B. `h=R_(e )`
C. `h=2R_(e )`
D. `h=4R_(e )`

Answer

To solve the problem of finding the height (h) of an artificial satellite moving in a circular orbit around the Earth with a speed equal to half the escape velocity, we can follow these steps:

### Step 1: Understand Escape Velocity
The escape velocity (v_e) from the surface of the Earth is given by the formula:
\[ v_e = \sqrt{2gR_e} \]
where:
- \( g \) is the acceleration due to gravity at the surface of the Earth,
- \( R_e \) is the radius of the Earth.

### Step 2: Determine the Speed of the Satellite
According to the problem, the speed (v) of the satellite is half the escape velocity:
\[ v = \frac{1}{2} v_e = \frac{1}{2} \sqrt{2gR_e} \]

### Step 3: Relate Centripetal Force and Gravitational Force
For the satellite in circular motion, the gravitational force provides the necessary centripetal force. The gravitational force (F_gravity) acting on the satellite is given by:
\[ F_{gravity} = \frac{GMm}{r^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite.

The centripetal force (F_c) required for circular motion is:
\[ F_c = \frac{mv^2}{r} \]

### Step 4: Set the Forces Equal
Setting the gravitational force equal to the centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{r^2} = \frac{v^2}{r} \]

### Step 5: Substitute for v
Substituting \( v = \frac{1}{2} \sqrt{2gR_e} \) into the equation:
\[ \frac{GM}{r^2} = \frac{\left(\frac{1}{2} \sqrt{2gR_e}\right)^2}{r} \]
This simplifies to:
\[ \frac{GM}{r^2} = \frac{\frac{1}{4} \cdot 2gR_e}{r} \]
\[ \frac{GM}{r^2} = \frac{gR_e}{2r} \]

### Step 6: Rearranging the Equation
Cross-multiplying gives:
\[ GM \cdot r = \frac{gR_e}{2} \cdot r^2 \]
Dividing both sides by \( g \):
\[ \frac{GM}{g} = \frac{R_e}{2} \cdot r \]

### Step 7: Substitute for g
We know that \( g = \frac{GM}{R_e^2} \), so substituting gives:
\[ \frac{GM}{g} = R_e^2 \]
Thus:
\[ R_e^2 = \frac{R_e}{2} \cdot r \]
Solving for \( r \):
\[ r = 2R_e \]

### Step 8: Calculate Height (h)
The distance \( r \) is the distance from the center of the Earth to the satellite:
\[ r = R_e + h \]
Substituting \( r = 2R_e \):
\[ 2R_e = R_e + h \]
Solving for \( h \):
\[ h = 2R_e - R_e = R_e \]

### Final Answer
The height \( h \) of the satellite above the Earth's surface is:
\[ h = R_e \]