A satellite is orbiting the earth in a circular orbit of radius `r`. Its

Options

A. 1. Kinetic energy varies as r
B. 2. angular momentum varies as `(1)/(sqrt(r ))`
C. 3. linear momentum varies as `(1)/( r)`
D. 4. frequency of revolution varies as `(1)/(r^(3//2))`

Explanation

To solve the problem regarding a satellite orbiting the Earth in a circular orbit of radius \( r \), we will analyze the forces acting on the satellite and derive relevant equations step by step.

### Step 1: Understand the Forces Acting on the Satellite
The satellite in a circular orbit experiences two main forces:
1. The gravitational force acting towards the center of the Earth.
2. The centripetal force required to keep the satellite in circular motion.

### Step 2: Write the Gravitational Force Equation
The gravitational force \( F_g \) acting on the satellite can be expressed as:
\[ F_g = \frac{G M m}{r^2} \]
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.

### Step 3: Write the Centripetal Force Equation
The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the orbital speed of the satellite.

### Step 4: Set the Forces Equal
For the satellite to maintain its circular orbit, the gravitational force must equal the centripetal force:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]

### Step 5: Cancel Out Mass and Rearrange
Since \( m \) (mass of the satellite) appears on both sides, we can cancel it out:
\[ \frac{G M}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \) gives:
\[ \frac{G M}{r} = v^2 \]

### Step 6: Solve for Orbital Speed \( v \)
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{G M}{r}} \]

### Step 7: Calculate Kinetic Energy
The kinetic energy \( KE \) of the satellite is given by:
\[ KE = \frac{1}{2} m v^2 \]
Substituting \( v^2 \) from the previous step:
\[ KE = \frac{1}{2} m \left(\frac{G M}{r}\right) = \frac{G M m}{2r} \]

### Step 8: Calculate Potential Energy
The gravitational potential energy \( PE \) of the satellite is given by:
\[ PE = -\frac{G M m}{r} \]

### Step 9: Calculate Total Energy
The total energy \( E \) of the satellite is the sum of its kinetic and potential energy:
\[ E = KE + PE = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r} \]

### Step 10: Analyze Angular Momentum
The angular momentum \( L \) of the satellite is given by:
\[ L = m v r \]
Substituting \( v \):
\[ L = m \left(\sqrt{\frac{G M}{r}}\right) r = m \sqrt{G M r} \]

### Step 11: Analyze Frequency of Revolution
The frequency \( f \) of revolution can be derived from the time period \( T \):
\[ T = \frac{2 \pi r}{v} = \frac{2 \pi r}{\sqrt{\frac{G M}{r}}} = 2 \pi \sqrt{\frac{r^3}{G M}} \]
Thus, the frequency \( f \) is:
\[ f = \frac{1}{T} = \frac{1}{2 \pi} \sqrt{\frac{G M}{r^3}} \]

### Summary of Results
- Kinetic Energy \( KE \) is inversely proportional to \( r \).
- Angular Momentum \( L \) is directly proportional to \( \sqrt{r} \).
- Linear Momentum \( p \) is inversely proportional to \( \sqrt{r} \).
- Frequency \( f \) is inversely proportional to \( r^{3/2} \).