In circular orbit of a satellite

Options

A. orbital speed is `sqrt((GM)/(r ))`
B. time period `T^(2) prop r^(3)`
C. kinetic energy is `(GMm)/(2r)`
D. potential energy is `-(GMm)/(2r)`

Answer

To solve the question regarding the properties of a satellite in circular orbit, we will analyze each statement provided in the options and derive the necessary formulas step by step.

### Step 1: Understanding the Forces Acting on the Satellite
In a circular orbit, a satellite is subject to two main forces:
1. The gravitational force acting as the centripetal force.
2. The centripetal force required to keep the satellite in circular motion.

### Step 2: Equating Gravitational Force and Centripetal Force
The gravitational force \( F_g \) acting on the satellite is given by:
\[ F_g = \frac{G M m}{r^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the planet to the satellite.

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.

Setting these two forces equal gives us:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]

### Step 3: Solving for Orbital Speed
We can simplify this equation by canceling \( m \) (assuming \( m \neq 0 \)) and multiplying both sides by \( r \):
\[ \frac{G M}{r} = v^2 \]
Taking the square root of both sides, we find the orbital speed \( v \):
\[ v = \sqrt{\frac{G M}{r}} \]
This confirms that the first option, \( v = \sqrt{\frac{G M}{r}} \), is correct.

### Step 4: Deriving the Time Period
Using Kepler's Third Law, we know that the square of the time period \( T \) of a satellite is proportional to the cube of the radius of its orbit:
\[ T^2 \propto r^3 \]
This means that the second option is also correct.

### Step 5: Calculating Kinetic Energy
The kinetic energy \( KE \) of the satellite can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the expression for \( v^2 \) from earlier:
\[ KE = \frac{1}{2} m \left(\frac{G M}{r}\right) = \frac{G M m}{2r} \]
Thus, the third option is correct.

### Step 6: Calculating Potential Energy
The gravitational potential energy \( U \) of the satellite is given by:
\[ U = -\frac{G M m}{r} \]
However, the option states it as \( -\frac{G M m}{2r} \), which is incorrect. Therefore, the fourth option is incorrect.

### Conclusion
The correct options are:
- Option A: Correct
- Option B: Correct
- Option C: Correct
- Option D: Incorrect