Orbital velocity of a satellite revolving in a circular path close to the planet

Options

A. is directly propotional to the density of the planet
B. is directly propotional to the square root of density of the planet
C. is directly propotional to the square of density of planet
D. is inversely proportional to the square root of density of planet

Answer

To determine the orbital velocity of a satellite revolving in a circular path close to a planet, we can follow these steps:

### Step 1: Understand the Forces Acting on the Satellite
A satellite in orbit experiences two main forces: gravitational force pulling it towards the planet and the centrifugal force due to its circular motion. For a satellite to maintain a stable orbit, these two forces must be balanced.

### Step 2: Write the Expression for Gravitational Force
The gravitational force \( F_g \) acting on the satellite can be expressed using Newton's law of gravitation:
\[ F_g = \frac{G M m}{r^2} \]
where:
- \( G \) is the 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.

### Step 3: Write the Expression for Centrifugal Force
The centrifugal force \( F_c \) acting on the satellite can be expressed as:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the orbital velocity of the satellite.

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

### Step 5: Simplify the Equation
We can cancel the mass of the satellite \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \) gives:
\[ \frac{G M}{r} = v^2 \]

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

### Step 7: Relate Mass to Density
The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
For a spherical planet, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the planet. Thus, we can write:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 8: Substitute Mass into the Velocity Equation
Substituting this expression for \( M \) back into the equation for \( v \):
\[ v = \sqrt{\frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{r}} \]
Since \( r \) is approximately equal to \( R \) when the satellite is close to the planet, we can simplify this to:
\[ v \propto \sqrt{\rho} \]

### Conclusion
Thus, the orbital velocity of a satellite revolving close to a planet is directly proportional to the square root of the density of the planet. Therefore, the correct option is:
**Option 2: directly proportional to the square root of density of planet.**