If the period of revolution of an artificial satellite just above the earth's surface is T and the density of earth is p, then `pT^(2)` is

Options

A. a universal constant whose value is `(3pi)/(G)`
B. a universal constant whose value is `(3pi)/(2G)`
C. proportional to radius of earth R
D. proportional to square of the radius of earth `R^(2)` Here, G=universal gravitational constant

Answer

To solve the problem, we will derive the relationship between the period of revolution \( T \) of an artificial satellite just above the Earth's surface and the density \( \rho \) of the Earth.

### Step 1: Understand the Forces Acting on the Satellite
The gravitational force acting on the satellite provides the necessary centripetal force for its circular motion. The gravitational force \( F_g \) is given by:

\[ F_g = \frac{G M m}{R^2} \]

where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( R \) is the radius of the Earth.

The centripetal force \( F_c \) required for the satellite to maintain its circular orbit is given by:

\[ F_c = \frac{m v^2}{R} \]

where \( v \) is the orbital velocity of the satellite.

### Step 2: Set the Gravitational Force Equal to the Centripetal Force
Since these two forces are equal, we can set them equal to each other:

\[ \frac{G M m}{R^2} = \frac{m v^2}{R} \]

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ \frac{G M}{R^2} = \frac{v^2}{R} \]

Multiplying both sides by \( R \):

\[ \frac{G M}{R} = v^2 \]

### Step 3: Express the Velocity in Terms of \( G \), \( M \), and \( R \)
From the above equation, we can express the velocity \( v \):

\[ v = \sqrt{\frac{G M}{R}} \]

### Step 4: Relate the Period \( T \) to the Velocity
The period \( T \) of revolution is the time taken to complete one full orbit. The circumference of the orbit is \( 2 \pi R \), so:

\[ T = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2 \pi R}{v} \]

Substituting the expression for \( v \):

\[ T = \frac{2 \pi R}{\sqrt{\frac{G M}{R}}} \]

### Step 5: Simplify the Expression for \( T \)
Simplifying the expression for \( T \):

\[ T = 2 \pi R \cdot \sqrt{\frac{R}{G M}} = 2 \pi \sqrt{\frac{R^3}{G M}} \]

### Step 6: Express Mass \( M \) in Terms of Density \( \rho \)
The mass \( M \) of the Earth can be expressed in terms of its density \( \rho \):

\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 7: Substitute Mass \( M \) into the Period Equation
Substituting \( M \) into the equation for \( T \):

\[ T = 2 \pi \sqrt{\frac{R^3}{G \left(\rho \cdot \frac{4}{3} \pi R^3\right)}} \]

### Step 8: Simplify the Expression for \( T \)
This simplifies to:

\[ T = 2 \pi \sqrt{\frac{3}{4 \pi G \rho}} \]

### Step 9: Square the Period \( T \)
Squaring both sides gives:

\[ T^2 = \frac{4 \pi^2}{4 \pi G \rho} \cdot 3 = \frac{3 \pi}{G \rho} \]

### Step 10: Rearranging for \( \rho T^2 \)
Multiplying both sides by \( \rho \):

\[ \rho T^2 = \frac{3 \pi}{G} \]

Thus, we find:

\[ \rho T^2 = \frac{3 \pi}{G} \]

### Final Answer
The final result is:

\[ \rho T^2 = \frac{3 \pi}{G} \]