A setellite os revolving very close to a planet of density p. The period of revolution of satellite is

Options

A. `sqrt((3pirho)/G)`
B. `sqrt((3pi)/(2rhoG))`
C. `sqrt((3pi)/(Grho))`
D. `sqrt((3piG)/rho)`

Answer

To find the period of revolution of a satellite revolving very close to a planet of density \( \rho \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for the period of a satellite**:
The period \( T \) of a satellite in orbit is given by the formula:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
where \( r \) is the distance from the center of the planet to the satellite, \( G \) is the gravitational constant, and \( M \) is the mass of the planet.

2. **Identify the distance \( r \)**:
Since the satellite is revolving very close to the surface of the planet, we can take \( r \) to be equal to the radius of the planet \( R \). Therefore, we have:
\[ r = R \]

3. **Express the mass \( M \) in terms of density \( \rho \)**:
The mass of the planet can be expressed in terms of its density and volume. For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass \( M \) of the planet can be expressed as:
\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) = \frac{4}{3} \pi \rho R^3 \]

4. **Substitute \( M \) into the period formula**:
Now we can substitute this expression for \( M \) back into the period formula:
\[ T = 2\pi \sqrt{\frac{R^3}{G \left(\frac{4}{3} \pi \rho R^3\right)}} \]

5. **Simplify the expression**:
Simplifying the equation gives:
\[ T = 2\pi \sqrt{\frac{R^3}{\frac{4}{3} \pi G \rho R^3}} = 2\pi \sqrt{\frac{3}{4\pi G \rho}} \]
This simplifies to:
\[ T = \frac{2\pi}{\sqrt{4\pi G \rho}} \cdot \sqrt{3} = \frac{3\pi}{\sqrt{4\pi G \rho}} \]

6. **Final expression**:
Thus, the period of revolution of the satellite is:
\[ T = \frac{3\pi}{\sqrt{G \rho}} \]

### Final Answer:
The period of revolution of the satellite is:
\[ T = \frac{3\pi}{\sqrt{G \rho}} \]