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

Options

A. `sqrt((3pi)/(DG))`
B. `((3pi)/(DG))^(3//2)`
C. `sqrt((3pi)/(2DG))`
D. `sqrt((3piG)/(D))`

Answer

To solve the problem of finding 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. **Understanding the Time Period Formula**:
The time 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. **Identifying the Distance**:
Since the satellite is revolving very close to the surface of the planet, we can set:
\[ r = R \]
where \( R \) is the radius of the planet.

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

4. **Substituting Mass into the Time Period Formula**:
Now, substituting \( M \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{R^3}{G \left(\rho \frac{4}{3} \pi R^3\right)}} \]

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

6. **Final Expression for the Time Period**:
Therefore, the period of revolution of the satellite is:
\[ T = 2\pi \sqrt{\frac{3}{4G\rho}} = \frac{3\pi}{\sqrt{G\rho}} \]

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