A satellite is revolving very close to a planet of density D. The time period of revolution of that planet is

Options

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

Answer

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

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

### Step 2: Determine the radius of the orbit
Since the satellite is revolving very close to the surface of the planet, we can take:
\[ r = R \]
where \( R \) is the radius of the planet.

### Step 3: Express the mass of the planet in terms of its density
The mass \( M \) of the planet can be expressed in terms of its density \( D \) and volume. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass \( M \) can be expressed as:
\[ M = D \cdot V = D \cdot \frac{4}{3} \pi R^3 \]

### Step 4: Substitute the mass into the time period formula
Substituting the expression for \( M \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{R^3}{G \left( D \cdot \frac{4}{3} \pi R^3 \right)}} \]

### Step 5: Simplify the expression
This simplifies to:
\[ T = 2\pi \sqrt{\frac{R^3}{\frac{4}{3} \pi G D R^3}} = 2\pi \sqrt{\frac{3}{4\pi G D}} \]
This can be further simplified to:
\[ T = \frac{3\pi}{\sqrt{4\pi G D}} \]

### Final Answer
Thus, the time period of revolution of the satellite is:
\[ T = \frac{3\pi}{\sqrt{4\pi G D}} \]