If G is the uciversal gravitational constant and p is the uniform density of a spherical planet. Then shortest possible period o0f rotation around a planet can be

Options

A. `sqrt((piG)/(2p))`
B. `sqrt((3piGP)/(p))`
C. `sqrt((pi)/(6Gp))`
D. `sqrt((3pi)/(Gp))`

Answer

To find the shortest possible period of rotation around a spherical planet with uniform density \( \rho \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Orbital Motion**:
The shortest possible period of rotation (or orbital period) for an object around a planet occurs when the object is at the surface of the planet. This is because the gravitational force is strongest at the surface.

2. **Use the Gravitational Force Equation**:
The gravitational force acting on a satellite of mass \( m \) at a distance \( R \) (the radius of the planet) from the center of the planet is given by:
\[ F = \frac{G M m}{R^2} \]
where \( M \) is the mass of the planet and \( G \) is the universal gravitational constant.

3. **Relate Mass to Density**:
The mass \( M \) of the planet can be expressed in terms of its uniform density \( \rho \) and volume \( V \):
\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

4. **Substitute Mass into the Gravitational Force Equation**:
Substitute \( M \) into the gravitational force equation:
\[ F = \frac{G \left(\rho \frac{4}{3} \pi R^3\right) m}{R^2} = \frac{4 \pi G \rho m R}{3} \]

5. **Set Gravitational Force Equal to Centripetal Force**:
For an object in circular motion, the centripetal force required is given by:
\[ F_c = \frac{m v^2}{R} \]
where \( v \) is the orbital speed. Setting the gravitational force equal to the centripetal force gives:
\[ \frac{4 \pi G \rho m R}{3} = \frac{m v^2}{R} \]

6. **Cancel Mass and Rearrange**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{4 \pi G \rho R}{3} = \frac{v^2}{R} \]
Rearranging gives:
\[ v^2 = \frac{4 \pi G \rho R^2}{3} \]

7. **Relate Orbital Speed to Period**:
The orbital speed \( v \) is also related to the orbital period \( T \) by:
\[ v = \frac{2 \pi R}{T} \]
Squaring both sides gives:
\[ v^2 = \frac{4 \pi^2 R^2}{T^2} \]

8. **Set the Two Expressions for \( v^2 \) Equal**:
Set the two expressions for \( v^2 \) equal to each other:
\[ \frac{4 \pi^2 R^2}{T^2} = \frac{4 \pi G \rho R^2}{3} \]

9. **Cancel \( R^2 \) and Solve for \( T^2 \)**:
Cancel \( R^2 \) (assuming \( R \neq 0 \)):
\[ \frac{4 \pi^2}{T^2} = \frac{4 \pi G \rho}{3} \]
Rearranging gives:
\[ T^2 = \frac{3 \pi}{G \rho} \]

10. **Take the Square Root to Find \( T \)**:
Finally, taking the square root gives:
\[ T = \sqrt{\frac{3 \pi}{G \rho}} \]

### Final Answer:
The shortest possible period of rotation around the planet is:
\[ T = \sqrt{\frac{3 \pi}{G \rho}} \]