A spherical planet has uniform density `pi/2xx10^(4)kg//m^(3)`. Find out the minimum period for a satellite in a circular orbit around it in seconds (Use `G=20/3xx10^(-11) (N-m^(2))/(kg^(2))`).

Explanation

To find the minimum period for a satellite in a circular orbit around a spherical planet with uniform density, we can follow these steps:

### Step 1: Calculate the Mass of the Planet
The mass \( M \) of the planet can be calculated using the formula:
\[ M = \rho \cdot V \]
where \( \rho \) is the density and \( V \) is the volume of the planet. For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass becomes:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 2: Substitute the Density
Given the density \( \rho = \frac{\pi}{2} \times 10^4 \, \text{kg/m}^3 \), we substitute this into the mass equation:
\[ M = \left(\frac{\pi}{2} \times 10^4\right) \cdot \frac{4}{3} \pi R^3 \]
This simplifies to:
\[ M = \frac{2\pi^2}{3} \times 10^4 R^3 \]

### Step 3: Gravitational Force and Centripetal Force
For a satellite in circular orbit, the gravitational force provides the necessary centripetal force. The gravitational force \( F_g \) is given by:
\[ F_g = \frac{G M m}{R^2} \]
where \( G \) is the gravitational constant, \( m \) is the mass of the satellite, and \( R \) is the radius of the planet.

The centripetal force \( F_c \) required for circular motion is:
\[ F_c = \frac{m v^2}{R} \]
where \( v \) is the orbital speed of the satellite.

### Step 4: Equate Gravitational Force and Centripetal Force
Setting the gravitational force equal to the centripetal force:
\[ \frac{G M m}{R^2} = \frac{m v^2}{R} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{R^2} = \frac{v^2}{R} \]
This simplifies to:
\[ v^2 = \frac{G M}{R} \]

### Step 5: Substitute for Mass \( M \)
Substituting the expression for \( M \):
\[ v^2 = \frac{G \left(\frac{2\pi^2}{3} \times 10^4 R^3\right)}{R} \]
This simplifies to:
\[ v^2 = \frac{2\pi^2}{3} \times 10^4 R^2 G \]

### Step 6: Calculate the Period
The orbital period \( T \) is related to the speed \( v \) and the radius \( R \) by:
\[ T = \frac{2\pi R}{v} \]
Substituting \( v \):
\[ T = \frac{2\pi R}{\sqrt{\frac{2\pi^2}{3} \times 10^4 R^2 G}} \]
This simplifies to:
\[ T = \frac{2\pi R}{\sqrt{\frac{2\pi^2}{3} \times 10^4 G} \cdot R} \]
\[ T = \frac{2\pi}{\sqrt{\frac{2\pi^2}{3} \times 10^4 G}} \]

### Step 7: Substitute Values
Now substituting \( G = \frac{20}{3} \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \):
\[ T = \frac{2\pi}{\sqrt{\frac{2\pi^2}{3} \times 10^4 \times \frac{20}{3} \times 10^{-11}}} \]
Calculating this gives:
\[ T = \frac{2\pi}{\sqrt{\frac{40\pi^2}{9} \times 10^{-7}}} \]
\[ T = \frac{2\pi}{\frac{2\pi}{3} \times 10^{-3.5}} = 3 \times 10^{3} \text{ seconds} \]

### Final Answer
Thus, the minimum period for the satellite in a circular orbit around the planet is:
\[ T \approx 3000 \text{ seconds} \]