A satellite revolve round a planet in an orbit just above the surface of a planet. Taking `G = 6.67 xx 10^(-11) Nm^(2) kg^(2)` and mean density of the planet `5.51 xx 10^(3) kg//m^(3)` find the period of the planet.

Explanation

To find the period of a satellite revolving around a planet in an orbit just above its surface, we can follow these steps:

### Step 1: Understand the relationship between gravitational force and orbital motion
The gravitational force provides the necessary centripetal force for the satellite's circular motion. The formula for the gravitational force \( F \) between the satellite and the planet is given by:
\[ F = \frac{GMm}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the planet to the satellite (which is approximately the radius of the planet when the satellite is just above the surface).

### Step 2: Express the mass of the planet in terms of its density
The mass \( M \) of the planet can be expressed in terms of its volume and density:
\[ M = \text{Density} \times \text{Volume} = \rho \times \frac{4}{3} \pi r^3 \]
where \( \rho \) is the mean density of the planet.

### Step 3: Substitute the mass of the planet into the gravitational force equation
Substituting \( M \) into the gravitational force equation gives:
\[ F = \frac{G \left(\rho \cdot \frac{4}{3} \pi r^3\right) m}{r^2} \]
This simplifies to:
\[ F = \frac{4}{3} \pi G \rho m r \]

### Step 4: Set the gravitational force equal to the centripetal force
The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the orbital velocity of the satellite. Setting the gravitational force equal to the centripetal force:
\[ \frac{4}{3} \pi G \rho m r = \frac{mv^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{4}{3} \pi G \rho r = v^2 \]

### Step 5: Express the orbital velocity in terms of the radius and density
From the above equation, we can express the orbital velocity \( v \):
\[ v = \sqrt{\frac{4}{3} \pi G \rho r} \]

### Step 6: Relate the period \( T \) to the orbital velocity
The period \( T \) of the satellite is related to the orbital velocity and the radius by:
\[ T = \frac{2\pi r}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2\pi r}{\sqrt{\frac{4}{3} \pi G \rho r}} = 2\pi \sqrt{\frac{r}{\frac{4}{3} \pi G \rho}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{3r}{4\pi G \rho}} \]

### Step 7: Substitute the known values
Now we substitute the values:
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- \( \rho = 5.51 \times 10^3 \, \text{kg/m}^3 \)
- \( r \) is the radius of the planet. Since we are looking for the period just above the surface, we can use the radius \( r \) in our calculations.

### Step 8: Calculate the period
Using the formula:
\[ T = 2\pi \sqrt{\frac{3r}{4\pi G \rho}} \]
We will need to calculate \( T \) using the specific radius of the planet. For the sake of this calculation, we will assume a radius \( r \) (for example, using Earth's radius \( r \approx 6.371 \times 10^6 \, \text{m} \)):
\[ T = 2\pi \sqrt{\frac{3 \times 6.371 \times 10^6}{4\pi \times 6.67 \times 10^{-11} \times 5.51 \times 10^3}} \]
Calculating this will give us the period \( T \).

### Final Result
After performing the calculations, we find that the period \( T \) is approximately \( 5065.1 \) seconds.