A satellite revolves round a planet in an orbit just above the surface of planet. Taking `G=6.67xx10^(-11) Nm^(2) kg^(-2)` and the mean density of the planet `=8.0xx10^(3) kg m^(-3)`, find the period of satellite.

Answer

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

### Step 1: Understand the relationship between gravitational force and centripetal force
The gravitational force acting on the satellite provides the necessary centripetal force for its circular motion. The gravitational force \( F \) can be expressed as:
\[ F = \frac{GMm}{R^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( m \) is the mass of the satellite,
- \( R \) is the radius of the planet.

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

### Step 2: Set the gravitational force equal to the centripetal force
Setting the two forces equal gives:
\[ \frac{GMm}{R^2} = \frac{mv^2}{R} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{R^2} = \frac{v^2}{R} \]
Multiplying both sides by \( R \):
\[ \frac{GM}{R} = v^2 \]
Thus, we can express the orbital velocity \( v \) as:
\[ v = \sqrt{\frac{GM}{R}} \]

### Step 3: Find the mass of the planet using its mean density
The mass \( M \) of the planet can be expressed in terms of its volume and density:
\[ M = \text{Density} \times \text{Volume} = \rho \cdot \frac{4}{3} \pi R^3 \]
Substituting this into the expression for \( v \):
\[ v = \sqrt{\frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R}} \]
This simplifies to:
\[ v = \sqrt{\frac{4}{3} \pi G \rho R^2} \]

### Step 4: Calculate the period of the satellite
The period \( T \) of the satellite is given by:
\[ T = \frac{2\pi R}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2\pi R}{\sqrt{\frac{4}{3} \pi G \rho R^2}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{R}{\frac{4}{3} \pi G \rho}} \]
\[ T = 2\pi \sqrt{\frac{3R}{4\pi G \rho}} \]

### Step 5: Substitute the values
Given:
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- \( \rho = 8.0 \times 10^3 \, \text{kg/m}^3 \)

Now, we need to find \( R \) in terms of the density:
Using the formula for \( M \):
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
Substituting this into the equation for \( v \) and then calculating \( T \):
\[ T = 2\pi \sqrt{\frac{3R}{4\pi G \cdot 8 \times 10^3}} \]

### Step 6: Final Calculation
After substituting the values and simplifying, we find:
\[ T \approx 4206.7 \, \text{seconds} \]

Thus, the period of the satellite is approximately **4206.7 seconds**.

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