Periodic time of a satellite revolving above Earth's surface at a height equal to R, radius of Earth, is : [g is acceleration due to gravity at Earth's surface]

Options

A. `2 pi rsqrt((2R)/(g))`
B. `4 sqrt(2)pi sqrt((R)/(g))`
C. `2 pi sqrt((R)/(g))`
D. `8 pi sqrt((R)/(g))`

Explanation

To find the periodic time of a satellite revolving above the Earth's surface at a height equal to the radius of the Earth (R), we can follow these steps:

### Step 1: Understand the Forces Acting on the Satellite
The gravitational force acting on the satellite provides the necessary centripetal force for its circular motion. The gravitational force \( F_g \) can be expressed as:
\[ F_g = \frac{GMm}{r^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Set Up the Centripetal Force Equation
The centripetal force needed 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.

### Step 3: Equate the Forces
Setting the gravitational force equal to the centripetal force gives us:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{r^2} = \frac{v^2}{r} \]

### Step 4: Solve for Orbital Velocity \( v \)
Rearranging the equation, we find:
\[ v^2 = \frac{GM}{r} \]
Taking the square root gives us the orbital velocity:
\[ v = \sqrt{\frac{GM}{r}} \]

### Step 5: Determine the Periodic Time \( T \)
The periodic time \( T \) is the time taken to complete one full revolution. The circumference of the orbit is \( 2\pi r \), so:
\[ T = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2\pi r}{v} \]
Substituting for \( v \):
\[ T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

### Step 6: Substitute for \( r \)
Since the satellite is at a height equal to the radius of the Earth, the total distance \( r \) from the center of the Earth is:
\[ r = R + R = 2R \]
Substituting this into the equation for \( T \):
\[ T = 2\pi \sqrt{\frac{(2R)^3}{GM}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{8R^3}{GM}} = 2\pi \cdot 2\sqrt{\frac{2R^3}{GM}} = 4\pi \sqrt{\frac{2R^3}{GM}} \]

### Final Result
Thus, the periodic time of the satellite is:
\[ T = 4\pi \sqrt{\frac{2R^3}{GM}} \]