One projectile after deviating from itspath starts movnig round the earth in a circular path of radius equal to nine times the radius of earth R.

Options

A. `2pisqrt((R)/(g))`
B. `27xx2pisqrt((R)/(g))`
C. `pisqrt((R)/(g))`
D. `0.8xx10xx3pisqrt((R)/(g))`

Explanation

To solve the problem of a projectile moving in a circular path around the Earth at a radius equal to nine times the radius of the Earth (R), we will follow these steps:

### Step 1: Identify the parameters
- Let \( R \) be the radius of the Earth.
- The radius of the circular path of the projectile is \( r = 9R \).

### Step 2: Apply the centripetal force equation
The centripetal force required to keep the projectile in circular motion is provided by the gravitational force acting on it. The centripetal force can be expressed as:
\[ F_c = m \cdot \omega^2 \cdot r \]
where \( m \) is the mass of the projectile, \( \omega \) is the angular velocity, and \( r \) is the radius of the circular path.

### Step 3: Write the gravitational force equation
The gravitational force acting on the projectile is given by:
\[ F_g = \frac{G \cdot M \cdot m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth.

### Step 4: Set the centripetal force equal to the gravitational force
Equating the centripetal force and the gravitational force gives:
\[ m \cdot \omega^2 \cdot (9R) = \frac{G \cdot M \cdot m}{(9R)^2} \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \omega^2 \cdot (9R) = \frac{G \cdot M}{81R^2} \]

### Step 5: Solve for \( \omega^2 \)
Rearranging the equation:
\[ \omega^2 = \frac{G \cdot M}{729R^2} \]

### Step 6: Relate \( G \cdot M \) to \( g \)
We know that the acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{G \cdot M}{R^2} \]
Thus, we can express \( G \cdot M \) as:
\[ G \cdot M = g \cdot R^2 \]

### Step 7: Substitute \( G \cdot M \) into the equation for \( \omega^2 \)
Substituting this into our equation for \( \omega^2 \):
\[ \omega^2 = \frac{g \cdot R^2}{729R^2} = \frac{g}{729} \]

### Step 8: Find \( \omega \)
Taking the square root:
\[ \omega = \sqrt{\frac{g}{729}} = \frac{1}{27} \sqrt{g} \]

### Step 9: Relate \( \omega \) to the time period \( T \)
The angular velocity \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Thus, we can express \( T \) as:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{1}{27} \sqrt{g}} = 27 \cdot 2\pi \cdot \frac{1}{\sqrt{g}} \]

### Final Result
The time period \( T \) for the projectile moving in a circular path of radius \( 9R \) is:
\[ T = 27 \cdot 2\pi \cdot \sqrt{\frac{R}{g}} \]