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 step by step, we need to analyze the motion of the projectile that is moving in a circular path around the Earth. The radius of this circular path is given as nine times the radius of the Earth (R).

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

### Step 2: Apply the centripetal force condition
For an object moving in a circular path, the net centripetal force required to keep it in circular motion is provided by the gravitational force acting on it. Therefore, we can write:

\[ \frac{m v^2}{r} = \frac{G M m}{r^2} \]

Where \( v \) is the tangential velocity of the projectile.

### Step 3: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ \frac{v^2}{r} = \frac{G M}{r^2} \]

Rearranging gives:

\[ v^2 = \frac{G M}{r} \]

### Step 4: Substitute the value of \( r \)
Since \( r = 9R \), we substitute this into the equation:

\[ v^2 = \frac{G M}{9R} \]

### Step 5: Relate velocity to angular velocity
The relationship between tangential velocity \( v \) and angular velocity \( \omega \) is given by:

\[ v = \omega r \]

Substituting \( r = 9R \):

\[ v = \omega (9R) \]

### Step 6: Substitute \( v \) in the equation
Now substituting \( v \) in the equation \( v^2 = \frac{G M}{9R} \):

\[ (\omega (9R))^2 = \frac{G M}{9R} \]

This simplifies to:

\[ 81R^2 \omega^2 = \frac{G M}{9R} \]

### Step 7: Solve for \( \omega^2 \)
Rearranging gives:

\[ \omega^2 = \frac{G M}{729R^3} \]

### Step 8: Calculate the period \( T \)
The period \( T \) of the circular motion is given by:

\[ T = \frac{2\pi}{\omega} \]

Substituting \( \omega \):

\[ T = \frac{2\pi}{\sqrt{\frac{G M}{729R^3}}} \]

This can be simplified to:

\[ T = 2\pi \sqrt{\frac{729R^3}{G M}} = 27 \cdot 2\pi \sqrt{\frac{R^3}{G M}} \]

### Final Answer
Thus, the period \( T \) of the projectile moving in a circular path of radius \( 9R \) is:

\[ T = 27 \cdot 2\pi \sqrt{\frac{R^3}{G M}} \]